calculus integration algebra polynomial function

[quote]$ a+2b+3c+4d+5e=2007$

Find the number of nonnegative integer solutions $ (a,b,c,d,e)$.[/quote]

I've tried finding a one to one relation to a bitstring without much luck. Any ideas welcome.

[hide]
Let e = e; f = e + d; g = e + d + c; h = e + d + c + b; i = e + d + c + b + a;

Find all nonnegative integral solutions to

e + f + g + h + i =  2007 such that e ≤ f ≤ g ≤ h ≤ i.

I imagine this can be done with some case analysis and PIE, but I don't have time to work on it now.
[/hide]

have u tried using the balls and urns theorem...im new to it myself  :oops: , but from i understand, to solve this problem, u would do
$ \dbinom{2011}{4}$
im probably wrong, but..ehh..at least i tried...i think im write in thinking that this problem can be solved with balls and urns, however, right?

Not directly; balls and urns calculates the number of unordered solutions to a + b + ... + n = x.

so, then how would u do it?
i think this is a good problem, and Im surprised that no other people want to try and solve it.. :|  :huh:

There isn't going to be a "nice" (like, say, polynomial) answer.  Even the number of solutions to $ a+2b = n$ in non-negative integers as $ n$ varies behaves somewhat oddly, and it just gets worse after that.

method 1:
We're gonna use the generating function.
f(x)=(1+x+x^2+x^3+....)(1+x^2+x^4+...)(1+x^3+x^6+...)...(1+x^5+x^10...)
     =1/((1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^5)
Now try to determine the coefficient of x^2007
According to Taylor's expansion
f(x)=.....+f2007'(0)/2007!x^2007+...
answer is f2007'(0)/2007!
Well,it's too hard.Who can find a equation for that?
method 2:
Let f(k,n) be the function equaling the number of pairs(x1,x2,x3..,xk) such that x1+2x2+3x3+...+kxk=n(n>=0,k is a positive integer)
It's easy to get f(1,n)=1
f(k,0)=1,f(k,1)=1
f(k,n)=f(k-1,n)+f(k-1,n-k)....f(k-1,n-[n/k]k)
So f(2,n)=[n/2]+1=n/2+3/4+(-1)^n/4
f(3,n)=f(2,n)+f(2,n-3)+f(2,n-6)+....+f(2,n-3[n/3])
To express [n/3] we will need  a complex function
let g(n)=[n/3] (n is a nonnegative integer)
g(n)=g(n-3)
So g(n)=c1*x1^n+c2*x2^n+c3
where x1,x2 are the two roots of x^3=1 other than 1.
g(n) will be complex function which is quite complex.
If we go on with it,the problem can be solved.
But it's both time-consuming and brain-exhausting. :wink:


inequalities
trigonometry
algebra unsolved
algebra


Could someone please tell me how to solve it?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261085]This is the AMC 12 Problem 18 too[/url]


calculus
integration
logarithms
trigonometry
calculus computations

[1] Tokyo Univ. of Science: $\int \frac{\ln x}{(x+1)^2}dx$

[2] Saitama Univ.: $\int \frac{5}{3\sin x+4\cos x}dx$

[3] Yokohama City Univ.: $\int_1^{\sqrt{3}} \frac{1}{\sqrt{x^2+1}}dx$

[4] Daido Institute of Technology: $\int_0^{\frac{\pi}{2}} \frac{\sin  ^ 3 x}{\sin x +\cos x}dx$

[5] Gunma Univ.: $\int_0^{\frac{3\pi}{4}} \{(1+x)\sin x+(1-x)\cos x\}dx$

Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin ^ 3 x}{\sin x +\cos x}dx$
Let $t=\pi/2 - x$ then
$I = \int_0^{\frac{\pi}{2}} \frac{\cos ^ 3 t}{\sin t +\cos t}dt$
Hence
$2I = \int_0^{\frac{\pi}{2}} \frac{\sin ^ 3 x+\cos^3 x}{\sin x +\cos x}dx$

$\sin^3 x + \cos^3 x = (\sin^2x+\cos^2x)(\sin x+\cos x) - \sin x\cos x(\sin x+\cos x)$
Hence
$2I = \int_0^{\frac{\pi}{2}} (1-\sin x\cos x)dx = \frac{\pi}{2}-\frac{1}{2}$
and
$I=\frac{\pi-1}{4}$

[quote="liyi"]I have posted a general method for [2] before.

And for [5], integrating by parts, we see that
$\int x\sin xdx = \sin x - x\cos x + c$
$\int x\cos xdx = \cos x + x\sin x + c$.

From this, we can easily get the value of the integral $=2(1+\sqrt 2)$

btw, do you like computing integrals?[/quote]

I remember your posting a general method.

Yes, I like computing integral, because it makes speeding calculation. :)

No one has done [b][2][/b] , here it is :

$I=\int \frac{5}{3\sin x+4\cos x}dx$

put $t=\tan \frac{x}{2}$

then $\sin x=\frac{2t}{1+t^{2}}$ and $\cos x=\frac{1-t^{2}}{1+t^{2}}$

and $\frac{dt}{dx}=\frac{1}{2}\left( 1+t^{2}\right) \Rightarrow dx=\frac{2dt}{1+t^{2}}$

Hence $I=-\frac{5}{2}\int \frac{dt}{t^{2}-\frac{3}{2}t-1}$

$I=\int \frac{-\frac{5}{2}}{\left( t-2\right) \left( t+\frac{1}{2}\right) }dt$

$=\int \left( \frac{1}{t+\frac{1}{2}}-\frac{1}{t-2}\right) dt$

$=\ln \left| \frac{2t+1}{2t-4}\right| =\ln \left| \frac{2\tan \frac{x}{2}+1}{2\tan \frac{x}{2}-4}\right| $

You can have a look at
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=general&t=26765


inequalities
geometric inequality
geometry proposed
geometry

Let $ M$ be an interior point of the triangle $ ABC$ and let $ P,Q ,R$ be the mid points of the sides $ BC,CA,AB$ then prove that $ (MA \plus{} MB \plus{} MC)\geq2(MP \plus{} MQ \plus{} MR)$.

[color=purple] In my opinion I think it is not true, if i do not remember false, the best form we can give is :
                                     $ MA \plus{} MB \plus{} MC\ge 2(w_1 \plus{} w_2 \plus{} w_3)$
with $ w_1,w_2,w_3$ are three internal bisectors of three angles $ BMC,CMA,AMB$ respectively.And a similar estimation is $ MA\plus{}MB\plus{}MC\ge MP\plus{}MQ\plus{}MR$.
   Of course, we can generalized it to n-gon more information you can read in the book : " Recent Advances In Geometric Inequality"
 :) [/color]


Find all positive integers $ x,y,z$ satisfying $ (2x^2\minus{}y^2)^2\equal{}z(z\plus{}6xy)$.

There are two families of integer solutions: $(x,y,z) = (t,0,2t^2)$ and $(0,t,t^2)$.

The case $y=0$ corresponds to the first family. If $y\ne 0$ Let $a=x/y$ and $b=z/y^2$ to get a curve:
\[(2a^2-1)^2 = b(b+6a)\]
that under rational transformation
\[\begin{cases} u = \frac{5a^2-18a+6-6b}{3a^2} \\
v = \frac{2(5a^2-6a+2-2b)}{a^3}
\end{cases}\]
is equivalent to the elliptic curve:
\[v^2 = u^3 - \frac{73}{3} u - \frac{1190}{27}.\]
This curve has zero rank over the rationals and its torsion points give raise only to $[a,b]=\pm [0,1]$ corresponding to the second family of solutions.


let $ x$ , $ y$ ,  $ z$ $ \in ]0,\plus{}\infty[$ , prove that :

$ xyz \geq (x\plus{}y\minus{}z)(z\plus{}x\minus{}y)(y\plus{}z\minus{}x)$

Substitutions  $ x\equal{}a\plus{}b$, $ y\equal{}b\plus{}c$, $ z\equal{}a\plus{}c$
$ (x\plus{}y\minus{}z)(z\plus{}x\minus{}y)(y\plus{}\minus{}x)\equal{}8abc$ and using AM>GM we got
$ xyz\equal{}(a\plus{}b)(b\plus{}c)(a\plus{}c)>\equal{}8abc$

nice solutions

this is my prof

$ xyz \geq (x \plus{} y \minus{} z)(z \plus{} x \minus{} y)(y \plus{} z \minus{} x)$

$ \Leftrightarrow x^3\plus{}y^3\plus{}z^3\plus{}3xyz \geq xy(x\plus{}y)\plus{}yz(y\plus{}z)\plus{}zx(z\plus{}x)$

which is true by schur's inequality

[quote="y-a-s-s-i-n-e"]hi!

do we have this inequality  :?:
$ (x \plus{} y \minus{} z)(z \plus{} x \minus{} y)(y \plus{} z \minus{} x) \geq 0$[/quote]

not really supposing $ z \equal{} \infty$ so $ (x \plus{} y \minus{} z)(z \plus{} x \minus{} y)(y \plus{} z \minus{} x)$<0

we can say that it true in condition that a.b.c sides of triangle

thank's ibn rochde

i think this solution is wrong

[quote="bunhiacovski"][quote="y-a-s-s-i-n-e"]let $ x$ , $ y$ ,  $ z$ $ \in ]0, \plus{} \infty[$ , prove that :

$ xyz \geq (x \plus{} y \minus{} z)(z \plus{} x \minus{} y)(y \plus{} z \minus{} x)$[/quote]
by AM-GM  $ (x \plus{} y \minus{} z)(z \plus{} x \minus{} y)\le x^2$
$ (z \plus{} x \minus{} y)(y \plus{} z \minus{} x)\le z^2$
$ (x \plus{} y \minus{} z)(y \plus{} z \minus{} x)\le y^2$
done~[/quote]

because if $ (x \plus{} y \minus{} z)(z \plus{} x \minus{} y)(y \plus{} z \minus{} x) \le 0$

we can't do$ \sqrt{((x \plus{} y \minus{} z)(z \plus{} x \minus{} y)(y \plus{} z \minus{} x))^2}$


algebra
polynomial
number theory open
number theory

Is it known whether every irreducible polynomial $ P$ such that $ \forall p \exists n| P(n) \not\equiv 0 mod p$ assumes a prime value? Is there some relationed result?

It's not know for any polynomial of degree greater than $ 1$; for degree $ 1$ it's just Dirichlet's theorem on primes in arithmetic progression.


AMC
USA(J)MO
USAMO
geometry
combinatorial geometry

Well, which ones did people like the most?  Personally, I liked 4 and 6 the most, although it took me a while to figure out what was even being asked by number 6.  Probably my least favorite was 5, which even though it was a good question, I had never seen any of the convex hull type stuff used to solve it and spent hours trying induction.

I liked 4 and 6 the most as well. I also didn't know about the other solution found in the paper mentioned in other threads dealing with the convex hull, but now that I took a look at it, it is a pretty elegant solution. Due to my inherent fear of geometry, I liked number three the least.

I liked 5 alot; even though they used all that 'convex hull' whatever in their write-up, i just called the set of points for which it was possible to draw a line through them such that all other points are on one side of the line the boundary polygon and showed that if it wasn't all one color, the solution was trivial, and if it was all one color, you just do that slowly rotating trick that i think the amc site used. Unfortunately, I did all this while half asleep after the usamo was over. Stupid 4 & 6 took up all my time to write.

No one likes the easiest problem?  :lol:

yeah... 1 was just sort of annoying, and didn't require much thought. I thought 4 was the cutest, though also not very hard. I liked 3 after I saw the solution.

I liked number 6 the best... it was a very interesting problem that required no advanced mathematical knowledge to solve, except for very basic logs and some common sense divisibility stuff.  My solution was totally different than the official one though.

With respect to #5, I had never even heard of a convex hull before... so on my paper, I ended up defining what one was and I actually proved that it existed!  I made up my own name for what it was and everything... ie "let the UBERGON represent the convex polygon containing all of the points, etc etc."  1/2 a page later, I proved that it existed and was able to proceed with my solution.  So all of you that had never seen a convex hull before have no excuses for not getting #5!

Too bad I was unofficial because I never got to write any other american contests.

In deference to how awesome that is, I will never again use the term, "convex hull."  To me, it will always be an Ubergon.

I liked the two problems authored by Zuming the most (#1, #3).  Particularly the geometry one.

I thought #3 was nice, even though I did not solve it.  #1 and #4 were a little too easy, although #4 was indeed kind of cute.  #5 really was very easy, and I had already gotten about 90% of the solution but didn't realize it and gave up.  #6 just plain scared me, but it was actually ok.  Ahh well, thankfully this was not a critical USAMO for me.

Oh yes, the problems were so easy, I did them in my sleep!! The solutions were left as an exercise for the grader.

Actually, I liked #4 a lot because it was very cute.  I'm sure 3 was a great problem, but I didn't have a clue as to how to approach it, so whatever.

My favorite USAMO questions wasn't used.  And I don't think it ever will be.  Here it is -

Proove that 1+1=2  :D

[quote="h_s_potter2002"]Prove that 1+1=2[/quote]

Number theory definition?

I think every question was pretty appropriate for where it was placed (1, 4 easy but still they were 1, 4) except #3. Honestly. How did that get there? Half the people that didn't get it were just afraid.

[quote="darktreb"]Half the people that didn't get it were just afraid.[/quote]

 :?

