Prove that for every irrational real number $a$, there are irrational real numbers $b$ and $b'$ so that $a+b$ and $ab'$ are both rational while $ab$ and $a+b'$ are both irrational.

[quote="mecrazywong"]one of (1-a)a,(2-a)a must be irrational; one of a+1/a,a+2/a must be irrational.[/quote]
 :D  Our solutions are exactly the same!

[quote="erdos"][quote="mecrazywong"]one of (1-a)a,(2-a)a must be irrational; one of a+1/a,a+2/a must be irrational.[/quote]
Can you please post a more detailed solution. I didn't understand yours very well.  :lol:[/quote]
Isn't that obvious?
(2-a)a-(1-a)a=a, which is irrational, so (2-a)a and (1-a)a cannot be both rational. Similar approach for the other one.

My solution is different.
1. I take b=1-a
2. I showed that if a(1-a) is rational, it must be of the form p+sqrt(q), where p,q are rational.
3. I take b=-sqrt q if a is of the form p+sqrt(q).

Similar for b'.

My solution is almost same to the crazy. But a bit generalized.

p, 1/kp

p*1/kp=1/k Rational.

If p+1/kp is irrational then we will be done. So let for all rational k p+1/kp is rational.

Then:
p+1/kp – p-1/tp=1/p*(1/k-1/t) which is irrational. But rational-rational should be rational. So we get a contradiction.

So we can get such rational k for which p+1/kp is irrational.

Now take, p, k-p
p+k-p=k Rational.

Say for all rational k p(k-p) is rational.

Then:
pk-p^2-pt+p^2=p(k-t) contradiction!

That ends my task!

Siuhochung, can you show yours?

I have showed mine.

1. I take b=1-a
2. I showed that if a(1-a) is rational, it must be of the form p+sqrt(q), where p,q are rational.
3. I take b=-sqrt q if a is of the form p+sqrt(q).

Similar for b'.

All these steps are easy to be shown.

Take $b=z-a : z \in Z$
if $a(z-a) \in Q \forall z \in Z$
 then $a(z-a) -a(z+1-a) \in Z\implies a \in Q $ a contradiction
Similar argument for$ b'= \frac{z}{a}$

Let $b=n_1-a$ and $b'=\frac{n_2}{a}$ where $n_1,n_2\in \mathbb{Z}-\{0\}$ (it is obvious that $b,b'\not\in \mathbb{Q}$), we claim that $\forall n_1,n_2\in \mathbb{Z}-\{0\}$ we can find a pair $(n_1,n_2)$ that makes $a+b, ab'\in \mathbb{Q}$ and $ab, a+b' \not\in \mathbb{Q}$.
Assume for contrary, we then have $a+b, ab'\in \mathbb{Q}$, $ab=a(n-a)\in \mathbb{Q} \ \forall n\in \mathbb{Z}-\{0\}$, let $s,t \in \mathbb{Z}-\{0\}$ such that \[a(s-a), a(t-a) \in \mathbb{Q}\] subtracting one from the another gives \[a(s-t)\in \mathbb{Q} \implies a\in \mathbb{Q}\] a contradiction. Similarly, we have $a+\frac{n}{a}\in \mathbb{Q} \ \forall n\in \mathbb{Z}-\{0\}$, let $u,v \in \mathbb{Z}-\{0\}$ such that \[a+\frac{u}{a}, a+\frac{v}{a} \in \mathbb{Q}\] subtracting one from another gives \[\frac{u-v}{a}\in \mathbb{Q}\implies a\in \mathbb{Q}\] also a contradiction.
Thus, we have that there must exist $(n_1,n_2)$ where $b=n_1-a$ and $b'=\frac{n_2}{a}$ such that $a+b, ab'\in \mathbb{Q}$ and $ab, a+b' \not\in \mathbb{Q}$. $\blacksquare$

Gosh, my solution was more sofisticaded, don't have much sure if it is correct, but here it is:

We can take $q$ irrational and $p$ rational, such that $p^2-4q$ is not a square of another rational. Hence we can create a second degree equation as $x^2-px+q=0$, with roots $a$ and $b$, which satisfies our claim. Same for $a$ and $b'$, $x^2-nx+m$, with $n$ irrational and $m$ rational.

Posting for storage
[hide=Solution]
$\textbf{Part 1)}$ \\
We shall firstly prove the existence of some irrational $b$ satisfying the condition. \\
Notice that if $a < 0 $, we can take $a \rightarrow -a$, $b \rightarrow -b$ which once again satisfies the condition so assume that $a>0$. \\
Then let $a = \sqrt{x}$ for some $x \in \mathbb{R}$. \\
If $x$ is irrational, take $b = -a$ and notice that $a+b = 0$ is rational, meanwhile; $ab = -a^2 = -x$ is irrational and consequently, $b=-a$ satisfies the condition. \\
If on the other hand $x$ is rational, take $x=a^2-a$, then $a+b = a^2 = x$ is rational, meanwhile; $ab = a^2(a-1)$ which must be irrational $a^2$ is rational and $a-1$ is irrational. $\square$ \\
$\textbf{Part 2}$ \\
We claim that either $b' = \frac{2}{a}$ or $b' = \frac{1}{a}$ works. \\
Assume that neither do, we clearly have that $ab' = a \cdot \frac{2}{a} = 2$ or $ab' = a \cdot \frac{1}{a} = 1$ both of which are rationals, on the other hand $a+b' \in \frac{a^2+1}{a}, \frac{a^2+2}{a}$. Assume that both of these reals are also rationals, we then have that their difference which is $\frac{a^2+2}{a} - \frac{a^2+1}{a} = \frac{1}{a}$ must also be a rational which is a contradiction. $\blacksquare$[/hide]

Claim: Either $b=1-a$ or $b=2-a$ works. Suppose not. Then, $a(1-a)$ and $a(2-a)$ are both rational. However, they differ by $a$, an irrational number, contradiction.\\

Claim 2: Either $b'=1/a$ or $b'=2/a$ works. Suppose not. Then, $a+1/a$ and $a+2/a$ are both rational, but they differ by $1/a$, which is irrational, contradiction.\\

Hence, we are done.


A bag contains 12 red marbles and 6 blue marbles. Two marbles are selected at random and without replacement. What is the probability that one marble is red and one is blue? Express your answer as a common fraction.