#4 did catch me...... from going to MOP....*sigh*


"Disjunction, together with negation, provide sufficient means to define all other truth-functions."(from [url=http://plato.stanford.edu/entries/disjunction/#7]Standford Encyclopedia of Philosophy[/url])
Note: "Disjunction" here means "Inclusive disjunction", or simply "OR". They are the same thing. [i]"XOR", however, is a different thing.[/i]

Then I started to wonder: how do you define "AND" using "OR" and "NOT"? How do you define "XOR"?
Can anyone explain it for me?

See:
http://www.answers.com/topic/logical-connective
http://www.answers.com/topic/functional-completeness

Related question(s) if you have more time:
Is {NAND, NOR} functional complete?
Symmetrical difference is the set-theoretic equivalent of XOR. How beautiful is that! So, what's the logic-theoretic equivalent of [url=http://www.answers.com/topic/complement-set-theory]complement[/url]?
How do you pronounce XOR?
How many logic connectives are there? "Implication" is a logic connective, but it doesn't have a [url=http://www.answers.com/topic/logic-gate]"gate"[/url]? And what's with the "BUF" gate?

Anyway, the following is true:
I will solve all puzzles above XOR I will be very unhappy.

More questions that I don't plan to solve, but also am interested in:
Can boolean functions have two operands? How can "AND" function [i]not[/i] have two operands? But how can a function have two operands anyway?

That was a lot of questions.  Here are answers and non-answers to a few of them.

[quote]Then I started to wonder: how do you define "AND" using "OR" and "NOT"? How do you define "XOR"? [/quote]
You should be able to get this by experimentation, either with truth tables or by logical reasoning.  For example, if you already know how to get "AND," getting XOR is easy: "a XOR b" means, "exactly one of a and b," which also means "a and not-b, or b and not-a."  And then you make things upper-case  :wink:

[quote]So, what's the logic-theoretic equivalent of complement? [/quote] Negation, of course.

[quote]How many logic connectives are there?[/quote]
A logical connective taking $n$ arguments (for NOT, $n = 1$; for the other "usual" connectives, $n = 2$) is equivalent to a function $F: \{0, 1\}^{n}\to \{0, 1\}$.  (For every possible truth value of its arguments, it must assume some truth value.)  Hopefully now this is a combinatorial question you can answer.

I found that it was actually simpler to derive XOR, then use it to derive AND, rather than the reverse process.

I will use NOT(a) to show that NOT is operating specifically on a, and [a OR b] that OR is operating specifically on b, with a similar convention for other binary operators.  I realize that these strings are not the easiest to understand, but it was the best way I could think of to show the order of operations unambiguously.

[hide="x XOR y"][NOT([x OR NOT(y)]) OR NOT([y OR NOT(x)])][/hide]

[hide="x AND y"]NOT([[x XOR y] OR NOT([x OR y])])[/hide]

Be sure to verify for yourself that these definitions do indeed give XOR and AND.

From now on, I want the answer to only one question:
How do you pronounce XOR?
And I'll figure out the rest myself.


linear algebra
matrix
number theory proposed
number theory

Proof:for any positive integers $ a$ $ d$ and $ n$
        $ \frac{1}{a}\plus{}\frac{1}{a\plus{}d}\plus{}\frac{1}{a\plus{}2d}\plus{}...\plus{}\frac{1}{a\plus{}nd}$  is not an integer
        I think it is a really nice problem but also a problem maybe posted somewhere :maybe:

$ \bullet$ $ (a,d) \equal{} s$ being $ s\not \equal{} 1$. Then substitute $ a \equal{} a_1s,\ d \equal{} d_1s$ (note that $ (a_1,d_1) \equal{} 1$):
\[ W \equal{} \frac {1}{s}\bigg(\frac {1}{a_1} \plus{} \frac {1}{a_1 \plus{} d_1} \plus{} \cdots \plus{} \frac {1}{a_1 \plus{} nd_1}\bigg) \equal{} \frac {T}{s}
\]
If $ T\not \in \mathbb{N}\Rightarrow W\not \in \mathbb{N}$. In the next case is explained why $ T\not \in \mathbb{N}$.

$ \bullet (a,d) \equal{} 1$. We have the following:

[b]Lemma[/b]_ Let $ A \equal{} \{a,a \plus{} d,\dots a \plus{} nd\}$. If $ hp\in A$ being $ h,p\in \mathbb{N},\ p > n \minus{} 1$ and $ p$ prime then it doesn't exist any $ k\in \mathbb{N}$ such that $ k\not \equal{} h, kp\in A$.

[u]Proof[/u]
Note first that it can't be $ hp \equal{} a$ and $ kp \equal{} a \plus{} nd$ because it would be: $ p\vert a\Rightarrow p\vert d$ which is in contrast to the hypothesis $ (a,d) \equal{} 1$.

Suppose the lemma is not true. Then $ hp \equal{} a \plus{} id,\ kp \equal{} a \plus{} jd$. Subtracting each member: $ p(h \minus{} k) \equal{} (i \minus{} j)d$.
Now, if the equality is true $ \displaystyle \frac {(i \minus{} j)d}{p}$ must be an integer. If $ (p,d)\not \equal{} 1 \Rightarrow p\vert d \Rightarrow p\vert a$ while $ (a,d) \equal{} 1$ so this is an absurd. So $ (p,d) \equal{} 1$.

Because of this, if $ \displaystyle \frac {(i \minus{} j)d}{p}$ is an integer then $ p\vert (i \minus{} j)$. This is impossible being $ i \minus{} j\le n \minus{} 1 < p$. So, the lemma is true.

$ \star$ For $ d,a>1$ it is always possible to find a $ p$ prime such that $ hp \equal{} a \plus{} id$. By the lemma we know there won't be any other $ j\not \equal{} i$ such that $ p\vert a \plus{} jd$.

Then consider $ R \equal{} W[a(a \plus{} d)\cdots (a \plus{} id \minus{} d)(a \plus{} id \plus{} d)\cdots (a \plus{} nd)]$.

This is a sum of factors which are integers except from $ \displaystyle \frac {a(a \plus{} d)\cdots (a \plus{} id \minus{} d)(a \plus{} id \plus{} d)\cdots (a \plus{} nd)}{a \plus{} id}$ so $ W$ is not an integer.

$ \star$ If $ d\equal{}a\equal{}1$ then we have $ W_{1,1}\equal{}1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\cdots \plus{}\frac{1}{n}$. Multiply by $ M\equal{}lcm(1,2,\dots ,2^q\minus{}1,2^q\plus{}1,\dots n)$ where $ q$ is such that $ 2^{q}\le n<2^{q\plus{}1}$ (note that $ \frac{n!}{2^q}$ is an integer):

$ \displaystyle MW_{1,1}\equal{}\frac{M}{2^{q}}\plus{}\sum _{i\equal{}1,\\ i\not \equal{} 2^{q}}^{n} \frac {M}{i}$. So $ MW_{1,1}$ is not an integer and the proof is finished  :) .

[quote="EUCLA"]$ \star$ For $ d,a > 1$ it is always possible to find a $ p$ prime such that $ hp \equal{} a \plus{} id$. [quote]
Can you explain this step which I think it is the main difficulty of this problem? :)
I am wondering if we can use Sylvester Theorem to prove it.

[quote="EUCLA"]Yes, you're totally right  :oops: . Maybe we need to find an other way to solve the problem.  Or, could you give me a link to this theorem, since I totally ignore it?
On google I found one about matrix, but I don't think it is the right one.[/quote]
OK  http://www.mathlinks.ro/viewtopic.php?t=259889


$ f(1)>0$

for every non negative integer m,n,
$ f(m^2\plus{}n^2)$=$ {f(m)}^2$+$ {f(n)}^2$
find all f functions which denotes this equation.

[hide]Plug in (m,0) and we get $ f(m^2)\equal{}f^2(m)\text{ (*)}$. Thus $ f(m^2\plus{}n^2)\equal{}f(m^2)\plus{}f(n^2)$ and we now have Cauchy's functional equation. So $ f(m)\equal{}cm$ for all m (and some constant $ c$). How ever, plugging this into $ (*)$ we get $ cm^2\equal{}c^2m^2$, so we see that c=1.

Note that this is also different from Cauchy's in that a constant function won't work unless we have $ f(m)\equal{}1$, since otherwise we have $ f(m)\equal{}c\Rightarrow c\equal{}c^2$ from $ (*)$. $ c$ cannot equal zero because $ f(1)>0$.[/hide]

Is function $ \mathbb{N}_0 \rightarrow \mathbb{N}_0$ ?
$ (m,n)\equal{}(m,0); f(m^2\plus{}n^2)\minus{}2f(0)\equal{}f(m^2)\minus{}2f(0)\plus{}f(n^2)\minus{}2f(0)$
Let $ g(x)\equal{}f(x)\minus{}2f(0) \therefore g(x\plus{}y)\equal{}g(x)\plus{}g(y)$
$ \therefore g(x)\equal{}ax \rightarrow  f(x)\equal{}ax\plus{}b$
plugin to $ f(m^2\plus{}n^2)\equal{}f(m)^2\plus{}f(n)^2$
we get $ a\equal{}b\equal{}1$
$ \therefore f(x)\equal{}x$

[quote="gauss202"][quote="Anaxerzia"][hide]Plug in (m,0) and we get $ f(m^2) \equal{} f^2(m)\text{ (*)}$.[/hide][/quote]

[hide]You have that $ f(m^2) \equal{} [f(m)]^2 \plus{} [f(0)]^2$, but how can you be certain that $ f(0) \equal{} 0$?[/hide][/quote]

Hmm.. I had plugged (0,0) in and forgot the squares...  :oops:

If not $ f(0)\equal{}0$..
[hide]
We have from $ (0,0)$ that $ f(0)\equal{}f^2(0)\plus{}f^2(0)\Rightarrow f(0)\equal{}1/2$. Thus $ f(m^2)\minus{}1/4\equal{}f^2(m)$ and we have $ f(m^2\plus{}n^2)\equal{}f(m^2)\minus{}1/4\plus{}f(n^2)\minus{}1/4\equal{}f(m^2)\plus{}f(n^2)\minus{}1/2$. Thus $ f(m^2\plus{}n^2)\minus{}1/2\equal{}f(m^2)\minus{}1/2\plus{}f(n^2)\minus{}1/2$. So we use the standard $ g(x)\equal{}f(x)\minus{}1/2$ and we get Cauchy's so $ f(x)\equal{}ax\plus{}1/2$ for some $ a$. We plug in to original to get $ a\equal{}0$ and $ \boxed{f(x)\equal{}1/2}$ as another solution...
[/hide]

As far as I can tell, every solution posted so far has gone from $ g(m^2 \plus{} n^2) \equal{} g(m^2) \plus{} g(n^2)$ for all $ m, n$ to $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ for all $ x, y$ for some function $ g$.  However, as our domain is the integers, this is totally invalid -- there's no way to conclude, for example, that $ g(3) \equal{} g(2) \plus{} g(1)$.

Also, it would be nice to know if we were looking at functions from the integers to the integers, or integers to reals, or what.

[quote="Anaxerzia"]We plug in to original to get $ a \equal{} 0$ and $ \boxed{f(x) \equal{} 1/2}$ as another solution...[/quote]

(Edit:  Misread  :oops: )

[quote="t0rajir0u"]Unfortunately, this solution clearly doesn't satisfy the original equation.[/quote]

Doesn't seem that clear to me... $ \frac{1}{2} \equal{} \left( \frac{1}{2} \right)^2 \plus{} \left( \frac{1}{2} \right)^2$  :maybe:

There are a lot of mistakes going on in this thread...

Let $ m$ be nonnegative integer, and plug in $ (m,0)$ for $ (m,n)$, this gives
\[ f(m^2) \equal{} [f(m)]^2
\]
In the original equation, we can place this in to get something weaker, namely,
\[ f(m^2 \plus{} n^2) \equal{} f(m^2) \plus{} f(n^2)
\]
where $ m$,$ n$ are non-negative integers...

That is to say,
\[ f(a \plus{} b) \equal{} f(a) \plus{} f(b)
\]
where $ a$,$ b$ are [b]perfect squares[/b].

Now I wonder if we can characterize the squares in this way. The problem is that we don't have the sum of two squares equal to a square, so we cannot use the idea of the cauchy functional equation.

Another annoying thing about this function is that any number that is not an integer or expressable as the sum of two squares (for example, primes in the form $ 4k \plus{} 3$) is defined in terms of numbers much larger than themselves since if we plug in $ m \equal{} p$, then we have it is equal to $ f(p^2 \plus{} n^2) \minus{} [f(n)]^2$. So I believe that this is hard. It would be nice to know if it was N to N.

I also noticed that if $ f(m^2\plus{}n^2)\equal{}f(m^2)\plus{}f(n^2)$, then $ m,n$ don't even have to be nonnegative...

Not that it helps much though...

but note that it is not cauchy's its different because you can not replace $ m^2$ by $ x$
$ a$ must be a perfect-square,you suppose $ a$ can be all natural numbers in your proofs.
,we can find more functions than $ f(x)\equal{}x$ i think


No pair of nonnegative integers $ (x,y)$ can satisfy $ 9x\plus{}16y\equal{}78$. What is the largest integer $ n$ for which no pair of integers $ (x,y)$, neither of which is negative, can satisfy $ 9x\plus{}16y\equal{}n$?

[hide="hint"]$ (9,16)\equal{}1$, so all solutions of $ 9x\plus{}16y\equal{}n$ are in the form $ \left(r,\frac{n\minus{}9r}{16}\right)$, where $ ar\equiv c\pmod b$[/hide]

[hide]
Try various modulos of 9 and 16 as nos of n
[/hide]

AoPSers must know the Chicken McNuggets Theorem :)

[quote="123456789"]Oh, I didn't know about the Chicken McNuggets Theorem before. Now this makes this question a lot easier!

How do you do it without it?[/quote]

The only other way to do it that I know of is to essentially derive the Chicken McNuggets theorem.

Consider the equation $ \pmod9$.  Once you have shown that every possible combination, $ \pmod9$ has been made, you can simply add 9 to those results and thus get every number.

$ 16\equiv7\pmod9$, so adding 7 mod 9 until all possible residues are covered, we can find the largest number that cannot be written.

0,7,5,3,1,8,6,4,2

Clearly, no positive integer that is $ 2\pmod9$ less than 128 can be written in the form $ 9x \plus{} 16y$, so the largest number that cannot be written is 128-9=119.

This is very informal, but hopefully it will give you an idea of how you would solve this problem.

[quote="b-flat"]The only other way to do it that I know of is to essentially derive the Chicken McNuggets theorem.

...so the largest number that cannot be written is 128-16=112...[/quote] :huuh: This answer is contradicting Chicken McNuggets Theorem! Better double-check your work.

This is from [i]High School Math Contests[/i].
[hide="McDonalds"]
$ 9 \cdot 16 \minus{} 16 \minus{} 9 \equal{} 119$.
Which the solution in the book shows.
[/hide]

[quote="AstroPhys"]This is from [i]High School Math Contests[/i].
[hide="McDonald's"]
$ 9 \cdot 16 \minus{} 16 \minus{} 9 \equal{} 119$.
Which the solution in the book shows.
[/hide][/quote]

Does the book explain how they determined that the answer is $ 9 \cdot 16 \minus{} 16 \minus{} 9\equal{}119$?

[quote="b-flat"]Clearly, no positive integer that is $ 2\pmod9$ less than 128 can be written in the form $ 9x \plus{} 16y$, so the largest number that cannot be written is 128-16=112.[/quote]

Where did you get $ 128$ from?

Note that my previous post had a major typo in it (it should have been 128-9, not 128-16).

To clear things up:
$ 16\cdot0\equiv0\pmod9$, so any number congruent to $ 0\pmod9$ that is greater than or equal to $ 16\cdot0$ can be written in the required form.
$ 16\cdot1\equiv7\pmod9$, so any number congruent to $ 7\pmod9$ that is greater than or equal to $ 16\cdot1$ can be written in the required form.
$ 16\cdot2\equiv5\pmod9$, so any number congruent to $ 5\pmod9$ that is greater than or equal to $ 16\cdot2$ can be written in the required form.
$ 16\cdot3\equiv3\pmod9$, so any number congruent to $ 3\pmod9$ that is greater than or equal to $ 16\cdot3$ can be written in the required form.
$ \vdots$
$ 16\cdot8\equiv2\pmod9$, so any number congruent to $ 2\pmod9$ that is greater than or equal to $ 16\cdot8$ can be written in the required form.

Now, every number has been accounted for that is $ \geq16\cdot8\equal{}128$.  Thus, the largest integer NOT accounted for is $ 128\minus{}9\equal{}119$ because $ 119$ is the largest integer less than 128 that is $ 2\pmod9$.


Need help on the following problem.

Carbonic acid, H2CO3 is a diprotic acid with

Ka1= 2.5 X10^-4           Ka2= 5.61 X 10^-11

At 1.0 atm (101, 325 kPa) pressure and 25C a solution is saturated with H2CO3 at [H2CO3] = 0.0010M. If the pH = , what is [CO3^2-] in such a solution expressed in mole/litre?

Thanks in advance.

I think I have already explained this to you in another post. Try to find it.

Thanks Carcul! Lol. Silly me. I was meaning to find that page and when I had a second look, it was there  :D. You posted hundreds of posts and you still remember that question. You have great memory Carcul and thanks again for the help!


[quote="vuthanhtu_hd"]Let $ a,b,c > 0$  such that $ abc \equal{} 1$

1)Prove that$ \frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2} \geq \frac {3}{2}$

2) True or false

$ \frac {a}{b^3 \plus{} c^3} \plus{} \frac {b}{c^3 \plus{} a^3} \plus{} \frac {c}{a^3 \plus{} b^3} \geq \frac {3}{2}$[/quote]
Your inequalities are wrong. Try $ a \equal{} b \equal{} 10$ and $ c \equal{} 0.01.$ :wink:
But the following inequality is true already.

Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$
 Prove that:
\[ \frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2} \plus{} \frac {a \plus{} b \plus{} c}{2} \geq 2
\]
More one.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2} \geq \frac {6\sqrt3}{\sqrt {11(a^2 \plus{} b^2 \plus{} c^2) \plus{} 5(ab \plus{} ac \plus{} bc) }}
\]


In the diagram, ABCD is a square with sides of length 6. M is the midpoint of AB, and N is the midpoint of BC. What is the area of quadrilateral AOCD?

[hide="Solution."]
Draw a line from point $ O$ that is perpendicular to side below it, crossing it at point $ F$.  Call the distance from $ A$ to $ F$ $ 2x$; so the distance from $ O$ to $ F$ is then $ x$ by similar triangles.  By applying the same procedure to the right side, you form two right triangles and a square.  So you know that the distance from $ F$ to $ B$ must be $ x$.  Thus, $ x \equal{} 2$ and the area of the two triangles combined is $ 8$; the area of the square is $ 4$.  The area of everything else is then $ \boxed{24}$.[/hide]


geometry
circumcircle
perpendicular bisector
angle bisector

The internal bisector of angle $A$ in a triangle $ABC$ with $AC > AB$ meets the circumcircle $\Gamma$ of the triangle in $D$. Join$D$ to the center $O$ of the circle $\Gamma$ and suppose that $DO$ meets $AC$ in $E$, possibly when extended. Given that $BE$ is perpendicular to $AD$, show that $AO$ is parallel to $BD$.

[hide]$AOD$ is isosceles $(\widehat{OAD}=\widehat{ODA})$
So it's enough to prove that $\widehat{ODA}=\widehat{BDA}$

Let $K = AD \cap BE$

The triangles $AKB,AKE$ are congruent $(a-s-a)$ so $BK=KE$.

Hence $AD$ is the perpendicular bisector of $BE \Rightarrow DB=DE$
$(DBE$ isosceles $) \Rightarrow AD$ is angle bisector of $\angle BDE$

[img]http://i8.tinypic.com/24vtgr8.gif[/img][/hide]

[quote=jayme]Dear Mathlinkers,
can some one gives a construction of the figure with the given condition...

Sincerely
Jean-Louis[/quote]

Good, but easy task!
Notation as per figure above
Let $ABC$ be the triangle with the given properties and let $E$ on $AC$ so that $AE=AB$; further, let $BE$ intersect the circle for second time at $F$. In $\triangle BDF$ , $DK$ is an altitude, $DO$ a diameter, hence they are isogonal, consequently they trisect $\angle BDF$.
Now let $L=AO\cap BE$ and we need $BL=AB$, hence construction:
Take $A,B$ on a circle $O$ so that $\angle AOB<60^\circ$, take $L$ on $(OA)$ so that $BL=BA$, extend $(BL$ to intersect the circle at $F$. The perpendicular from $A$ to $BF$ intersect the circle again at $D$, and take $E$ the reflection of $B$ in $AD$; $AE$ intersects the circle at $C$.

Best regards,
sunken rock

The condition $BE \perp AD \implies \boxed{\angle B=3\angle C} \text{   (   }$[hide="proof"] $CE=BE \implies$ $\angle CBE=\angle C \implies \angle B-\angle C=90^{\circ} -\frac{\angle A}{2} \implies \angle B=3\angle C$ [/hide] $\text{)  }$ and, we have ,$$\angle OAD=\angle ODA =90^{\circ} -\frac{\angle A}{2} -\angle C =\frac{1}{2}( \angle B-\angle C) =\frac{3\angle C}{2}-\frac{\angle C}{2}=\angle C=\angle ADB \implies AO||BD$$

We will prove that $\angle BDA=\angle OAD$
Let $K=AD \cap BE$.
It is easy to see that $\triangle ABK \cong \triangle AEK$.
Now using $BK=EK$, notice that $\triangle DBK \cong \triangle DEK$
Thus we have $\angle BDK=\angle KDE=\angle OAD$ (As $\triangle OAD$ is $O$-isosceles).
Our proof is thus complete.


calculus
integration
trigonometry
function
calculus computations

Alternately, substitute $t=\sin \theta$ first; the integral becomes $\int \sin\theta\cdot t\cdot \cos\theta\,d\theta=\frac12\int \theta\cdot\sin 2\theta\,d\theta$. You still need to integrate by parts, but trig functions are more familiar than inverse trig.


A college class has exactly enough students to form eight equal rows.  On Monday a student is absent, and the professor is able to seat the students into five equal rows.  On Tuesday two students are absent, and the professor can seat the students into nine equal rows.  What is the least possible number of students in the class?

Taking each of the three facts into account, we have that the number of students is

8x=5y+1=9y+2, for x, y, and z being integers.

As a rather simple problem, the fastest way to solve this problem for many people may first be listing out multiples of 8 that end in 6 (as it cannot end in 8 because it must be an even number):
16, 56, 96, 136
we can easily recognize that 56=9(6)+2, and that the smallest number of students possilbe in this class is [b]56[/b]


geometry
3D geometry
function
calculus
derivative
algebra
polynomial

Prove that if $ n$ is an integer such that the equation
\[ x^3\minus{}3xy^2\plus{}y^3\equal{}n\]
has a solution in integers $ x$ and $ y$, then the equation has at least 3 such solutions.  Show that the equation has no solutions in integers for $ n\equal{}2891$.

The second part of the equation is easy; mod 9 is always a good thing to look at when cubes are involved.  However, it took me a long time to figure out the first part.  I knew that I needed to show that if $ (x,y)$ is a solution, then so is $ (f(x,y),g(x,y))$.  However, it took me a really long time to find a particular pair of functions, $ f$ and $ g$ (I had to find a set of solutions for $ n\equal{}3$ and then I had to guess and check until I found a function that would work).

None of the solutions I have seen provide a good motivation as to how to come up with $ f$ and $ g$ and since there are many possible "simple" functions to try.

Does anyone have a motivated solution to the problem?

Elliptic curves always come with a pair of such functions $ f, g$ given by what's called the doubling law.  Are you familiar with the geometric construction of the group law?

[b]Edit:[/b]  Here's the construction.  We know that generically a line and a cubic intersect in three locations.  Therefore a good way to find a third point given one point $ P$ is to construct the tangent line at $ P$ and look at its third intersection.  Implicit differentiation gives

$ 3x^2 \, dx \minus{} 3y^2 \, dx \minus{} 6xy \, dy \plus{} 3y^2 \, dy \equal{} 0 \Leftrightarrow$
$ \frac {dy}{dx} \equal{} \frac {y^2 \minus{} 2xy}{y^2 \minus{} x^2}$

so starting with a point $ (p, q)$ its tangent line is the set of points of the form $ (p \plus{} (q^2 \minus{} p^2)t, q \plus{} (q^2 \minus{} 2pq)t)$.  Substitute this into the original equation, which will give you a double root at $ t \equal{} 0$; the remaining root determines the third point.

(The group law isn't actually quite defined in this way, but this is good enough for finding points, although they won't in general be integer points.  Think of this as a generalization of Vieta jumping.)

Thanks; that helps a lot.  I usually try to avoid calculus if I can, so I didn't even consider using calculus on a Number Theory problem.  I'm not familiar with the geometric construction of the group law, but your explanation is fine.

Since this is an IMO problem, I assume there must be some way to come up with a solution without using calculus, but since its old, they might have just assumed it was fairly simple to find the other solutions.


i don't know latex yet so I'm going to use my own terms

SUM (i, j, expression) $\sum_{i}^{j}$ is the summatory from i to j of the expression that shows up,,,

Reading a book on algorithms I found that

$(1+x)^{n}$

(1+x)^(n+1)=SUM(0,n,C(n,k)*x^k)+SUM(0,n,C(n,k)*x^(k+1))
$(x+1)^{n+1}=\sum_{0}^{n}\binom{n}{k}\times x^{k}+\sum_{0}^{n}\binom{n}{k}\times x^{k+1}$
and then they said

(1+x)^(n+1)=SUM(0,n,C(n,k)*x^k)+SUM(0,n,C(n,k-1)*x^k))

$(1+x)^{n+1}=\sum_{0}^{n}\binom{n}{k}\times x^{k}+\sum_{0}^{n}\binom{n}{k-1}\times x^{k}$

which I don't find right 'cause the coefficients might be the same but the ditro of x's not...

in the first one I found x^(n+1) and in the next step I found that that term is missing...

Any suggestion..

[size=75][color=red]Try the $\text{\LaTeX}$ section of AoPS.
http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php
b-flat[/color][/size]

What does C(n,k) mean? (combinations? )

I'm not sure if I interpret the equations right, but they sure don't look too correct.

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php#sums

might help...

I tried to post your things in $\LaTeX$.  Please tell me if I messed any of them up or did not convert them correctly.

Yes, it does.  Could you look at where I edited your post to see if I fixed it correctly?  Thanks.

Do you mean this?