This is simply $ \frac{12\cdot6}{\dbinom{18}2}\equal{}\boxed{\frac8{17}}$.

first of all, i put this problem up here

now...
i noticed this problem has only 1 answer box even though the answer is a fraction
Do all fractional-answer-questions now only have 1 entry box?

secondly, i would like to propose a shorter solution than the one given

[hide="precise form"]
Suppose all marbles are distinct and that the red marbles are denoted $ r_1, r_2, r_3, \ldots r_{12}$ and the blue marbles are deoted similarly. A group of 1 red marble and 1 blue marble may be denoted $ r_k, b_l$ for all $ 1 \le k \le 12, 1 \le l \le 6$. Note that k and l are independent of eachother. There are $ 12 \cdot 6 \equal{} 72$ ways to assign values to k and l, so there are 72 groups that consist of 1 marble and 1 blue marble. We are to choose 2 marbles out of $ 12 \plus{} 6 \equal{} 18$, so there are $ \binom{18}{2} \equal{} 9 \cdot 17$ total ways to choose 2 marbles without regard to order. The number of ways to choose 1 red marble and 1 blue marble is $ 72 \equal{} 8 \cdot 9$, so the answer is $ \frac {9 \cdot 8}{9 \cdot 17} \equal{} \frac {8}{17}$
[/hide]
[hide="casual form"]
there are 12 red marbles and 6 blue marbles, so there are $ 12 \cdot 6 \equal{} 72$ ways to choose 1 red 1 blue marble. there are $ \binom{12 \plus{} 6}{2} \equal{} 153$ ways to choose 2 marbles. the aswer is then $ \frac {72}{153} \equal{} \frac {8}{17}$
[/hide]


1. How many times does the digit 1 appear in the list of whole numbers from 0 to 100 inclusive?

2. The value of $ \frac{12345678901234567890}{(12345678901234567891)^{2})\minus{}(12345678901234567890)(12345678901234567892)}$ is an integer. Find the sum of its digits.

3. If a is a positive integer, how many solutions are there to $ \sin{ax}\plus{}\cos{ax}\equal{}1$ on the interval $ 0\le x<\frac{2\pi}{a}$?

4. What's the value of $ {\Sigma}^{\plus{}\infty}_{k\equal{}3}\sin^{k}(\frac{\pi}{6})$?

5. How many 5 digit numbers are there satisfying the condition that every digit is greater than the digit to its right?

[hide="hint for 5"]No matter what 5 numbers you choose, there is only one way to arrange them.[/hide]

[hide="Problem 3"]
$ \sin ax\plus{}\cos ax \equal{} 1$
Square,
$ \sin^{2}ax\plus{}2\sin ax\cos ax\plus{}\cos^{2}ax \equal{} 1$
So,
$ \sin 2ax \equal{} 0$
This means,
$ 2ax \equal{}\pi n ,\ n\in\mathbb{Z}$.
Since $ a\not \equal{} 0$ we have,
$ x \equal{}\frac{\pi n}{2a}$ and we want $ x\in\left[0,\frac{2\pi}{a}\right)$
So, there are solutions for,
$ x \equal{} 0 ,\frac{\pi}{2a},\frac{2\pi}{2a},\frac{3\pi}{2a}$.
Now check if these solve the original equation.
[/hide]

1.

Count:
0-9: 1.
10-19: 10+1=11
20-29: 1
...
90-99: 1
100: 1

total=11+9*1+1=21 (times) Ans.

3.

By subsidiary angle, we can convert the equation to:

(sqrt1/2)*sin(ax+45)=1
sin(ax+45)=(sqrt2)
Because x belongs to [0,360/a), (ax+45) belongs to [0,405)
so,ax+45=45,135
x=0,90/a Ans.

4.

The series can be rephrased into:

(1/2)^3+(1/2)^4+...
=(1/2)^3*[1+1/2+(1/2)^2+...]
=1/8*[1/(1-1/2)]
=1/4 Ans.

5.

According to the given condition, we can only choose different numbers through 0 to 9, and they can only be arranged in decreasing number.Thus, the answer is: C(10,5)=252.Ans.

(Sorry, but I don't know how to hide text and type standard maths sentences)


How many real roots does $\sin{x} = \log_{10}{x}$ have?

I dunno if i did this right but:
[b]1[/b]
10 sinx = x
sin(x)log(10) = x
sin(x) = x
0

i think 3?

i graphed it on my ti-89, and the graphs intersect in  places.

kinda cheap tho

Graphing is sometimes the way to go when comparing functions.  I was more interested in people solving this one without a graphing calculator do it for them however...

If you're in degree mode, I'd expect it to intersect twice...

Should the mode you're expressing angles in matter?

I'unno... radian mode cycles about 57 times faster than degree mode...

I m pretty sure that you can't take the log of a number that is in degrees.  So it should be in radians.

I am  curious to know that  how many people  noticed  "right away" (that is not even without thinking)  that log (pi)  is about 0.5 , and log (5*pi/2) is about .9 ?

Thus the log curve meet the sine curve between pi/2 and pi  (sin goes from 1 to 0 and log goes from 0.2 to .5) and on the two sides of  (5 pi/2) .

Yes, not that long ago, before the age of calculators, all our calculation was done by hand ( in my math  and science tests, kind of now when "approved calculators" are allowed, we were allowed "log tables" )  and one comes to know these things by heart (even after many decades) that  pi^2 is about 10,  and thus log(pi) is about  half.   ...

Wonder how many people recognze now numbers like:

0.4343 (or its reciprocal 2.303)
0.6931 or .30103
etc...


Determine the least natural number $n$ for which the following holds: No matter how the numbers 1 to n is divided into two disjoint sets, in at least one of the sets, there exist four (not necessarily distinct) elements $w, x, y, z$ st $w+x+z=y$.

PS: corrected a very embarassing typo :blush:

[quote="zhaobin"]I am afraid I think there must be
$x+y+z=w$[/quote]

you are right, thanks for pointing that out.

I always wonder why an olympiad problem is posted in the pre-olympiad section and not in the olympiad ( formerly known as 'advanced') one.
Anyway, the minimal $n$ is $n=11$.

Assume that $A,B$ form a partition of $\{1, \cdots , 11 \}$ with no solution of the equation in one of the sets.
Wlog, we may assume that $1 \in A$. Thus $3 \in B \B$.
It follows that $9 \in A$ so that $11 \in B$.
Now :
- If $2 \in A$ then $4,5,6 \in B$. In that case $11 = 3+3+4$ gives a solution in $B$. A contradiction.
- If $2 \in B$ then $6,7,8 \in A$, but $8 = 6 + 1 + 1$ gives a solution in $A$. A contradiction.
It follows that the minimal $n$ satisfies $n \leq 11$.

In another hand, for $n=10$ and if $A = \{1,2,9,10 \}$ and $B = \{3,4,5,6,7,8 \}$, we have a partition into two parts with no solution of the equation. Hence result.

Pierre.

[quote="pbornsztein"]I always wonder why an olympiad problem is posted in the pre-olympiad section and not in the olympiad ( formerly known as 'advanced') one.

Pierre.[/quote]

because i feel it is really easy. i will put bmo 2 in the olympiad section


modular arithmetic
number theory proposed
number theory

Let $p$ be a prime such that $p^2$ divides $2^{p-1}-1$. Prove that for all positive integers $n$ the number $\left(p-1\right)\left(p!+2^n\right)$ has at least $3$ different prime divisors.