$(x+1)^{n+1}=\sum_{k=0}^{n}\binom{n}{k}x^{k}+\sum_{k=0}^{n}\binom{n}{k}x^{k+1}$

$(1+x)^{n+1}=\sum_{k=0}^{n}\binom{n}{k}x^{k}+\sum_{k=0}^{n}\binom{n}{k-1}x^{k}$

Since
$(1+x)^{n+1}=(x+1)^{n+1}$

and
$\sum_{k=0}^{n}\binom{n}{k}x^{k}=\sum_{k=0}^{n}\binom{n}{k}x^{k}$

then
$\sum_{k=0}^{n}\binom{n}{k-1}x^{k}=\sum_{k=0}^{n}\binom{n}{k}x^{k+1}$

But the last equation I typed is undefined, because you can't choose -1 things from x things, where x is a natural number.

If you actually meant
$\sum_{k=1}^{n}\binom{n}{k-1}x^{k}=\sum_{k=0}^{n}\binom{n}{k}x^{k+1}$

the equation isn't undefined.

Please tell me which equations you meant, and I will hels prove/disprove the equations for you.

[quote="nutz_for2.718281828"]Do you mean this?

$(x+1)^{n+1}=\sum_{0=k}^{n}\binom{n}{k}x^{k}+\sum_{0=k}^{n}\binom{n}{k}x^{k+1}$

$(1+x)^{n+1}=\sum_{0=k}^{n}\binom{n}{k}x^{k}+\sum_{0=k}^{n}\binom{n}{k-1}x^{k}$

Since
$(1+x)^{n+1}=(x+1)^{n+1}$

and
$\sum_{0=k}^{n}\binom{n}{k}x^{k}=\sum_{0=k}^{n}\binom{n}{k}x^{k}$

then
$\sum_{0=k}^{n}\binom{n}{k-1}x^{k}=\sum_{0=k}^{n}\binom{n}{k}x^{k+1}$

But the last equation I typed is undefined, because you can't choose -1 things from x things, where x is a natural number.

If you actually meant
$\sum_{1=k}^{n}\binom{n}{k-1}x^{k}=\sum_{0=k}^{n}\binom{n}{k}x^{k+1}$

the equation isn't undefined.

Please tell me which equations you meant, and I will hels prove/disprove the equations for you.[/quote]

Generally, you write it as $k=0$ on the bottom of the sigma.

All of that comes down to Pascal's identity $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$

I've got an idea! Let x be one! You plug it into the first equation and get $2^{n}*2=2^{n+1}$ which is correct!
The second equation isn't correct, though. I think you should find out why.

Well in your post nutz_for2.718281828 you said that the equation is undefine...

if we take C(n,k) = 1 when k < 0 just like so many reference point to be like that....then that equation is not undefine but not well define...

You can't just omit a term in that equation like saying that the value of k will start in 1 instead of 0...

but then is clear that this :

SUM (0, n, C(n,k)*x^k) = SUM(0, n, C(n, k-1)*x^k)

this isn't true...if you expand the terms in both equation you'll find that there are some of them that aren't in the other side...

and is clear that taking that true you get by inducction that:

(1 + x)^n = SUM (0, n, C(n, k)*x^k)

So how do we get this solved....??? :(

OK. Compute: $\binom{7}{-1}$

Se what I mean?


Does the exist a function $ f: \mathbb R^* \rightarrow \mathbb R$ so that for every $ x\ne 0$, we have
\[ f(f(x))=-\frac 1x \, ?\]

[quote="Virgil Nicula"]Isn't integer part ...[/quote]
Ok, so $ f(f(x))=-\frac{1}{x}$ $ \forall x\neq 0$

We have $ f(f(f(f(x))))=x$ and so we just have to organize all non null reals in sequences $ \{a,b,-\frac{1}{a},-\frac{1}{b}\}$

Here is a general solution :

Let $ \mathbb{A}$ and $ \mathbb{B}$ two equipotent subsets of $ \mathbb{R^*^+}$ such that $ \mathbb{A}\cup\mathbb{B}=\mathbb{R^*^+}$ and $ \mathbb{A}\cap\mathbb{B}=\emptyset$

Let $ \mathbb{C}=\{x$ such that $ -\frac{1}{x}\in\mathbb{A}\}$
Let $ \mathbb{D}=\{x$ such that $ -\frac{1}{x}\in\mathbb{B}\}$

Let $ h(x)$ any bijective function from $ \mathbb{A}\longrightarrow\mathbb{B}$
Let $ k(x)$ the reciprocal of $ h(x)$
Let $ e(x)$ any function from $ \mathbb{A}\longrightarrow\{-1,+1\}$

All solutions are of the form :

$ \forall x\in\mathbb{A}$ : $ f(x)=e(x)h(x)^{e(x)}$ (to be clear : either $ h(x)$, either $ -\frac{1}{h(x)}$)
$ \forall x\in\mathbb{B}$ : $ f(x)=-e(k(x))k(x)^{-e(k(x))}$
$ \forall x\in\mathbb{C}$ : $ f(x)=-e(-\frac{1}{x})h(-\frac{1}{x})^{-e(-\frac{1}{x})}$
$ \forall x\in\mathbb{D}$ : $ f(x)=e(k(-\frac{1}{x}))k(-\frac{1}{x})^{e(k(-\frac{1}{x}))}$

[quote="Virgil Nicula"]Thanks. Please, can you present us a concret example ? Thank you.[/quote]

Ok :

Let $ \mathbb{A}\equal{}\{x>0$ such that $ [\minus{}x]$ is odd$ \}$  (where $ [.]$ is the integer part).
Let $ \mathbb{B}\equal{}\{x>0$ such that $ [\minus{}x]$ is even$ \}$
Let $ h(x)\equal{}x\plus{}1$
Let $ k(x)\equal{}x\minus{}1$
Let $ e(x)\equal{}\plus{}1$

Then :

$ \forall x>0$ such that $ [\minus{}x]$ is odd, $ f(x)\equal{}x\plus{}1$

$ \forall x>0$ such that $ [\minus{}x]$ is even, $ f(x)\equal{}\minus{}\frac{1}{x\minus{}1}$

$ \forall x<0$ such that $ [\frac{1}{x}]$ is odd, $ f(x)\equal{}\frac{x}{1\minus{}x}$

$ \forall x<0$ such that $ [\frac{1}{x}]$ is even, $ f(x)\equal{}\minus{}\frac{x\plus{}1}{x}$

Verification :

1) $ \forall x>0$ such that $ [\minus{}x]$ is odd :
$ f(x)\equal{}x\plus{}1$ and $ [\minus{}f(x)]\equal{}[\minus{}x\minus{}1]\equal{}[\minus{}x]\minus{}1$.
So $ f(x)>0$ and $ [\minus{}f(x)]$ is even.
So $ f(f(x))\equal{}\minus{}\frac{1}{f(x)\minus{}1}\equal{}\minus{}\frac{1}{x}$

2) $ \forall x>0$ such that $ [\minus{}x]$ is even :
$ f(x)\equal{}\minus{}\frac{1}{x\minus{}1}$
So $ f(x)<0$ ($ [\minus{}x]$ even implies $ x>1$) and $ [\frac{1}{f(x)}]\equal{}[1\minus{}x]\equal{}1\plus{}[\minus{}x]$ is odd.
So $ f(f(x))\equal{}\frac{f(x)}{1\minus{}f(x)}\equal{}\frac{\minus{}\frac{1}{x\minus{}1}}{1\plus{}\frac{1}{x\minus{}1}}\equal{}\minus{}\frac{1}{x}$

3) $ \forall x<0$ such that $ [\frac{1}{x}]$ is odd
$ f(x)\equal{}\frac{x}{1\minus{}x}$
So $ f(x)<0$ and $ [\frac{1}{f(x)}]\equal{}[\frac{1\minus{}x}{x}]\equal{}[\frac{1}{x}]\minus{}1$ is even
So $ f(f(x))\equal{}\minus{}\frac{f(x)\plus{}1}{f(x)}\equal{}\minus{}\frac{\frac{x}{1\minus{}x}\plus{}1}{\frac{x}{1\minus{}x}}\equal{}\minus{}\frac{1}{x}$

4) $ \forall x<0$ such that $ [\frac{1}{x}]$ is even
$ f(x)\equal{}\minus{}\frac{x\plus{}1}{x}$
So $ f(x)>0$ ($ [\frac{1}{x}]$ even with $ x<0$ means $ \frac{1}{x}<\minus{}1$) and $ [\minus{}f(x)]\equal{}1\plus{}[\frac{1}{x}]$ is odd.
So $ f(f(x))\equal{}f(x)\plus{}1\equal{}\minus{}\frac{1}{x}$

Q.E.D.

[quote="Virgil Nicula"]Thanks. Please, can you present us a concret example ? Thank you.[/quote]

Here is another one, simpler than the previous, I think

Let $ \mathbb{A}\equal{}]0,1[$.
Let $ \mathbb{B}\equal{}[1,\plus{}\infty[$
Let $ h(x)\equal{}\frac{1}{2x}$ if $ x\equal{}2^{\minus{}n}$ for some integer $ n\geq 1$, and $ h(x)\equal{}\frac{1}{x}$ else.
Let $ e(x)\equal{}\plus{}1$

Then :

$ \forall x\in]0,1[$ such that $ x\equal{}2^{\minus{}n}$ for some integer $ n\geq 1$, $ f(x)\equal{}\frac{1}{2x}$

$ \forall x\in]0,1[$ such that $ x\neq 2^{\minus{}n}$ for all integers $ n\geq 1$, $ f(x)\equal{}\frac{1}{x}$

$ \forall x\in[1,\plus{}\infty[$ such that $ x\equal{}2^n$ for some integer $ n\geq 0$, $ f(x)\equal{}\minus{}2x$

$ \forall x\in[1,\plus{}\infty[$ such that $ x\neq 2^n$ for any integer $ n\geq 0$, $ f(x)\equal{}\minus{}x$

$ \forall x\in]\minus{}\infty,\minus{}1[$ such that $ x\equal{}\minus{}2^n$ for some integer $ n\geq 1$, $ f(x)\equal{}\frac{2}{x}$

$ \forall x\in]\minus{}\infty,\minus{}1[$ such that $ x\neq \minus{}2^n$ for any integer $ n\geq 1$, $ f(x)\equal{}\frac{1}{x}$

$ \forall x\in[\minus{}1,0[$ such that $ x\equal{}\minus{}2^{\minus{}n}$ for some integer $ n\geq 0$, $ f(x)\equal{}\minus{}\frac{x}{2}$

$ \forall x\in[\minus{}1,0[$ such that $ x\neq \minus{}2^{\minus{}n}$ for any integer $ n\geq 0$, $ f(x)\equal{}\minus{}x$

Verification is nearly immediate (I'll give it if some reader encounters problem)

Thank you, [b]Patrick[/b], for your effort to explain us on two nice examples.


quadratics algebra polynomial trigonometry function geometry calculus

[color=darkred]my conjectures [/color]

1)  No formula is completly generalized if it contains a number in it

2)All formulas contains atleast two or more numbers in it

3)therefore no formula is completly generalized and it can never be.

4) absolutly all formulas are related to each other,we can arrive from any one formula to another.

[color=green]Explaination:[/color] by completly generalized i mean , for example the quadratic formula $ x \equal{} \frac { \minus{} b \pm \sqrt {b^2 \minus{} 4ac}}{2a}$ we say it is a general formula but this is not completly generalized, because it contains the numbers $ \minus{} 1,1,2, \minus{} 4,$ i want to say that instead of those numbers the formula has another very important letters like the letters $ a,b,c$.and the letters instead of $ \minus{} 1,1,2, \minus{} 4,$ are related to the quadratic formula which is always equal to $ \minus{} 1,1,2, \minus{} 4,$ [size=150]in that formula[/size] so we use those numbers but they are something else in reality.

another example $ V \equal{} \frac {4}{3} \pi r^3$ is not completly generalized because it contains the numbers $ 4,3,3,1$ those numbers are something else like the $ r$ and $ \pi$

No formula is completly gerenalized and it can never be.because let $ U_n \equal{} n$ which is the simplest formula ,and can never get simpler than that. it still contains [size=150]numbers[/size] in it. the number $ 1$ is before the letter $ n$ and also for it's power. even if there was another letter for the $ 1$ before the $ n$. that letter would have $ 1$ as its power,therefore no formula is completly generalized.but its is good to generalize, because we will know more.

absolutly all formulas are related to each other, the more we generalize the more usefull the formulas will be.because we arrive at the other formulas that we already discovered or that we havnt.we see abit of this in proving formulas.

:lol:$ e \equal{} mc^2$ is not completly general, the $ 2$ is another very important letter which we dont know.

Write it $ E\equal{}mcc$...  :wink:

Instead of $ 5x$ you can write $ x\plus{}x\plus{}x\plus{}x\plus{}x$, instead of $ x^3$ you can write $ xxx$...
So what's the point¿ And please define your terms rigorously, giving examples is not a definition!

[quote="ZetaX"]Write it $ E \equal{} mcc$...  :wink:

Instead of $ 5x$ you can write $ x \plus{} x \plus{} x \plus{} x \plus{} x$, instead of $ x^3$ you can write $ xxx$...
So what's the point¿ And please define your terms rigorously, giving examples is not a definition![/quote]

ok why is $ cc$ or $ c^2$ why do you have to multiply the $ c$ $ \color{red}2$ times (twice) by itself?
why should $ 5x$ be added to itself $ \color{red}5$ times? where does $ 2$ and $ 5$ come from?
$ x^3 \equal{} xxx$ why do you have to multiply $ x$ by itself $ 3$?