[i]Aleksandar Ivanov[/i]

Clearly for all $p > 3$ we have $2| p - 1$ and $p - 1 > 2$.
Therefore $\omega ( p - 1 ) \geq 2$. Let $q > 2$ be a prime such that $q|p - 1$.
Suppose that $q|p! + 2^n$. Since $q \leq p - 1$ we have $q|p!$, hence $q| 2^n$. A contradiction.
Therefore $p - 1$ and $p! + 2^n$ may possibly share in common only the prime $p = 2$.
Therefore
\begin{eqnarray*} \left. \left. \omega ( ( p - 1 ) \cdot ( p! + 2^n \right) \right) & = & 2 \end{eqnarray*}
if and only if $q|p! + 2^n \Longrightarrow q \mbox { power of two}$.
Therefore $\omega ( ( p - 1 ) \cdot ( p! + 2^n ) ) \geq 3$ if and only if
\begin{eqnarray*} p! & = & 2^m - 2^n \end{eqnarray*}
has no integer solutions in positive integers. We have
\begin{eqnarray*} p! & = & \left. 2^n \cdot ( 2^{m - n} - 1 \right) \end{eqnarray*}
Therefore $n = v_2 ( p! ) \leq p$. We have
\begin{eqnarray*} 2^{m - n} & \equiv & 1 ( %Error. "tmop" is a bad command.
{mod} p ) \end{eqnarray*}
This implies that $m - n| \varphi ( p ) = p - 1$ or $p - 1| m - n$ (since $( 2, p ) = 1$)
If $p - 1| m - n$ or $m - n = p - 1$ then
\begin{eqnarray*} p^2 & | & 2^{m - n} - 1 \end{eqnarray*}
And since $p! \equiv - p ( %Error. "tmop" is a bad command.
{mod} p^2 )$ this implies that $p! = 2^n \cdot ( 2^{n - m} - 1 )$ cannot hold.
Therefore $m - n|p - 1$ with $p - 1 \neq m - n$. Whence
\begin{eqnarray*} m - n & \leq & \frac{p - 1}{2}\\ m & \leq & \frac{p - 1}{2} + n\\ & \leq & \frac{p - 1}{2} + p \end{eqnarray*}
Therefore
\begin{eqnarray*} p! & = & 2^m - 2^n\\ & \leq & 2^m\\ & \leq & 2^{\frac{p - 1}{2} + p} \end{eqnarray*}
A contradiction.

[quote="{x}"][...] Therefore $\omega ( p - 1 ) \geq 2$ [...] [/quote]
Why¿  :?
(You need that $p-1$ is not a power of two).

PS: $n!=2^a-2^b$ has only finetely many solutions, thus we could also check them :)

You are absolutely correct. Unfortunately by no means I can fix that today :(

Argh ! The "proof" is absolutely flawed. I leave you the pleasure of discovering the mistake :)

[quote="ZetaX"]
PS: $n!=2^a-2^b$ has only finetely many solutions, thus we could also check them :)[/quote]

Hm, how can we do this?
And why $n!=2^a-2^b$ has only finitely many solutions, I don't know?

The $2$ can be changed to any integer, see http://www.mathlinks.ro/Forum/viewtopic.php?t=55883 .

[quote="bilarev"]Let $p$ is such prime that $p^2$ divides $2^{p-1}-1$. Prove that for all positive integers $n$ the number $\left(p-1\right)\left(p!+2^n\right)$ has at least 3 different prime divisors.[/quote]
Here si my way to solve this problem.
If $p$ has the form $2^k+1$then
\[ 2^{p-1}-1=2^{{2^k}}-1=(2^{{2^{k-1}}}+1)(2^{{2^{k-2}}}+1)...(2+1) \]
So $2^{p-1}-1$ can't dividisible by $p^2$.
Now if ther existe such a $r$ that
1.$t<r$=>$2^t-1$ isn't div. by $p$
2.$||2^r-1||_p=1$.
then it is known that if $2^x-1|p^2$=>$x|rp$(it is not hard to prove)
but as we have that $2^{p-1}-1|p^2$ and $(p-1;p)=1$=>there isn't such a $r$=> for any $q$ that $2^q-1|p$ we should have that $2^q-1|p^2$.
Now if $p!+2^n=2^m$ then $2^n(2^{m-n}-1)|p$=>$2^{m-n}-1|p^2$
=>$2^{m-n}-1|p^2$=>$p!|p^2$=>$(p-1)!|p$ so we get to the contradiction. ;)

[quote="Jumbler"]I think you misunderstood something, Tiks, the problem says $p^{2}|2^{p-1}-1$, not the other way round.[/quote]
Just read my solution again,leting that $X|Y$ means $X=kY$ for a positive integer $k$. :P

[quote="Jumbler"]Hmm... I think there's still something strange.

When you suppose there exists such a $r$, you have condition (1) as well, that is, if $t<r$ then $p$ doesn't divide $2^{t}-1$.

But in your conclusion, you don't have this. I think it should be "for any $q$ that $p$ divides $2^{q}-1$ and $p$ doesn't divide $2^{t}-1$ when $t<q$, then $p^{2}$ divides $2^{q}-1$"

Please correct me if I'm wrong  :)[/quote]

Look,let me restate my solution in other wards.I hope this time everything will be clear.

At first I mentioned something like a lemma.

[color=blue]Lemma.If $r$ is the smallest natural number such that $p|2^{r}-1$ and $p^{2}$ doesn't divide $2^{r}-1$,then any natural number $x$ that $p|2^{x}-1$ have the following form $x=kr$.[/color]

 Afterward I said that $p^{2}|2^{p-1}-1$ and $(p;p-1)=1$,which means that the set of numbers $r$ from the lemma is empty,that is for any natural number that $p|2^{r}-1$ we have that $p^{2}|2^{r}-1$.
The last two lines of the solution are written rather clear :P .
So,I hope now you haven't got any questions,have you? :huh:

[quote="Tiks"]
So,I hope now you haven't got any questions,have you? :huh:[/quote]

No!

It's a great solution  :lol:

[quote="maky"]i agree with you, Tiks, regarding the fact that the order of $2$ in $\mathbb{Z}_{p}$ is the same with the order of $2$ in $\mathbb{Z}_{p^{2}}$.
however, i don`t understand your proof. can you please explain it a little more ? my proof has somewhat similar ideas, but it is not exactly the same[/quote]

Okay,let's proceed this way:you are telling me what exactly you don't understand in the proof(i.e. a statement,a formula and so forth), and I am trying to make it clear for you. Is that okay?

[quote="maky"]i don`t understand your proof that $r$ does not exist. more precisely, when you take $r$ the order of $2$ modulo $p$, and then you claim that if $2^{x}\equiv 1\pmod{p^{2}}$, then $x|rp$. i can`t understand this. what about $x=12345\cdot p(p-1)$ ? :blush:[/quote]

Oh,I see the problem.But, actually I have asked you (Posted: Sun Nov 19,see above :roll: ) to understand by $X|Y$,that $X$ is dividisible by $Y$,see?

So your conterexample does not work because $x=(12345(p-1))p$ is dividisible by $p$ ,and you should still define $r$ :P  :D.

oh, sorry then. i have read now and i understood. i think i did the same thing... if $r$ is the order of $2$ modulo $p$, then $2^{pr}\equiv 1\pmod{p^{2}}$ and because $\gcd(p,p-1)=1$ it follows that $r$ is the order of $2$ modulo $p^{2}$.

We are already done if $p-1$ has at least $3$ different prime divisors. Therefore we need to consider $2$ other possibilities $p -1=2^aq^b$ or $p-1=2^a$, where $q$ is an odd prime and $a$ and $b$ are positive the integers.

Consider the first case. Note that $q \mid p!$, meaning that $q$ does not divide $p!+2^n$. We are done if $p!+2^n$ is not a power of $2$. Therefore it remains to deal with the case, when for some positive integer $n$, the number $p!+2^n$ is equal to the power of $2$, say $2^m$. Then we have the following equation:
$$ p!+2^n = 2^m \implies p! = 2^n(2^{m-n}-1) $$
Observe that $p \mid 2^{m-n}-1 \implies 2^{m-n} \equiv 1 \pmod p$, but $p^2$ does not divide $2^{m-n}-1$. This means that $ord_p 2 \mid p-1$ and $ord_p 2 \mid m-n$. Assume that $ord_p 2 =d$. Then by LTE lemma, we have:
$$ \nu_p(2^p -1) =\nu_p( (2^{d})^{\frac{p-1}{d}} -1) = \nu_p(2^d-1) + \nu_p(\frac{p-1}{d}) =\nu_p(2^d-1)+\nu_p(p-1) - \nu_p(d) = \nu_p(2^d-1) \ge 2$$
This means that $2^d \equiv 1 \pmod {p^2} \implies (2^d)^{\frac{m-n}{d}} \equiv 2^{m-n} \equiv 1 \pmod {p^2}$ - contradiction.

Now it remains to deal with the case when $p-1=2^a$. We will prove that it is impossible for the condition $p^2 \mid 2^{p-1}-1$ to be satisfied. Note that:
$$ 2^{p-1}-1 =(2^{2^{a-1}}+1)(2^{2^{a-2}}+1) \ldots (2+1) $$

If we let $F_n =2^{2^{n}}+1$, then it is easy to prove that $F_n = F_0F_1 \ldots F_{n-1}+2$. This implies that $\gcd(F_i; F_j) =1$ for any $i \neq j$. Indeed suppose otherwise that there exists prime $p$ such that $p \mid F_i$ and $p \mid F_j$. WLOG assume that $i <j$. Then we $p \mid F_i \mid F_0F_1 \ldots F_j \implies p \mid 2$. But $F_n$ is an odd number - contradiction.

This means that $2^{2^{a-1}}+1; 2^{2^{a-2}}+1, \ldots, 2+1$ are all pairwise coprime, which gives us that there exist $i$ such that $p^2 \mid 2^{2^{i}}+1$. On the other hand it is easy to show that if $p-1 =2^a$, then $a=2^c$ for some positive integer $c$. From before established results this means that $i=c$, but then $p^2$ obviously does not divide $2^{2^{i}}+1$.

This exhausts all possible cases. Thus the problem is solved.


if $a<b<c$ prove that $[a]\leq [b]\leq [c]$.

[quote="amir2"]masalan $a=2.1<b=2.3<c=2.9$[/quote]
are dorost migi.eslahesh kardam. :D


calculus
integration
ratio
LaTeX
trigonometry
real analysis
real analysis unsolved

Hi guys,

I am trying to integrate the following expression:
u_ratio:=(4/omega_star)*(sin(omega*t)-(exp(-B)/sqrt(r_star))*sin(omega*t-B));

where
omega_star:=omega*r0^2/nu;
r_star:=r/r0;
B:=(1-r_star)*(sqrt(omega_star/2));

I would like to obtain integral of u_ratio with respect to r_star, and separately the integral of u_ratio over one period. When i try to use Maple with something like int(u_ratio,r_star) I only get back a large expression with an integral symbol still in it. What would be the best way to approach integrating this guy?

Thanks,
Randy

Can anyone type this problem with Tex? :maybe:

I tried to $\LaTeX$ this for you. Check whether the result is what you meant.
-----------------------------
$u_{ratio}: =\frac{4}{\omega_{*}}\cdot \left(\sin(\omega t)-\frac{exp(-B)}{\sqrt{r_{*}}}\sin(\omega t-B)\right)$;

where
$\omega_{*}: =\frac{\omega r_{0}^{2}}\nu\quad$; $r_{*}: =\frac{r}{r_{0}}\quad$; $B: =(1-r_{*})\cdot\sqrt{\frac{\omega_{*}}2}$;

I would like to obtain integral of $u_{ratio}$ with respect to $r_{*}$, and separately the integral of $u_{ratio}$ over one period.
-----------------------------
Now, there are a couple of points that need clarification:

1) What do you mean by the integral with respect to $r_{*}$? (If $r_{*}$ is considered as a free variable, what do you fix: $r$ or $r_{0}$? Also, do you want some definite integral or the antiderivative?)

2) The integral over the full period is just $0$ (and, at any rate, it is easy to find an untiderivative because you are dealing with a linear combination of two sine functions). So, what did you really mean for the second question?

rwilliams 06, you seem to have said that you want us to calculate $\int_{0}^{\frac{2\pi}{\omega}}u_{ratio}\ dr,$ though I think that we are to calculate $\int_{0}^{\frac{2\pi}{\omega}}u_{ratio}\ dt$ because this calculation seems to have some thing with physics problem such as alternative current.


Find the nth term of the sequence of positive integers such that each term is neither a perfect square nor a perfect cube.

Are you familiar with how to do this for squares only?

No... all I know is that $ a_n\minus{}\lfloor{\sqrt{a_n}}\rfloor\equal{}n$ but no idea how to solve for $ a_n$ explicitly...

[hide="For squares"] The idea is that $ a_n \equal{} n \plus{} e_n$ for some sequence $ e_n$ that should go up by $ 1$ roughly every time $ n$ is a square, which means $ e_n$ should grow roughly proportional to $ \sqrt{n}$.  Can you figure out exactly when $ e_n$ should increase and what modification you need to make to $ \sqrt{n}$ to get an exact formula for $ e_n$?  (Hint:  use floor functions.) [/hide]

So what's the way in general to adjust the frequency of each number (in this case from an odd number of times 2n+1 to an even number of times 2n)?
I just assume it has to be in the form $ a\lfloor{b\sqrt{n} \plus{} c}\rfloor \plus{} d$ and solve for the coefficients...