You ask why somethings exists¿
Answer one (mathematical): by axioms.
Answer two (philosophical): e.g. I exist, and I can image the idea of "$ 2$".

so you disprove all of this ?
no i didnt ask why it exists, i ask why it is that number, it can be another number, why not that number?
ok if its by axiom then i can say the axiom is not complete and general.

the $ \color{red}2$ in $ E = mc^\color{red}2}$

Well, can we please go away from physical formulas to mathematical ones¿ It's quite hard to reason constants in physics  :wink:

E.g. in the quadratic equation, it comes from reformulating $ ax^2\plus{}bx\plus{}c$ (and there is already a $ 2$ in it  :wink: ).

ok  :lol:

[quote]E.g. in the quadratic equation, it comes from reformulating $ ax^2 \plus{} bx \plus{} c$ (and there is already a $ 2$ in it  :wink: )[/quote]true, that is what i want to say,the $ 2$ in $ x \equal{} \frac { \minus{} b \pm \sqrt {b^2 \minus{} 4ac}}{2a}$ can be $ d$ for degree.i believe there is another formula for the degree (2) of the polynomials which we can substitute for the $ 2$ in that formula. and where does the $ 4$? come from? they are something else i dont know, may be you know.
the same is true for other formulas numbers comes from other formulas.

What is "generalized", to you, supposed to mean?

No expression can possibly be such that it can't be written using constants according to your standards. for example, take $ n^{n^n}$. It could be written as $ n\uparrow \uparrow 3$. Or the innocuous $ \sin x$. It could be written $ \sin 1x$, $ \sin \frac{2x}{2}$, $ \sin \frac{3x}{3}$... etc. Your "conjectures" are making use of a definition that doesn't exist.

$ 4$ comes from $ 2\plus{}2$ for example. And the occurence of powers of two has indeed to do with $ 2$ being the degree of that equation. But nonetheless, for other problems, "unexpected" numbers can occure.

[quote="binomial_4eva"]it seems you didnt read my first post carefully.read again and pay attention[/quote]

What? Temperal is saying you are relying on undefined terms, and he is correct. You are not defining anything, just naming definitions that don't even make logical sense.

Things that are not defined: formula, generalized, contains, number

[quote]But nonetheless, for other problems, "unexpected" numbers can occure[/quote]

Yes , that is what my conjecture says , that those "unexpected numbers" that occur are not unexpected, they are another formula that we can substitute in the quadratic formula, like the $ 2$ for the degree of the quadratics .the $ 4$ and $ 1$ are also another formula that we dont know.the $ 4$ and $ 2$ and $ 1$ have another formula like quadratic formula, that we can substitute for the $ 4,2,1$ in the quadratic formula.

Are you saying that the quadratic formula is a special case of a formula that solves arbitrary polynomials? This is untrue. See [url]http://en.wikipedia.org/wiki/Abel%27s_impossibility_theorem[/url]

[quote="fedja"][quote="Temperal"]
I'm very surprised that this thread hasn't been moved/locked yet.
[/quote]
We had a short discussion about it in Moderator Lounge ;) The consensus was that despite it is not the most informative thread on AoPS, it doesn't violate any rules, so let it be. [/quote]

I think I know that, seeing as I initiated the discussion.  :P But I thought that unhelpful threads that clogged up the forum were gnerally locked after a few days... and this definitely doesn't belong in Pre-Olympiad.

binomial_4eva, I think I finally understand what you're talking about (though, as 10000th user said, you expressed it extremely poorly in your first post). The thing is, that not all constants signify things. We can see where they come from quite often, but why must they signify something? One might assume that in Heron's formula, the $ 2$ in the square root might signify the number of dimensions we're operating, but we can see, if we attempt to prove it, that there's no correlation. (Pardon the bad example, but I'm typing this in a hurry)

fedja's example is an instance where the constants do signify something, but those cases are not all of them, nor even the majority.

[quote="binomial_4eva"]

[color=green]Example 1:[/color]In physics we studied this formula $ s_{v} \equal{} ut \plus{} \frac {1}{2}at^{2}$
In further maths we then studied the same formula as $ s_{v} \equal{} utsin{\theta} \minus{} \frac {1}{2}gt^2$

where did $ sin{\theta}$ come from then ? Isnt this my conjectures? [/quote]

they are not the same '$ u$'!

$ u\sin\theta$ is the vertical component of the initial velocity.  Once you realize this, it is just a matter of substituting into the familiar first formula.  There is no more generality, just combining a familiar formula with simple geometry.

[quote="fedja"][quote="Temperal"]and this definitely doesn't belong in Pre-Olympiad.
[/quote]
I agree 100%, but where exactly do you suggest to move it? :?[/quote]

What about the Round Table?

[quote="Meddle"]But also, there is a standard of quality to maintained on internet forums. Though it is arguable that all knowledge is for better, I believe this thread is inappropriately placed and should have been more thought out by the poster.[/quote]

I don't know what us on the moderating staff ever would have done about this thread before you, Meddle, signed up and informed us that you did not feel that this thread was appropriately placed!  Thanks for helping us out of our bind!

I've some feeling we are going to need some "Philosophical Mathematics" (or better "Esoteric Mathematics", at least for some other dicussions) forum...  :wink:

Yes, I had the same thought when I read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1019389802&t=28607]this[/url]. I would gladly move it from Advanced Fields, but where? In defence of the current thread, I should say that it makes infinitely much more sense.

[quote="fedja"][quote="Temperal"]and this definitely doesn't belong in Pre-Olympiad.
[/quote]
I agree 100%, but where exactly do you suggest to move it? :?[/quote]

Well, I was thinking intermediate, since it's technically a problem and it seems okay for intermediate students, but it's the mod's choice, I guess.

EDIT: About the round table... well, I don't know. It's technically a problem, as I said.

I think this thread belongs to The Round Table because, as fedja suggested, there are some points worth discussing about generality and such in mathematics. Even if it is esoterical or philosophical or whatnot, as binomial_4eva insists that he is serious, The Round Table accepts "topics for serious discussions." A good fit I'll say.

i dont think this should be moved in round table, in round table people talk of everything, like ",weed,marijuana,Israeli-Palestinian Peace Treaty,terrorism,+... ,in my opinion this thread shouldnt be there.
but it is up to the moderator to where he likes to move it,
i have an opinion,
isnt it possible to make another section for new mathematical ideas,or "mathematical philosophy" as Zetax mentioned, i say new mathematical idea because it encourages people to look for discovering new things in mathematics.a section like "intermediate topic" "Pre-olympiad" and high school basic, etc.I love discovering new things.thats the joy of mathematics.

people this Idea is an important Idea and its right believe me
to find the unknowns or formula for the numbers in a formula we need to find a few examples to do with that formula.

I believe you but it's not really getting anywhere...
Now that I think about it, I'm not too sure about your definitions... they are not clear enough to illustrate something... it feels like this whole work is like a junkyard, where you just put random things in the yard and literally 'looking' at the mess.
Sorry but this is just not worth my time...   :surrender:

The theoretical and practical definitions and explanations of generality have been discussed and discussed and discussed in this thread. Beyond what is stated here, generality really is just a matter of taste.

So yes, it's an important idea, and is intimately involved in decision making by mathematicians when they choose which direction to take their work in. But no, I don't think it's worth continuing this discussion, and no, I don't think it merits reviving this thread after 4 months.

On a sidenote, try not to take others' criticisms too personally. We don't dislike you or anything like that (I hope). However, understand that it can be quite frustrating to write out an explanation and see the original poster completely miss the point. That multiplied by whatever number of people that responded to this thread.

Please don't revive non-mathematical old threads just for the sake of bumping them. Locked.


[quote="b-flat"]The number $-\frac{63}{5}$ is between two consecutive odd integers.  What is the product of these two integers?[/quote]

[hide]$-\frac{63}{5}=-12\frac{3}{5}$

That is between $-13$ and $-11$

$-13*-11=143$[/hide]

[quote="b-flat"]The number $-\frac{63}{5}$ is between two consecutive odd integers.  What is the product of these two integers?[/quote]
[hide]
-63/5 is between -65/5 and -55/5, -13*-11=143[/hide]


Pogo has a Lotto game.  There are 20 chances to land on 80 squares.  What are the odds of 10 of the 20 balls landing on all 10 of the predetermined squares?  I knew how to figure this out in high school but that was decades ago.  Thanks. [list]
[/list]

The chances are awfully low (about 1 in 10 million).
$\binom{20}{10}\frac{10\cdot9\cdot8\cdots2\cdot1}{80\cdot79\cdot78\cdots71}\approx .0000001$

Can I ask how you did your solution?

I had a different approach, but I think it will result with a different probability.

Heres what I did,
[hide]
20 chances to land on 80 squares computes to 80 choose 20, or $\binom{80}{20}$, possible combinations.

If 10 of the 20 chances land on the 10 winning square, then there are 10 chances left and 70 possible square left, so there are $\binom{70}{10}$ ways of winning.

Therefore the probability is $\frac{\binom{70}{10}}{\binom{80}{20}}$, which is $\frac{(70\cdot69\cdot68\cdot\cdot\cdot62\cdot61)(20\cdot19\cdot18\cdot\cdot\cdot11)}{80\cdot79\cdot78\cdot\cdot\cdot62\cdot61}$

Have I taken a wrong approach?[/hide]

No, our solutions are equivalent.  My solution can be written as $\binom{20}{10}\frac{70!10!}{80!}$.  This expands to $\frac{20!70!}{80!10!}$.  Similarly, your final solution can be written as the same expression.  I was thinking of this as 20 chances to land on the 10 squares.  Assuming that there cannot be more than one ball on each square the probabilities are $\frac{(10\cdot9\cdot8\cdots1)(70\cdot69\cdot68\cdots61)}{(80\cdot79\cdot78\cdots61)}$.  The second half of the numerator represents the probabilities of the other 10 balls NOT landing on a winning square.  This is trivial since after the 10 winning squares are chosen, it doesn't matter where the rest of the balls land.   The order doesn't matter either, the result would be the same if we assumed 10 misses and then 10 wins.

Your approach is correct also.  There is actually a fairly well known combinatorial formula which is basically what you used.  Its known as a [url=http://en.wikipedia.org/wiki/Hypergeometric_distribution]Hypergeometric distribution[/url]. It's given by $\frac{\binom{K}{x}\binom{M-K}{n-x}}{\binom{M}{n}}$
where $M =$ population(80), $K=$ number of type I (success), in our case winning squares. $x=$ number of type I chosen (10), $n=$ number of samples (type I and II)(20).  Writing it out in this form would give $\frac{\binom{10}{10}\binom{70}{10}}{\binom{80}{20}}$ which is equivalent to our solutions.


function ARML AMC AIME USA(J)MO USAMO geometry

What if this thread is deleted for being irrelevant, immaterial, spam-ridden, yet nostalgic?

then we delete it ourself

What if we can spam on forever....lol???

Then there would be a lot of spam?

What if tomorrow was Saturday?

Then I would be really happy!

What if I actually had a good question to put here?

then you would post it

what if they really really wanted to but were caught by the gendarmes

Then they would be in prison

What if ppl stop using big words???

Then we have a boring conversation

What if we beat College Station this week?

Then we go out and party.

What if the world ends tommorow?

then I die tomorrow

What if I die tomorrow?

then the world has ended tomorrow.
what if we can stop the word association game from having more posts than this topic?

Then this topic would always have more posts than the word association (I love stating the obvious ^_^)

What if I didn't launch my water rocket into my face?

Then you'll keep your nose.

What if rruczyk decides to destroy this spam?

Then Bob's post count would go down.

What if he doesn't?

Then Bob's post count will vary according to the function $p=t^t$, where $t$ is measured in seconds and $p$ is posts.

What if Tare makes a return and spots this thread?

You would be bored.
What if summer vacation ends early?

Then we'd have less time for math during the summer.

What if the AMC were cancelled?

I'd cry.

What if my lachrymal ducts were dysfunctional?

You would take eye drops.

What if the eye drops were suspiciously replaced with hydrochloric acid or another compound severely detrimental to your eyes?

then the world would go blind.

what if it weren't an eye for an eye, but rather an eye for a nose?

then the whole world would be disfiguredly wild.

What if $E=mc^2 \text{ actually was }E^2=mc^5$?

Then atom bombs would be a heck of a lot more powerful.

What if gravity strength followed the inverse fifths law, rather than the inverse square law?

Then the following identity would hold: Gravity = Weak Sauce.

What if "What If...? version 2" had a post in it with the question "What if 'What if...? version 2' had a post in it with..........

Then my answer to this question would be "X", where X is my answer to this question.

Actually, never mind.

What if you slackers actually started signing up for NOML 2005 like you're supposed to be doing?   :P

then we wouldn't have anything to look forward to doing

what if I answered my own question

Then the rules of this thread would be violated.
What if there were no rules on this planet?

Then I wouldn't post another question to continue this game.

What if there were a rule that you violated by knowing the rules?

Then i'd just feign ignorance (loophole!  :) )

What if there's nobody to enforce the rules?

i dont think there is anyone to enforce the rules (i just broke the rules!)

what if $\mathbb{Z}\subseteq\mathbb{N}$?

then everyone would disregard it

what if art of problem solving didn't exist?


[code]Rating: 2.3[/code]
[color=blue]¿Alguien puede explicarme su utilidad?[/color]

[color=blue]Ahora cambió a $2.4$. ¿Quién me puede explicar qué indica esta función, por favor?[/color]

mmmm es q el numero varia de acuerdo al rating q los usuarios del foro le dan a tus posts. si el numero ha cambiado de 2.3 a 2.4 quiere decir q alguien ha "rateado" :rotfl: algunos de tus posts con ratings altos, como "excellent" o "very good". en cambio si el numero disminuye quiere decir q los usuarios les han dado ratings bajos

el maximo es seis, el minimo es uno.
los siento, mi espanol es mal.