Actually n inside the sqrt function still needs modifications...
How about this?
[hide]$ \lfloor{\frac {1 \plus{} \sqrt {4n \minus{} 3)}}2}\rfloor$[/hide]


function
limit
logarithms
trigonometry
complex numbers
real analysis
real analysis unsolved

Some question about $(-1)^\pi$

-Can it be defined in $\bf R$?
-Can we define it in $\bf C$? How can we?
-If so, is it a complex number?
-Why? Isn't it strange that operations with real numbers end up in a larger number system?
-Can someone define it correctly? (apparently I only know the definition of $log$ in $\bf R$)
-Is it transcedental?

thanx :)

[quote="Jacobson"]-Why? Isn't it strange that operations with real numbers end up in a larger number system?[/quote]You mean like $(-1)^{1/2}$?  :wink:

Work in the complex numbers: in $\bf C$, if one raises any nonzero complex number to a real rational power, $z^{p/q}$ where $(p, q) = 1$, this expression actually represents $q$ different complex numbers, yes?  This is true even if $z$ is real, of course -- it's just that when $z$ is positive real, there is always one of these numbers which is a positive real, and that's the reason we are used to thinking of these functions as, well, functions.  If you want to raise a positive real to an irrational real power, you can do it just by defining $x^{p}= \lim_{n \to \infty}x^{a_{n}}$ where $\lim_{n\to\infty}a_{n}= p$ is a sequence of rationals.  (This limit will converge always to the same value, irrespective of your choice of sequence.)  In the larger setting of the negative reals, this doesn't work at all, because some sequences will have elements with even denominators (and thus only complex values), even numerators and odd denominators (positive real values) or odd numerators and odd denominators (negative real values).

In the complex numbers, we have something else which happens entirely.  One can do it by a limit definition, but something strange happens: as the sequence $a_{n}$ approximates our irrational exponent $p$, the denominator will grow larger and larger, so the number of roots will increase without bound -- you get infinitely many values for $z^{p}$.  This is not a mistake.  The complex logarithm is very similar to the real logarithm, but recall that $e^{2\pi i}= 1$.  It follows that the complex $\log$ should really be a multi-valued function, where $\log z$ is a collection of values which differ by integer multiples of $2\pi i$.  So, how to make sense of $(-1)^\pi$, then?  $-1 = e^{\pi i}= e^{(2n+1)\pi i}$ for integers $n$, so $(-1)^\pi = e^{(2n+1)\pi^{2}i}$.  These have imaginary part $\sin (2n+1)\pi^{2}$ which is never zero, so it has no meaning as a real number.  (In fact, that set of values is a countable dense family of values on the unit circle.)

Incidentally, this all should be covered in a first course in complex analysis.


Prove that $ 4 \cdot 10^{n} \plus{} 9$ is never a perfect square for all natural $ n \ge 2$.
(The question has been edited)

This isn't true.  Let n = 1.  We get 49 which is 7^2.

Lets prove by contradiction .....
Assume $ 4 * 10^{n} \plus{} 9 \equal{} k^{2}$
$ 4* 10^{n} \equal{} (k\plus{}3)(k\minus{}3)$
clearly $ k$ is odd .....
So let it be $ 2m \plus{} 1$
$ 4* 10^{n} \equal{} (2m \plus{} 4)(2m \minus{}2)$
$ 10^{n} \equal{} (m\plus{}2)(m\minus{}1)$
now take a case where $ m \equal{}2l$ and a case where $ m \equal{} 2l \plus{} 1$
you get for the case 1 .....$ 2^{n} > 5^{n}$ which is not possible for $ n \ge 2$
for the second case ....you get $ 5^{n} \equal{} 2^{n} \plus{} 3$... which is not possible for $ n \ge 2$

I am pretty sure that the given expression is true for $ n \ge 2$
harekrishna might have missed it......hope so  :roll:

I am not too sure I understand srinath.r 's solution.
Here's an easy [hide="proof"]

    $ 4 \cdot 10^{n} \plus{} 10 \minus{} 1 \equal{} k^{2}$
$ \rightarrow 10 (4 * 10^{n\minus{}1} \plus{} 1 ) \equal{} k^{2} \plus{} 1$
$ \Rightarrow 10|k^{2}\plus{}1$

Therefore $ k^2$ ends with 9
or $ k \equiv 10a\plus{}7$ or $ k \equiv 10a\plus{}3$ for natural $ a$.

Now,
$ 4* 10^{n}\equal{} (k\plus{}3)(k\minus{}3)$
Substituting $ k \equal{} 10a \plus{} 7$, we get
$ 2 * 10^{n\minus{}1} \equal{} (a\plus{}1)(5a\plus{}2)$

$ a$ is odd $ \Rightarrow 5a\plus{}2$ ends with 7.
But $ 2*10^{n\minus{}1}$ does not have any such factor.

$ a$ is even $ \Rightarrow a\plus{}1$ is odd.
Except when $ a\equal{}4$, this can never divide $ a*10^{n\minus{}1}$
$ a \equal{} 4 \Rightarrow 11 | 2*10^{n\minus{}1}$ which is false.

Similarly, one can substitute $ k \equal{} 10a\plus{}3$
to get $ 2*10^{n\minus{}1} \equal{} (5a\plus{}3)(a)$
Observe that (5a+3) and (a) are of opposite parity. Thus except when $ a \equal{} 5$ we are done.
$ a \equal{} 5 \Rightarrow 7|2*10^{n\minus{}1}$ which is false.

[/hide]

[quote="harekrishna"]

$ a$ is even $ \Rightarrow a \plus{} 1$ is odd.
Except when $ a \equal{} 4$, this can never divide $ a*10^{n \minus{} 1}$
$ a \equal{} 4 \Rightarrow 11 | 2*10^{n \minus{} 1}$ which is false.[/quote]

Sorry if i am being dumb.......
But $ a \plus{} 1$ can divide $ 2 * 10^{n \minus{} 1}$ if it is of the form $ 5^{\alpha} ,\alpha \leq n \minus{} 1$
Cant it?

srinath.r is correct. My proof is wrong. It can still be proved but it is complicated.

I agree with harekrishna....the srinath.r 's proof isn't too much clear at all.


inequalities
number theory
least common multiple
greatest common divisor
number theory proposed

hi,
determine the set
\[ A\equal{}\{n \in \mathbb{N} : \forall a \in \mathbb{N}, \text{ odd }, a^{2}<n \Rightarrow a|n\}.\]

For $ k\in\mathbb{N}_0$ set $ t_k \equal{} 2k \plus{} 1$ and let $ P_k \equal{} [t_0,t_1,\ldots,t_k]$ be the least common multiple. Using $ t_{k \plus{} 1}^2 \minus{} 1 \equal{} t_k^2 \plus{} 8k \plus{} 7$ we define a (disjoint) covering $ \{I_k\}$ of $ \mathbb{N}$ via $ I_k \equal{} \{t_k^2,t_k^2 \plus{} 1,\ldots,t_k^2 \plus{} 8k \plus{} 7\}$, i.e. $ \mathbb{N} \equal{} \bigcup_{k\in\mathbb{N}_0}I_k$ . Obviously, we have the description $ A\cap I_k \equal{} \{a\in I_k\,|\,P_k|a\}$.