Let K and L be metric compact spaces. and f: K X L $ \longrightarrow$ $ \mathbb{R}$ is continuous. For each y $ \in$ L, define G(x) = sup f(x,y) with x $ \in$ K. Prove that G: L $ \longrightarrow$ $ \mathbb{R}$ is continuous.

Could you explain what metric space you mean for KXL ?That may be for example :
d(a,b)=max{d(a_1,b_1),d(a_2,b_2)  ,
d(a,b)=(d(a_1,b_1)^2+d(a_2,b_2)^2)^(1/2)  or something else.
a=(a_1,a_2)  ,  b=(b_1,b_2)

It doesn't matter as long as the topology is the same (i.e., the standard product topology).

A brief solution :
It is provable that KXL is compact too. So f: KXL----> R is uniformly continuous and bounded .
A(y)={f(x,y) | x is a member of K }
A(y_0)={f(x,y_0) | x is a member of K }
For each e there is a p such that :
If d_L(y,y_0) <=  p  then d_(KXL) ( (x,y),(x,y_0) )= ((d_K(x,x))^2+ ((d_L(y,y_0) )^2)  )^0.5 <= p
 (I used this definition ,I think many of other definitions work too)
Then:
|f(x,y)-f(x,y_0) | <e/2
 Because f is uniformly continuous.
Two sets A(y) and A(y_0) have corresponding members which are closer than e/2 one by one ,on the other hand both are bounded so sup of them is closer than e. It shows that G is continuous .


i sense another 24 hour lockdown of this forum

so do I
they are invariably preceeded by a topic with bubka in its name
that is generally accepted as the threshold of tolerance

[quote="trigonometry1.01"]I AM NOT NEALTH

I AM NOT DELONG[/quote]
me neither, we have so much in common
this is off topic but who cares


superior algebra
superior algebra unsolved

Let $ 0\minus{}>L\minus{}>M\minus{}>N\minus{}>0$ be a split exact seuence of $ A$- modules.Then show that there is a bijection between $ Hom(N,L)$ and the set of all splittings $ s: N\minus{}>M$

Look at the difference of two splittings. :wink:

But how does the difference between 2 splittings give me a homomorphism. Any way I have to show that the 2 sets are in bijective correspondence

fix a splitting $ s : N \to M$ of $ p : M \to N$. if $ t$ is another splitting, we have $ p(s \minus{} t)\equal{}0$, i.e. $ s \minus{} t : N \to M$ factors uniquely through $ L \to M$. this yields a map from the splittings to the homomorphisms $ N \to L$, which is clearly bijective.


A number is created by dividing 36 by $ \frac{1}{2}$ and adding 6. What is the number?


I was sent instructions for publishing a paper. Quoting:

[b]4. The manuscript should be prepared in basic font Roman 10pt size,
 and
     must remain legible after small reduction. The papers should be in
     English and formatted with text width/height=13/20 cm on
 A4-format.[/b]

The question is how to make all these possible, I mean:

1) Basic Roman font
2) 10pt size
3) width/height = 13/20 cm
4) A4 format
5) what is "legible after small reduction"?

Thanks
Jean

What class are you using? For example 1, 2, 4 are sastisfied by using
\documentclass[10pt,a4paper]{article}
The font is Computer Modern Roman ie seriffed.

3. You can set textwidth and textheight by themselves eg \setlength{\textheight}{20cm} but it can mess up other things so I would use [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=geometry]the geometry package[/url] to set the margins correctly.

5. This means that the text remains legible even if the page is viewed at say 80-90% (this may happen as part of the publisher's printing process). LaTeX is excellent for this and you can check it by zooming out in Adobe Reader. For example, I have the geometry manual open at 77% and it's perfectly readable.


Find the number of ordered quadruples $ (x,y,z,w)$ of integers with $ 0\leq x,y,z,w\leq 36$ such that $ x^{2}+y^{2}=z^{3}+w^{3}(mod37)$

Where can I find Turkish Math Olympiad 2001 remaining problems from the national round?

in fact, the problem proposed in december, 1999. The exam is one of the steps of imo 2000. So some books refer the date as 2000. But not 2001. The reason of my post is that the problems of turkish math olymp. in december, 2000 are missing on the internet.

Another solution is in
http://www.mathlinks.ro/viewtopic.php?t=112191


number theory
prime numbers
number theory unsolved

Prove that the period of the number 1+1/2+1/3+...1/(p-1) is less or equal to the period of number 1/p (where p is a prime number) or find the counterexample if such exists and disprove the statement.

Period?  You mean the period of the repeating decimal in base $ 10$?

Yes. I mean the period of the repeating decimal in base 10.

It would be even better to say in the statement of the problem "length of a period" than a "period".

Take $ p \equal{} 11$. The period of $ \frac {1}{11} \equal{} 0.090909\ldots$ is $ 2$. But $ 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} \cdots \plus{} \frac {1}{10} \equal{} \frac {7381}{2520}$ has a $ 7$ in its denominator so its period is certainly greater than $ 2$.

(Note: you don't have to actually calculate that sum in order to see that it has a $ 7$ in the denominator. I added the actual sum just for completeness).

So, the statement is not valid for all prime numbers but... can anybody prove that there exist infinitely many prime numbers such that length of a period of 1+1/2+1/3+...+1/(p-1) is less or equal to the length of a period of a number 1/p  (p is prime number).

[quote="t0rajir0u"]That will be the case if there exist infinitely many prime numbers $ q$ such that $ 10$ is a primitive root $ \bmod q$, in which case we can take $ p$ to be the smallest prime greater than $ q$.  I am not sure if this is a known result, though.  A weaker result can be used about the orders of $ 10$ mod various primes, but it's less nice.[/quote]

t0rajir0u, what does it mean that [b]10[/b] is a primitive root [b]mod q[/b]? I do not know what does the term primitive root mean? Can you explain me your statement in simpler terms? And are there infinitely many prime numbers [b]q[/b] such that [b]10[/b] is a primitive root [b]mod q[/b] ?


Let $ a,b,c\geq 0$. Prove
$ a^5\plus{}b^5\plus{}c^5\plus{}abc(a^2\plus{}b^2\plus{}c^2)\geq ab^3(a\plus{}b)\plus{}bc^3(b\plus{}c)\plus{}ca^3(c\plus{}a)$
 :roll:

When $ a\equal{}0,b\equal{}1$
The inequality become:
$ c^5\plus{}1 \ge c^4\plus{}c^3$
$ \iff (c^4\minus{}c^2\minus{}c\minus{}1)(c\minus{}1) \ge 0$
Which is wrong :lol:

[quote="mufc"]Let $ a,b,c\geq 0$. Prove
$ a^5 \plus{} b^5 \plus{} c^5 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2)\geq ab^3(a \plus{} b) \plus{} bc^3(b \plus{} c) \plus{} ca^3(c \plus{} a)$
 :roll:[/quote]

It is equivalent to :

$ [5,0,0]\plus{}[3,2,1]\ge [4,1]$ it's true by Muirhead.

[quote="quangpbc"][quote="mufc"]Let $ a,b,c\geq 0$. Prove
$ a^5 \plus{} b^5 \plus{} c^5 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2)\geq ab^3(a \plus{} b) \plus{} bc^3(b \plus{} c) \plus{} ca^3(c \plus{} a)$
 :roll:[/quote]

It is equivalent to :

$ [5,0,0] \plus{} [3,2,1]\ge [4,1]$[/quote]
Are you sure?  :wink:

I'm sorry.  :blush: This is right ineq
Let $ a,b,c\geq 0$. Prove
$ a^5 \plus{} b^5 \plus{} c^5 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2)\geq\frac{1}{2}\sum ab(a\plus{}b)(a^2\plus{}b^2)$
 :)

[quote="mufc"]
Let $ a,b,c\geq 0$. Prove
$ a^5 \plus{} b^5 \plus{} c^5 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2)\geq\frac {1}{2}\sum ab(a \plus{} b)(a^2 \plus{} b^2)$
 :)[/quote]
It's equivalent to $ \sum_{cyc}(2a^5\minus{}a^4b\minus{}a^4c\minus{}a^3b^2\minus{}a^3c^2\plus{}2a^3bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^5\minus{}a^4b\minus{}ab^4\plus{}b^4)\minus{}\sum_{cyc}(c^3a^2\minus{}2c^3ab\plus{}c^3b^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2(a^3\plus{}a^2b\plus{}ab^2\plus{}b^3\minus{}c^3)\geq0.$
Let $ a\geq b\geq c.$ Then
$ \sum_{cyc}(a\minus{}b)^2(a^3\plus{}a^2b\plus{}ab^2\plus{}b^3\minus{}c^3)\geq$
$ \geq(a\minus{}c)^2(a^3\plus{}a^2c\plus{}ac^2\plus{}c^3\minus{}b^3)\plus{}(b\minus{}c)^2(b^3\plus{}b^2c\plus{}bc^2\plus{}c^3\minus{}a^3)\geq$
$ \geq(b\minus{}c)^2(a^3\plus{}a^2c\plus{}ac^2\plus{}c^3\minus{}b^3\plus{}b^3\plus{}b^2c\plus{}bc^2\plus{}c^3\minus{}a^3)\geq0.$

There is a simpler way . By Schur and Chebychev we have

$ a^5 \plus{} b^5 \plus{} c^5 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2)\geq \sum ab(a^3 \plus{} b^3)\geq \sum \frac {1}{2}ab(a \plus{} b)(a^2 \plus{} b^2)$

qed :D


Only the hours are visible on a digital clock. During what fraction of a 12-hour period is at least one ''1'' visible?

All times can be written in the form AB:CD.

For 120 minutes, B = 1.

For 180 minutes, A = 1.

However, we counted 11:00 twice, so we subtract out 60 minutes to 240, so far.

Now we turn our attention to C and D. There are 480 minutes where D = 1 and A, B aren't 1. Then for C, there are also 480 minutes. However, we counted the ones where both C and D are 1 twice, so we have to correct that by subtracting 8, so we have 952.

So in total, there are $ 240 \plus{} 952 \equal{} 1192$ minutes where we have at least 1, and there are 1440 minutes in a day. Therefore, we have $ \frac{149}{189}$.

Hmm...I seem to get a different answer.
I used complementary counting.
Since the time must either have at least one "1" visible or no "1" visible at all, the sum of the two probabilities is 1.  Therefore, the probability that there is at least one "1" visible=1- the probability that there is no "1" visible.
Now, we solve for the no "1" case before subtracting from 1 (this is easier than solving for at least one "1")
The possible hours are 1,2,3,4,5,6,7,8,9,10,11, and 12.  1,10,11, and 12 (4 numbers out of 12) have a 1, so the probability that the hour doesn't have a 1 is 8/12=2/3.
Now, we look at the minutes.  Using the minutes as "CD" as izzy uses, there are 6 possible values of C (0,1,2,3,4, and 5).  Only 1 has a "1", so the probability that C is not 1 is 5/6.
Now, we look at "D".  The values of D are 0,1,2,3,4,5,6,7,8, and 9.  Here, only 1 has a "1", so the probability that D is not 1 is 9/10.
Thus, the probability that there is no "1" visible in the time is 2/3*5/6*9/10=1/2
Thus, using complementary counting, the probability that there is at least one "1" visible is 1-the probability that no "1"'s are visible=1- 1/2= [b]1/2[/b]

Although I don't know how to solve this problem, I'm pretty sure it said only the hours are available to see.

The solution is pretty simple. It asks what fraction of hours have at least one "1" in it. Since it only shows the hours in a 12 hour period (the numbers 1-12), only 4 numbers have at least one "1" digit in them:1, 10, 11, and 12. $ \frac {4}{12} \equal{} \frac {1}{3}$. The correct answer is$ \frac {1}{3}$.

You're forgetting about minutes, which only go up to 59. Jongao's answer is correct - I messed up in my solution.

[quote="GameBot"][b]Only the hours are visible on a digital clock.[/b] During what fraction of a 12-hour period is at least one ''1'' visible?[/quote]

I hate when I misread problems, too.  Also, Look at the difficulty level: School Sprint.  There's no way the question would have gone as in-depth as Izzy and Jongao took it.  Therefore, Cal Culator's solution is correct.


A white line is drawn in the direction of the north and south in the schoolyard. At first, a person is standing on a point which is away 5 meteres in the direction of west from the $A$ point on the line. When he toss up a coin there, if the head appeares, he will proceed by 1 meter in the direction of the east ,  or the tail appeares, he will proceed by 1 meter in the direction of the north. Note that he arrives on the line, he must stop there.Let $p_{n}$ be the probability such that the person arrives in the point of $n$ meter toward the direction north from the $A$ point. Find the value of $n$ for which $p_{n}$ is maximized.

I think no one has responded because the problem is unreadable, did you use some sort of translator service? I think you should proofread it a little... it looks like a ninteresting problem.