[b]Claim[/b]: For $ k\geq 3$ we have $ P_k > t_{k \plus{} 1}^2$ and as a consequence $ A\cap I_k \equal{} \emptyset$

[i]Proof[/i]: First note $ P_3\equal{}105$ and $ t_4^2\equal{}81$, i.e. we can assume $ k> 3$. Observe the following identities of greatest common divisors for $ k>2: \;(t_k,t_{k\minus{}1})\equal{}(t_k,t_{k\minus{}2})\equal{}1$, i.e. for $ k>3$ we get the inequality $ P_k\geq t_kt_{k\minus{}1}t_{k\minus{}2}\geq 5t_kt_{k\minus{}1}>\frac{4k\plus{}2}{2k\plus{}3}\cdot\frac{4k\minus{}2}{2k\plus{}3}\cdot t_{k\plus{}1}^2> t_{k\plus{}1}^2$

The claim yields $ A \equal{} A\cap (I_0\cup I_1\cup I_2) \equal{} \{1,\ldots, 8\}\cup \{9,12,15,18,21,24\}\cup\{30,45\}$.

[quote="hjbrasch"]......$ I_k \equal{} \{t_k^2,t_k^2 \plus{} 1,\ldots,t_k^2 \plus{} 8k \plus{} 7\}$, i.e. $ \mathbb{N} \equal{} \bigcup_{k\in\mathbb{N}_0}I_k$ . Obviously, we have the description $ A\cap I_k \equal{} \{a\in I_k\,|\,P_k|a\}$.
.....
 [/quote]
i think that wa have $ A\cap I_k \equal{} \{a\in I_k\,|\,gcd(t_1,t_2,...,t_k)|a\}$. and $ P_k\neq gcd(t_1,t_2,...,t_k)$
for example $ A\cap I_4 \equal{} \{a\in I_k\,|\, 5.7.9|a\}$ but $ P_4\equal{}3.5.7.9$
but i think that you'r idea gives the solution, you have just to prove that for $ k\ge 4: \ gcd(t_1,...,t_k)>t_{k\plus{}1}^2$

you are right  :blush: I have edited my original post

P.S you probably meant lcm instead of gcd


OK, this is my NEW AND IMPROVED game!!!

Instructions:

You must create a logic chain like this:

Original post: I am nobody
-------------------------------------------------
nobody is perfect

Therefore, I am perfect
-------------------------------------------------
God is perfect

Therefore, I am God
-------------------------------------------------
God is Love

Therefore, I am Love
-------------------------------------------------
Love is blindfolded

Therefore, I AM BLINDFOLDED!!!!!!! :o:rotfl:
-------------------------------------------------

When the logic chain is enough funny, boring or hard, you can post FORFEIT and start again.

Original post: Soccer is English

soccer is english
------------------
english is confusing

just like this?

i think u have to connect the original to the one u put, like

soccer is english
------------------------------
english is confusing
therefore soccer is confusing

is that correct?

ooooooohhh so that's how it works? where's pacman?

1=1

Therefore, a=b if a=1=b
Therefore, ${ a^2-b^2=a-b}$
Therefore, $ \frac {a^2-b^2} {a-b} = a+b$
Therefore, 1=2

[quote="dannyhamtx"]1=1

Therefore, a=b if a=1=b
Therefore, ${ a^2 - b^2 = a - b}$
Therefore, $ \frac {a^2 - b^2} {a - b} = a + b$
Therefore, 1=2[/quote]

wait
a=b=1, sooo

$ \frac {a^2 - b^2} {a - b} = \frac{1 - 1} {1 - 1} = \frac {0} {0}$
which is undefined and doesn't equal $ a+b = 1 + 1 = 2$....
so either i'm wrong or your logic is a little off

[quote="modeler"][quote="pacman2812"]OK, this is my NEW AND IMPROVED game!!!

Therefore, I am perfect
-------------------------------------------------
God is perfect

Therefore, I am God[/quote]

Better check your logic :P

I am in this thread
-------------------------------------------------
pacman is in this thread

Therefore, I am pacman :lol:[/quote]

:yup:

Following the original:

Therefore, soccer is confusing

BOGTRO is confusing

Therefore, soccer is BOGTRO

Therefore, soccer is confusing

BOGTRO is confusing

Therefore, soccer is BOGTRO

BOGTRO is an AoPS user

therefore, soccer is an AoPS user

You just must post the new part of the chain.

oh ok

Therefore, soccer is BOGTRO

BOGTRO is an AoPS user

therefore, soccer is an AoPS user

A tuna sandwich is better than nothing.
Nothing is better than infinite wealth, health, and happiness.
Therefore, a tuna sandwich is better than infinite wealth, health, and happiness.

soccer is an AoPS user
Pacman is an AoPS user
therefore,
Soccer is pacman

FORFEIT? Didn't know who's to use lol.

Crazychinesemaniac is insane
insanity is bliss
therefore crazychinesemaniac is bliss
bliss is dangerous
therefore crazychinesemaniac is dangerous...

love is dangerous
therefore crazychinesemaniac is love
love is romance
romance is stupid
therefore crazychinesemaniac is stupid :D

im sorry was that too mean?

JUST POST ONE SKIP!!!

FORFEIT :rotfl:

CTC is from mewto55555

More information about CTC in the FTW other stuff forum


So I realize that a) the official dates haven't been announced yet and b) more likely than not, I won't make it in the first place, but....
does anyone have any idea what the dates will be like? I've been invited to apply to another summer program (in kind of unrelated stuff) and it would possibly overlap with the last part of MOP. So, if they do the same weeks this year that they have in the past, what will the dates be?
thanks

I believe the dates depend on the IMO dates and what the US team plans to do for that, so they might not be set yet...

I believe it tends to end the first/second week of July or thereabouts.

Tentative dates at this time are June 11 -- July 1, 2007.

Steve Dunbar
AMC Director

[quote="MysticTerminator"]would the IMO team go one week in advance or are they spending an extra week in Vietnam?[/quote]

Oh yeah what about that@@ Although I think it was originally China... man jet lag/falling onto Zeb was pretty annoying.

[quote="AMCDirector"]Tentative dates at this time are June 11 -- July 1, 2007.[/quote]

Thanks for the preliminary word on tentative dates for MOSP/MOP 2007. Within that span of dates, is there any distinction among the likely schedules for the Black, Blue, and Red groups?

Thanks for all that you do administering the AMC contests. I'm very grateful that the students on my math team have the opportunity to participate.

[quote="MysticTerminator"]would the IMO team go one week in advance or are they spending an extra week in Vietnam?[/quote]

hmm. has it not yet been decided?


for every $m\in\mathbb{N}$ we define $f_m$ such that:
$f_1=1, f_2=11, f_3=111, ..., f_m=1111....1$($m$ times)
 how many of these $f_m$'s can be expressed as sum of two perfect natural squares?

That is only if you take all square factors out first. For example $f_{9}$ is divisable by 3 squared...

[quote="paladin8"][hide]
None of them? Because $f_m \equiv 3 \mod{4}$ except for $m=1$ which has no solutions.[/hide][/quote]
 :yup:


Instead of saying something like Class transcript __/__, how bout something like:
Lesson 12: Addition.

It would help when trying to sift through the transcripts to find something.

Well, Dojo:
If you want to save it into your own Word files, you can name it whatever you want.  :lol:

I actually meant the My Classes section. It just says what it says above.


On one of the forums, it says that the USAMTS scores will be posted in early November. Since it already is early November, does anybody know the exact date or approximate date that the solutions will be posted?

Read the last post in this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=307703]thread[/url]

[quote="DPatrick"]Update: due to the unexpectedly large number of participants this year (up over 20% from last year), the grading is taking longer than expected.  We are aiming for the week of Nov 16-20 for the Round 1 scores and comments to be available.

Please keep in mind that one of the unique features of the USAMTS is that you get written comments to your solutions, not just numeric scores, and this takes time.[/quote]


1. $\sin{(\frac{\pi}{4}-x)}=\frac{5}{13}$. Find the value of $\frac{\cos2x}{\cos{(\frac{\pi}{4}+x)}}$.

2. The equation $\sqrt{2}\sin{x}+\sqrt{6}\cos{x}=\frac1k$ has root. Find $k$.

2.
The equation:  $\sqrt 2 \sin x + \sqrt 6 \cos x = \frac{1}{k},k \in \mathbb{N}$  has root.  Find  $k \in \mathbb{N}$

[u]Solution:[/u]
$\sqrt 2 \sin x + \sqrt 6 \cos x = \frac{1}{k} \Leftrightarrow \frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x = \frac{1}{{2\sqrt 2 k}} \Leftrightarrow $

$ \Leftrightarrow \sin x\cos \frac{\pi }{3} + \sin \frac{\pi }{3}\cos x = \frac{1}{{2\sqrt 2 k}} \Leftrightarrow \sin \left( {x + \frac{\pi }{3}} \right) = \frac{1}{{2\sqrt 2 k}}$ has root only if  $ - 1 \le \frac{1}{{2\sqrt 2 k}} \le 1$

1. Let’s draw a right triangle ABC such that < ABC =pi/4 –x . If we take a point D on the side BC such that AD=BD=a . So < BAC=pi/4 + x , <ADC=pi/2 – 2x and AB=13,AC=5,BC=12, AD=169/24 …(for 0<x<45)
    Since the triangle ABC is a right triangle,
                cos2x =sin(pi/2 -2x)=120/169 and cos(pi/4 + x)=sin(pi/4 - x)=5/13

Hence cosx / cos(pi/4 + x)    =  24/13

Oh, i forgot something...
For problem one, there should be $0 <x <\frac{\pi}{4}$.

If there is no bound of $x$,matematikator's solution was not correct, but if we add this, he's final result will be the right answer.

I created a 5-12-13 right triangle, then extended the sides to create a 5-12 rectangle with diagonal 13.  On one of the corners, the two angles (angles created by the sides and the diagonal) are 90 - x, and x in degrees.  So, cos x = 5/13, sin x = 12/13, and tan x = 12/5.  Using cos(2x) = cos^2 x - sin^2 x, cos(2x) = -119/169.  Using the formula for sum of cosine, cos(90 + x) = cos 90 * cos x - sin 90 * sin x = - sin x = -12/13.  Dividing -119/169 by -12/13, we get 119/156.


algebra polynomial function quadratics ratio

Leaders
Mystic: 16
Cats: 3
Osiris: 1

Problems
1992 3)
a) Write down a fourth degree polynomial P(x) such that P(1) = P(-1) but P(2) does not equal P(-2)
b) Write down a fifth degree polynomial Q(x) such that Q(1) = Q(-1) and Q(2) = Q(-2) but Q(3) does not equal Q(-3)
C) Prove that, if a sixth degree polynomial R(x) satisfies R(1) = R(-1), R(2)=R(-2) and R(3) = R(-3), then R(x) = R(-x) for all x.

1985 2)
A) Find a triple of primes (p,q,r) such that p=q+2 and q=r+2
B) Prove that there is only one such triple of primes

1985 3) The function g is defined recursively on the positive integers by g(1) = 1, and for n>1, g(n) = 1+ g(n-g(n-1)).
A) Find g(1), g(2), g(3) and g(4)
B) Describe the pattern formed by the entire sequence g(1), g(2), g(3), ...

1985 4) Let x, y and z be real numbers such that x+y+z=1 and xyz=3.
A) Prove that none of x,y nor z can equal 1.
B) Determine all values of x that can occur in a simultaneous solution to these two equations (where x, y and z are real numbers)

1984 1) For what integers n is 2^6 + 2^9 + 2^n the square of an integer?

1983 1) Find the largest integer which is a factor of all numbers of the form n(n+1)(n+2) where n is any positive integer with unit digit 4. Prove your claims.

1983 3) A and B are two points on a circular race track. One runner starts at A running counter clockwise, and, at the same time, a second runner starts from B running clockwise.

They meet first 100 yards from A, measured along the track. They meet a second time at B and a third time at A. Assuming constant speeds, how long is the track?

1982 3) Determine the exact sum of the series
1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + ... + 1/(98*99*100)

1. [hide]x^4-2x^3-x^2+2x; (x:^2:-1)(x:^2:-4)(x-3), whatever that multiplies to; let f(x) be this sixth degree polynomial and let g(x)=f(x)-f(-x). Note that as this as f was originally 6th degree, the highest terms will cancel and g will be (at most) 5th degree. However, g has roots :pm:1, :pm:2, :pm:3, i.e. 6 roots, so g must be everywhere 0 and f(x)=f(-x)[/hide]

2. [hide]clearly all of these primes are odd, else we would have 3 even primes. So, they are equivalent to 1, 3, and 5 mod 6 (in some order). However, there is only one prime congruent to 3 mod 6, namely 3. Thus, the only possible triple is (7,5,3)[/hide]

6. [hide]the largest it can be is 120, as 4*5*6=120. Indeed, it is 120 since if n ends with a 4, both n and (n+2) are even, and thus one must be a multiple of 4 and the other a multiple of 2, giving us that n(n+1)(n+2) is a multiple of 8. In addition, (n+1) ends with a 5, giving us a factor of 5, and the product of any 3 consecutive numbers is a multiple of 3.[/hide]

3. [hide]g(n) follows the following pattern: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, .... This can be proven by induction. The base case is given. Assume it follows this pattern up to n(n-1)/2. (meaning there are (n-1) (n-1)s right at the end of this sequence). For n(n-1)/2+1, 1+g(n-g(n-1))=1+g(n(n-1)/2+1-n+1))=1+g((n-1)(n-2)/2+1)=1+(n-1)=n.

Now, for any i such that n(n-1)/2+1<i:le:n(n+1)/2, 1+g(i-g(i-1))=1+g(i-n)=1+(n-1)=n (the penultimate step follows from the fact that i-n will be somewhere in the range of [(n-1)(n-2)/2+1, n(n-1)/2]. Thus, it now holds true up to n(n+1)/2.

by the way, g(n) can be expressed as closed form. I don't remember what off the top of my head though and I don't want to derive it. [/hide]

4.

[hide]

WLOG, assume x=1.  then y=-z, and -z^2=3, which is false.

WLOG, assume x=y=1.  then z=-1 in order to satisfy x+y+z=1, so this is false.