I have just edited.  How about this time? :)

Sorry for complaining to you about your wording... :(
It is much better this time, anyway here is how I would say it:
[quote]A north-south white line is drawn in a schoolyard. At first, you are 5 meters due west from a point $A$ on the line. You toss a coin, and if it is heads move 1 meter east. If it is tails, move 1 meter north. Note that once you reach the line you must stop. Let $p_{n}$ be the probability that you stop at the point $n$ meters north from $A$. Find the value of $n$ for which $p_{n}$ is maximized.[/quote]

[quote="kunny"]A white line is drawn in the direction of north and south in the schoolyard. At first, a person is standing on a point which is 5 meteres west of a point $A$, on the line. He tosses a coin; if the head appears, he will walk 1 meter east ,  or if tails appears, he will walk 1 meter north. When he arrives on the line, he must stop there. Let $p_{n}$ be the probability such that the person arrives on the line $n$ meters north of point $A$. Find the value of $n$ for which $p_{n}$ is maximized.[/quote] [hide] $p_{n}$ is the number of ways to arrange $HHHHHTT...T$ with $n$ Ts (and T cannot be at the end), divided by the number of ways he could have tossed $n+5$ coins. Or $\frac{\binom{n+4}{4}}{2^{n+5}}$. Don't know how to maximize that...  :maybe: [/hide]

Your answer is correct.

[hide]By your result, we have $p_{1}<p_{2}<p_{3}=p_{4}>p_{5}>p_{6}>p_{7}>\cdots.$[/hide]

[quote="kunny"]Consider the range of $n$ such that $p_{n+1}>p_{n}.$ :)[/quote] Thanks for the new method! :D


That's not true if $ n\in N$. Take for an example $ n\equal{}1$.

i'm checking for $ 2^{n}| [n\plus{}\sqrt{5}]^{n}$

...

err... it's not true either

Note that for positive $ n$, $ a^n | b^n \iff a | b$. So, $ 2^n | (n \plus{} \sqrt{5})^n \iff 2 | n \plus{} \sqrt{5}$. Assuming that we intend for $ n$ to be an integer, there are no solutions apart from the trivial case $ n\equal{}0$.

[quote="The Zuton Force"]Note that for positive $ n$, $ a^n | b^n \iff a | b$. [/quote]

Wrong !

Eg : $ (\sqrt{2})^2\vdots 1^2$ but $ \sqrt{2}\not\vdots 1$

You cant really apply divisibilty to reals.

My best guess was that this is supposed to apply to $ \mathbb{Z}[\sqrt{5}]$... but this still fails for several $ n$.

As for divisibility in the reals, basically for any field (rationals, reals, complex numbers) everything is divisible by everything, with the exception that 0 divides only itself. So yeah, it doesn't have much meaning.


inequalities
geometry
parallelogram
vector
summer program
PROMYS

Ok the wording might be bad, but screw it.

$\frac{a}{b}$ and $\frac{c}{d}$ are irreducible fractions with $a, b, c, d$ all integers, $a<b$, $c<d$, and $\frac{a}{b}<\frac{c}{d}$. Prove that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$.

Actually, I'm not entirely sure this is true, but try to prove it.

[hide]
It is for any a, b, c, d as long as $\frac{a}{b}<\frac{c}{d}$ and $a, b, c, d, a+c, b+d$ are positive. \[\frac{a}{b}< \frac{a+c}{b+d}\iff ab+ad<ba+bc \iff ad<bc \iff \frac{a}{b}< \frac{c}{d}\]  \[\frac{c}{d}>\frac{a+c}{b+d}\iff cb+cd > da+dc \iff bc>ad \iff \frac{c}{d}> \frac{a}{b}\] [/hide]

If the fraction is negative, let the numerator be negative, so that cross multiplying does not affect the sign. Then \[\frac{a}{b}>\frac{c}{d}\] implies \[ad<bc,\]  \[ab+ad<ab+bc,\]  \[a(b+d)<b(a+c),\]  \[\frac{a}{b}>\frac{a+c}{b+d}.\] Use a similar argument for the other side by adding $cd$ to both sides.

[quote="t0rajir0u"]It's obvious geometrically.  Consider the parallelogram formed by the vector addition of $\left< a, b \right>+\left< c, d \right>$.[/quote]

Heh...yeah, you could observe this in continued fractions...did that at promys


AMC
AIME
USA(J)MO
USAMO
ARML
MATHCOUNTS
AMC 10

How did this AIME go for Georgia (or at least, people who visit this forum)?  You can talk here, or I've added a poll if you want to be anonymous.

Didn't even take an AMC. Poor, sad me.  :|

Miles, you got a 14=15?  Sigh, such amazing people.  As you can see from the poll, I failed miserably and have a high, nay, astronomical chance of not making USAMO :( .

EDIT: More people polled, so yeah, I'll admit that I got a 6 on the AIME.  I know, it's bad, huh?

[quote="Math Geek"]Miles, you got a 14=15?  Sigh, such amazing people.  As you can see from the poll, I failed miserably and have a high, nay, astronomical chance of not making USAMO :( .[/quote]
Oh, no, I got a 12; I don't know who voted that.  Anyhow, assuming the score distribution parallels the distribution from 2006, you did better than 90% of the AIME qualifiers.  Several solid ARML people got sixes on the AIME that year.  (I'm comparing the two years because many people are saying they were about the same in difficuly.)  So you did not fail miserably.  You are in middle school, yes?  I could never have gotten a 6 on the AIME in middle school (I didn't even hear about it until too late).  In 2006, Georgia's [i]only[/i] middle-school AIME score was a 5.  On the other hand, it is true that your USAMO chances are slim, for this year.  But take this as motivation to work harder for next year.  Do not coast.  Aim for MOP.

Sitan, I got the same score last year that you got this year. Don't be disappointed. (Well, I was, but that's behind me now.)
I got an 8, from what other people are posting.
"I must work harder." (Quoth Boxer from [u]Animal Farm[/u])

nice miles. for some reason i can never seem to break double digits on the aime. rather sad, but oh well. doubt i made usamo

As an impartial observer (one who did not officially take the AIME I), I think the index will be about 220 with a floor of 8 or a floor of 7 with severe tie breaking. Wish me luck on the AIME II!

[quote="Arvind_sn"]OMG Dood it's the first poll in GA forum history!!! :clap2:

But yea...ultimate fail, i got a 5!!!! NOO!!! :cursing:

I guess I can remove perfect at USAMO and IMO off my "goals for 2008" list... :P[/quote]

And your "make 7 or 8" on AIME." :P  Ah well, I shouldn't be poking fun, seeing as I failed almost as much.

[quote="Goldentide"]I completely messed up; I only got 4.

I didn't know that I could make so many careless errors on one test...[/quote]

Felt the same way last year with the same score.

This year I got a 9: 1,2,3,4,5,6,8,9,12. Guessed $ \sqrt[3]{867}\rightarrow 870$ for #15. That's probably the closest I'll come to guessing correctly.

In my opinion, 1-5 were way too easy and didn't deserve being on the AIME.  Then there was the one with the crates which I thought was pretty easy.  That's all I got right, so yeah, I'm pretty nubbish. :lol:

First, Congratulations to everyone who reached their goal, and best wishes for those who are taking AIME II. (You can ignore the two paragraphs below, which is just some babbling of one who failed his goal...)

I only got a 6, much lower than my expected 9ish from practice tests. I did 1-6, 8-9 and then spent one and an half hour on #7 (counting/NT...) and #10...but unfortunately I got 8 and 9 wrong because of last-step careless mistakes...

Looking back the test again, I had to admit it's actually a little easier than the past two years' AIME I. #14 and #15 were surprisingly simple and straightforward, but I didn't attempt them during the test...Well, you can't really get all you are able to do on the actual test, especially when you are not that solid, like me...

Okay, I still have chance since I'm a freshman, to look at the positive side. Besides, I have an unrelated question: can anybody tell me the process of making Georgia ARML team? Thanks.

I scored a five... should've been a six, but 225+25+64 != 316. Oh well, a six wouldn't have qualified me for the USAMO anyway so I'm alright with my five.

[quote="timwu"][quote="Boy Soprano II"][quote="timwu"]Okay, I still have chance since I'm a freshman, to look at the positive side. Besides, I have an unrelated question: can anybody tell me the process of making Georgia ARML team? Thanks.[/quote]
There is no automatic process for making ARML.  Beat ARML people at local competitions; attend local competitions.  (Unfortunately, there aren't that many left, except JV state for you guys, ARML local, the Walton math fest, and then varsity state.)  Do well on the AMC contests (also a little late, but it looks like you did pretty well for a ninth grader.)  Finally, distinguish yourself at the state contest.  As a ninth grader, you get more benefit of the doubt than a senior would, but you're still selected "not . . . on potential, but on demonstrated ability."  See the [url=http://paideiaschool.org/TeacherPages/Steve_Sigur/resources/GAARML/GEORGIAARML/Welcome.html]2006 Georgia ARML website[/url] for more.

Anyhow, at the state math competition, the ARML coaches take everything into account and select the people for the teams.  They announce them at the awards ceremony.[/quote]

Thanks. However, due to some accidental misinformation, our school will not be able to go to JV State this Saturday. Can I do the Varsity State instead? I didn't see it appear on the GA tournament list, though. Also, how can one attend the GCTM state competition, or, how can one know if his school is qualified? I have some vague memory that somebody said that even if your school is not qualified, you might still get an individual invitation from the coach. Is that true?

BTW, I'm from Parkview High School in Gwinnett County.[/quote]

Yes, you can get invited individually even if your school is not invited.

[quote="timwu"]Thanks. However, due to some accidental misinformation, our school will not be able to go to JV State this Saturday. Can I do the Varsity State instead? I didn't see it appear on the GA tournament list, though. Also, how can one attend the GCTM state competition, or, how can one know if his school is qualified? I have some vague memory that somebody said that even if your school is not qualified, you might still get an individual invitation from the coach. Is that true?

BTW, I'm from Parkview High School in Gwinnett County.[/quote]
The GCTM tournament and the varsity state tournament are one and the same.  Your coach will know (eventually) whether your school has qualified or not.

From a post made earlier by an ARML coach in this forum, I believe this is how schools are invited to the GCTM tournament.  At all the local tournaments in the year (apart from the Walton Math Fest, which happens too late), the coaches give each school a score, which is the number of schools attending the tournament divided by the school's place (thank you, Santhosh).  The schools with the 30-40 best scores are invited.  I believe it was said in the past that a score of 5 is usually good enough.  But my memory could be faulty, and I can't find the particular post at the moment, so if an ARML coach says something else, (s)he is probably right and I am probably wrong.

The ARML coaches might also invite you as an individual if you will not be attending as a member of a team.  I'm not exactly sure how this works, because it's never happened to me, but I don't think such people are eligible for awards, but the coaches will consider their results when choosing the ARML team.

Anyhow, I urge you to go to JV state, even as an individual, because it's a good place for young people to distinguish themselves.  Many of the best people are out of the picture, but there are still enough ARML people around to compete against, and you won't be penalized for lack of knowledge like you might be at an ordinary varsity tournament.

If you really can't go to JV state, then there's still ARML local, but I imagine that it will be tough to do well enough there to merit an individual invitation.  But I imagine your school will be invited, although to be honest, I have no idea.

[quote="Boy Soprano II"]At all the local tournaments in the year (apart from the Walton Math Fest, which happens too late), the coaches give each school a score, which is the school's place divided by the number of schools attending the tournament.  The schools with the 30-40 best scores are invited.  I believe it was said in the past that a score of 5 is usually good enough.  But my memory could be faulty, and I can't find the particular post at the moment, so if an ARML coach says something else, (s)he is probably right and I am probably wrong.[/quote]

Mr. Slater told me that the score was the number of teams attending the tournament divided by the school's place (not the other way around), and it was top 50 teams which had approximately a score of slightly less than 4, but that is what I remember from two years ago. I also heard something about individuals being ineligible for awards, but I'm not entirely certain. Regardless, I believe that Boy Soprano's post is fairly accurate. You should try to go to the JV State tournament if you aren't busy with Science Olympiad or something else.

[quote="Xantos C. Guin"]Mr. Slater told me that the score was the number of teams attending the tournament divided by the school's place (not the other way around), and it was top 50 teams which had approximately a score of slightly less than 4, but that is what I remember from two years ago. I also heard something about individuals being ineligible for awards, but I'm not entirely certain. Regardless, Boy Soprano's post is fairly accurate. You should try to go to the JV State tournament if you aren't busy with Science Olympiad or something else.[/quote]
Yes, all that's correct; my math teacher enjoys reminding people that he's the one who proposed the current system. :wink: I wonder what JV state will be like this year, now that it's not at Woodward for the first time. Apparently they're keeping most of the format the same.

All of these posts are basically correct. We never know what the cutoff for an invitation is for state until the end of March. That's why the Walton MathFest doesn't have a team competition. (Math Bowl doesn't count.) If your team isn't invited, it is possible that you will receive an invitation from an ARML coach. This typically only happens if you have been identified as a "person of interest". You achieve this status by performing well at math competitions. AMC/AIME is a great way to get noticed.

(Sorry Karthik, the top 4 MATHCOUNTS finishers got invited this year. Of course you don't actually have to bring Arvind!  :D)

Tom

Thanks for all the help. After many tries, I eventually managed to make up a team of four going for the JV State yesterday afternoon; but the two of those seemed to never appear on the competition site today morning. Therefore, uh, when the test began, there were only two people on the Parkview Team; and of course we failed the team tests (we actually didn't do the relay officially and got a 0 on it, which was in favor of us considering our score was -15)...

However, I did do well on the individual test and ended up with a first place. With my another first place on the Ga Tech JV and a 144 on AMC 10A & 108 on AMC 12B (and the 6 on AIME as mentioned before), do you think that this can assure me an individual invitation for GCTM, if my school is not invited? I think my main problems are: 1, inadequate attendance of local tournaments this year (the only ones are UGA, Ga Tech, JV State, and some other competitions held in my school if they count); 2, lack of good performance on varsity-level competitions, like UGA and AMC 12 (which will probably also apply for the GCTM, provided I'm invited).

[quote="timwu"]I did do well on the individual test and ended up with a first place. With my another first place on the Ga Tech JV and a 144 on AMC 10A & 108 on AMC 12B (and the 6 on AIME as mentioned before), do you think that this can assure me an individual invitation for GCTM, if my school is not invited?[/quote]
You won JV state?  Congratulations.  Yes.

Thanks very much.  :)

A friend of mine (a sophomore) has the same question. He, too, got a 144 on AMC 10 but didn't do very well on AMC 12 (105ish). On local tournaments, as I know, he had a 13th place on UGA and top 10 on GA Tech JV Division. However, he was unable to attend the JV State because of the state science olympiad and thus lost a good chance to make a top 5. I hope he could get a decent score on the coming AIME II, but he was not confident himself. So, how much chance does he have to get invited for GCTM, not taking the AIME II into account? We both are looking forward to your reply.

Neither Miles nor I can say definitively whether or not your friend will be invited, but one of the criteria for invitation seems to be to beat former ARML members, and he beat about 10 former ARMLers at UGA alone. His AMC10 score should also catch significant attention from the ARML coaches. I would think he would be invited.

[quote="Twiz"][quote="Math Geek"]I failed miserably and have a high, nay, astronomical chance of not making USAMO :( .[/quote]
Interesting... :P[/quote]

Well in my opinion and the opinion of many, I did fail miserably, but I never said I had no chance whatsoever of making USAMO (okay, fine, I said that in my blog, but no one needs to no that :ninja: ).  My chances of qualifying were admittedly quite slim, but fate (and that guy who got a 144 on his AMC 12 and 6 on his AIME) smiled upon me, so yeah, interesting circumstances indeed, but my statement did not prove to be false. :wink:

PS: I'm still in shock. :|

GUESS WHO MADE CARELESS MISTAKES ON THE AIME AND MISSED MORE THAN 7 QUESTIONS DUE TO CARELESS MISTAKES!

not me, because i only made 3

BUT STILL

this officially sucks... i missed the cutoff again by like 7 or so points... if only i had gotten that arctan problem :wallbash_red:

[quote="Arvind_sn"]Guess who missed USAMO by one question:

MEEEEEE :surrender:[/quote]

Arvind, you had to make the cutoff USAMO index, too, I think, which was 202.5, so yeah. :wink:

I wish I didn't live in the middle of nowhere and my school actually did AMC... I sat down with the AIME I, and I can do eight or nine of them. Blarg.

Have invitations gone out for State yet? I'm worried that neither I nor my school has been invited, which would suck much. I would love an opportunity to go to ARML, and I feel like I could make it.

Just curious, but what's your real name?

Sorry. Should've introduced myself. I'm Stephen Brown. I'm in Pike County. Hello World.

Made USAMO with 9 AIME I and 121.5 AMC  :lol:
Yay!!
- Oliver

Steven, you completely owned Columbus State and Georgia Southwestern, so I think you will get invited.


trigonometry
calculus
integration
limit
complex analysis

Hi, can anyone help me with this problem?  Thanks.

Find $ \displaystyle\mathcal{L}^{\minus{}1}\left\{\frac{1}{s\cosh s}\right\}$

I believe:

$ \mathcal{L}^{\minus{}1}\left\{\frac{1}{s\cosh s}\right\} \equal{}2U[\minus{}\cos(\pi/2 t)]$

via the Residue Theorem where $ U$ is the unit step function.

May you please show your solution?

According to the back of the book the answer should be $ \displaystyle 1 \minus{} \frac {4}{\pi}\left[\cos\frac {\pi t}{2} \minus{} \frac {1}{3}\cos\frac {3\pi t}{2} \plus{} \frac {1}{5}\cos\frac {5\pi t}{2} \minus{} \cdots\right]$

I believe, although I've not rigorously proven below that
\[ 1 \minus{} \frac {4}{\pi}\left[\cos\frac {\pi t}{2} \minus{} \frac {1}{3}\cos\frac {3\pi t}{2} \plus{} \frac {1}{5}\cos\frac {5\pi t}{2} \minus{} \cdots\right] \equal{} 2U[ \minus{} \cos(\pi/2 t)].
\]
Your expression is the sum of all the residues right. For now, let's assume we use a square contour and the integral over the three excipient legs of the contour goes to zero although showing that is really an integral part of the problem but let's continue.  Your sum is:
\[ 1 \plus{} \frac {4}{\pi}\sum_{n \equal{} 0}^{\infty} \frac {( \minus{} 1)^{n \plus{} 1}\cos\left[(2n \plus{} 1)/2\pi t\right]}{2n \plus{} 1}.
\]
But
\[ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \plus{} 1}z^{2n \plus{} 1}}{2n \plus{} 1} \equal{} \minus{} \arctan(z),
\]
so if we let $ z \equal{} e^{\pi i t/2}$ and $ z \equal{} e^{ \minus{} \pi i t/2}$ and then add the sums up, the imaginary component $ \sin((2n \plus{} 1)/2 \pi t)$ cancels leaving
\[ \sum_{n \equal{} 0}^{\infty}\frac {( \minus{} 1)^{n \plus{} 1}\cos\left[(2n \plus{} 1)\pi t/2\right]}{2n \plus{} 1} \equal{} \minus{} \frac {1}{2}\left(\arctan\left(e^{\pi i t/2}\right) \plus{} \arctan\left(e^{ \minus{} \pi i t/2}\right)\right).
\]
Now
\[ \arctan(x) \plus{} \arctan(1/x) \equal{} \begin{cases} \displaystyle\frac {\pi}{2} & x > 0 \\
\displaystyle \minus{} \frac {\pi}{2} & x < 0. \end{cases}
\]
At this point I'm not sure but I think the expression holds for $ Re(z) > 0$ as well.  Perhaps someone here can comment on this.  Assuming it does, then the sum of the residues leads to a Heaviside function.  So that
\[ \begin{aligned}\mathcal{L}\left\{\frac {1}{s\cosh(s)}\right\} & \equal{} 1 \plus{} \frac {4}{\pi}\sum_{n \equal{} 0}^{\infty} \frac {( \minus{} 1)^{n \plus{} 1}\cos\left[(2n \plus{} 1)/2\pi t\right]}{2n \plus{} 1} \\
& \equal{} 1 \minus{} \frac {2}{\pi}\left\{\arctan\left(e^{\pi i t/2}\right) \plus{} \arctan\left(e^{ \minus{} \pi i t/2}\right)\right\} \\
& \equal{} 1\pm 1 \end{aligned}
\]
depending on the real component of $ e^{\pi i t/2}$.  Obviously I'm having some problems with this as well but at this point I actually plotted that $ \arctan$ expression and noted that it was a square-wave from zero to two changing at $ (2n \plus{} 1)$ which is equivalent to $ 2U\left[ \minus{} \cos(\pi/2 t)\right]$.

Hi youriens, thanks for your reply.  I will put up what I did.  Hopefully, there would be other people here that would offer their opinions.

Let $ \displaystyle f(s)\equal{}\frac{1}{s\cosh s}$.  Then $ s\equal{}0$ is a simple pole as is $ \displaystyle s\equal{}s_k\equal{}(2k\plus{}1)\frac{i\pi}{2}$ for $ k\equal{}0,\pm 1\pm 2\cdots$.

The residue of $ \displaystyle e^{st}f(s)$ at $ s_k$ is found by $ \displaystyle\lim_{s\to s_k}(s\minus{}s_k)\left(\frac{e^{st}}{s\cosh s}\right )$
Then by L'Hospital's Rule, equals $ \displaystyle \lim_{s\to s_k}\left(\frac{1}{\sinh s}\right)\cdot\frac{e^{s_kt}}{s_k}$
Hence, the residue equals $ \displaystyle \frac{2(\minus{}1)^{k\plus{}1}e^{i(2k\plus{}1)\pi t/2}}{(2k\plus{}1)\pi}$
The residue at $ s\equal{}0$ is $ \displaystyle \lim_{s\to 0}(s\minus{}0)\left(\frac{1}{s\cosh s}\right)\equal{}1$

By the Laplace Inversion formula, $ \displaystyle\mathcal{L}^{\minus{}1}\left\{\frac{1}{s\cosh s}\right\}\equal{}1\plus{}\sum_{k\equal{}\minus{}\infty}^{\infty}\frac{2(\minus{}1)^{k\plus{}1}e^{i(2k\plus{}1)\pi t/2}}{(2k\plus{}1)\pi}$.
This simplifies to $ \displaystyle \boxed{1\plus{}\frac{4}{\pi}\sum_{k\equal{}0}^{\infty}\frac{(\minus{}1)^{k\plus{}1}}{2k\plus{}1}\cos\frac{(2k\plus{}1)\pi}{2}}$


geometry
circumcircle
trigonometry
projective geometry
geometry proposed

[color=darkblue]Let $ABC$ be a triangle with the incircle $i=C(I)$ and circumcircle $e=C(O)$.
Define the second intersections $A'$ , $A_{1}$ between the circle $e$ and the lines $AO$ , $AI$
Note the intersections $X\in AI\cap CA'$ and $Y\in AB\cap IA'$.
The circle $i$ touches the side $[BC]$ in the point $D$.
Prove that $D\in XY\ .$[/color]

Let the incircle touch CA, AB at E, F. Let CA' meet AB at Z. Use Menelaus theorem for $\triangle ZBC.$ Assuming Y, A, B follow on the line AB in this order, other cases are similar. First, $\frac{DB}{DC}= \frac{s-b}{s-c}.$ Next, AA' is a circumcircle diameter, $\angle ACA'$ is right, AX bisects $\angle CAZ$ in $\triangle ACZ,$ hence $\frac{XC}{XZ}= \frac{CA}{AZ}= \cos A.$ In the right $\triangle ABA',$ $\angle BA'A = \angle C,$ $A'B = \frac{AB}{\tan C}= \frac{c}{\tan C}$ and in the right $\triangle ZBA',$ $\angle ZBA' = \angle A,$ $BZ = A'B \tan A = c\ \frac{\tan A}{\tan C}.$ The right $\triangle A'BY \sim \triangle IFY$ are similar,

$\frac{YB}{A'B}= \frac{YF}{IF}= \frac{YB-FB}{IF},\ \ YB = \frac{A'B \cdot FB}{A'B-IF}= \frac{c(s-b)}{c-r \tan C}.$

$\frac{YZ}{YB}= 1+\frac{BZ}{YB}= 1+\frac{\tan A}{\tan C}\cdot \frac{c-r \tan C}{s-b}$

Substituting $c\ \frac{\tan A}{\tan C}= a\ \frac{\cos C}{\cos A}$ (by the sine theorem for $\triangle ABC$) and $r = (s-a) \tan \frac{A}{2}= (s-a)\ \frac{1-\cos A}{\sin A},$

$\frac{YZ}{YB}= 1+\frac{a}{s-b}\cdot \frac{\cos C}{\cos A}-\frac{s-a}{s-b}\cdot \frac{1-\cos A}{\cos A}$

Substituting $b = c \cos A+a \cos C$ to get rid of $\cos C$ (that's how the B-altitude cuts CA),

$\frac{YZ}{YB}= \frac{(s-b) \cos A+b-c \cos A-(s-a)+(s-a) \cos A}{(s-b) \cos A}=$

$= \frac{(2s-a-b-c) \cos A+(a+b-s)}{(s-b) \cos A}= \frac{s-c}{s-b}\cdot \frac{1}{\cos A}$

As a result, $\frac{DB}{DC}\cdot \frac{XC}{XZ}\cdot \frac{YZ}{YB}= \frac{s-b}{s-c}\cdot \cos A \cdot \frac{s-c}{s-b}\cdot \frac{1}{\cos A}= 1,$ which means that X, D, Y are collinear.

[quote="Virgil Nicula"][size=117][color=darkred]Let $ABC$ be a triangle with the incircle $i=C(I)$ and circumcircle $e=C(O)$.
Define the second intersections $A'$ , $A_{1}$ between the circle $e$ and the lines $AO$ , $AI\ .$
Note the intersections $X\in AI\cap CA'$ and $Y\in AB\cap IA'$.
The circle $i$ touches the side $[BC]$ in the point $D$. Prove that $D\in XY\ .$[/color][/size][/quote]

[color=darkblue][size=117][b]Proof.[/b] From the lemma presented at the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=111734

obtain that the intersection $S\in DA_{1}\cap IA'$ belongs to the circumcircle of the triangle $ABC\ .$

Appear the following lines : $\overline{SYIA'}\ ,\ \overline{SDA_{1}}\ ,\ \overline{AIA_{1}X}\ .$

Appy the [u]Pascal's theorem[/u] to the hexagon $ABCA'SA_{1}\ :$ ${\{\begin{array}{c}Y\in AB\cap A'S\equiv AB\cap IA'\\\\ D\in BC\cap SA_{1}\\\\{ X\in CA'\cap A_{1}A}\equiv CA'\cap AI\end{array}\Longrightarrow D\in XY\ .$[/color][/size]


How many more distinguishable arrangements are there of the
letters TALK than of the letters TOOT?

There are 4! = 24 ways to arrange the letters in TALK. Similarly, there are $ \frac{4!}{2!2!} \equal{} 6$ for TOOT. Thus, our answer is 18.