Then letting x=y=z=1 is also false.

b)  z=3/xy, substitute.  x+y+3/xy=1.

x^2y+xy^2+3=xy, we factor/solve for y, get

y=((x-x^2)  :pm:  :sqrt: (x^2-x)^2-12x)))/(2x)  .  This has real values of x if the discriminant is 0, or if x^4-2x^3+x^2-12x  :ge: 0.  Factoring, we have x(x-3)(x^2+x+4)  :ge: 0.  But x cant be 0, or else xyz !=1.  All that matters is x(x-3) :ge: 0, since x^2+x+4 has vertex at x=-1/2, and when you plug it in it's positive.  Then x(x-3) :ge: 0.  This is true for (- :inf: , 0) U [3,  :inf: ) i think.  So all that junk works.[/hide]

5.

doing this as i go along; i might not have full solution.

[hide]

x^2 = 2^6+2^9+2^n = 64+512+2^n = 576+2^n.  Im pretty sure 576 is 24^2.

X^2-24^2 = 2^n

(x-24)(x+24)=2^n.  hmmm, so if we break up 2^n as 2^p*2^q, where p+q=n, we'll say p>q

then x-24=2^q, x+24=2^p

Subtract the 2 eqns to get 2^p-2^q = 48.  I think this is 64-16, or 2^6-2^4.  so p+q=10, n=10 ?  I think that should be the only one... as 2^x grows and grows, it's unlikely that there'll be more than one solution to 2^p-2^q=48.  [/hide]

5. [hide]n must be even. You can see this by looking at quadratic residues mod 3 or 4. So, we have 2^6+2^9+2^2k. This is bounded on either side by (2^k)^2 and (2^k+1)^2 for k:ge:9. Thus, we just need to check k<9 and find that n=10 gives the only solution.[/hide]

I'll grade these after somebody answers #7

7. [hide]let x be the distance between A and B and s the total distance (that we're looking for). then as the ratio of the distances these guys/girls travel at any given time is a constant, (x-100)/100=s/x=(s+x)/s => we try to solve for s in order to set the two expressions for x equal => s=(x:^2:-100x)/100, 100s+100x=sx-100s=>s=100x/(x-200) => (x-100)/100=x/(x-200) => x:^2:-300x+20000=100x => x=200+100:sqrt:(2) okay i'm starting to get big numbers which means i made a mistake [/hide]

7.[hide]  Let a be the speed of the runner who starts at A and b be the speed of the runner who starts at B.  Let C be the point where they first

meet.  (All the following arcs measured counter clockwise.)  Then arc AC = 100, arc CB = some distance = arc BA, since they have to meet

 at evenly spaced positions along the track.  Let CB = BA = d.  Then we have

100/a = d/b

(d + 100)/b = d/a

Multiplying those two equations together and cancelling out the ab in the denominator, we have

100(d + 100) = d2.  So letting 100 be x for a second to make writing easier,

d2 - dx - x2 = 0,

d = (x :pm: :sqrt:(x2 + 42))/2, from which we deduce

d = x*(1 + :rt5:)/2 = 50 + 50 :rt5:, and thus the entire length of the track is

200 + 100 :rt5:, or roughly 423 yards.[/hide]


function
floor function
trigonometry
number theory
prime factorization

Let $f(p) = 1$ for prime $p$ and $f(ab) = f(a)+f(b)$ for all integers $a, b$.  In other words, $f(p_{1}p_{2}p_{3}...p_{k}) = k$.  Is there a way to express this function in terms of elementary functions?

[hide="Slightly cheap"] ${f(n) = \sum_{i=1}^{\infty}\sum_{k=1}^{\infty}\lfloor | \frac{\cos \pi n}{p_{k}^{i}}}| \rfloor$ [/hide]

Can you post some sort of explanation for that? I have no idea where that is coming from.

[hide="Oh, it really is a cheap one..."] The function $\left\lfloor \left| \frac{\cos \pi n}{d}\right| \right\rfloor$ merely returns $1$ if $d | n$ and $0$ otherwise.  That's because if $d \not | n$ then $| \cos \frac{\pi n}{d}| < 1$.

So the function $f(n)$ that I posted merely counts $1$ every time a prime power divides $n$.  This gives the sum of the exponents in the prime factorization of $n$, which is the number of primes needed to multiply out to $n$ as stipulated in the problem. [/hide]

[quote="t0rajir0u"]$p_{i}$ is the sequence of primes, same as in your problem.  Like I said... slightly cheap.  There might be a better answer.[/quote]
I might be misunderstanding your answer, but my problem doesn't mention the sequence of primes.  In my problem it was just how many primes divide $n$ with repetitions allowed.  For example, $f(20) = f(5)+f(2)+f(2) = 1+1+1 = 3$.  I think your solution would answer $2$ because it does not account for the $2$ being multiplied twice.

EDIT: Is it okay to put an infinite summation sign inside any other sum?

[quote="miyomiyo"]it does not account for the $2$ being multiplied twice.[/quote]

Oh, but it does.  That's why I have the second summation outside the first, which counts prime powers.  So $f(20) = 3$ because $2 | 20, 2^{2}| 20, 5 | 20$.

[quote="miyomiyo"]Are you allowed to have an infinite summation sign inside another infinte summation sign?[/quote]

The summation is well-defined.  Perhaps you should be more specific about what "in terms of elementary functions" means  :P

well...obviously p_n is not an elementary function

[quote="t0rajir0u"][b]Altheman:[/b]  Oh, but it's a [i]sequence[/i] I'm referring to.  I could just as well explicitly write out

$f(n) = \left\lfloor \left| \frac{\cos \pi n}{2}\right| \right\rfloor+\left\lfloor \left| \frac{\cos \pi n}{3}\right| \right\rfloor+\left\lfloor \left| \frac{\cos \pi n}{4}\right| \right\rfloor+\left\lfloor \left| \frac{\cos \pi n}{5}\right| \right\rfloor+\left\lfloor \left| \frac{\cos \pi n}{7}\right| \right\rfloor+\left\lfloor \left| \frac{\cos \pi n}{8}\right| \right\rfloor+\left\lfloor \left| \frac{\cos \pi n}{9}\right| \right\rfloor+\left\lfloor \left| \frac{\cos \pi n}{11}\right| \right\rfloor+...$[/quote]

Since we are assuming positive integers, the only value of that equation is 0.

Hmm?  The last time I checked, $\cos \pi =-1$, so at the least we have $f(p) = 1$ for primes $p$.


Hey - anyone participated in them? Are any of them not so competitive and is prior research necessary?

It's very hard to get into most REUs before your junior year.  Since REUs are the primary route that undergraduates take to getting research experience, it would be rather odd if they required previous research experience, don't you think?  Like anything else, of course, having done more won't hurt your chances and could possibly help them.

What you say is comforting (that previous research experience isn't required), but I'm pretty sure that most of the applications I've seen do ask about your research experience.  Also, wouldn't prior research experience be the best indication of how one would perform at REU?  And if previous research experience isn't essential to getting in, why is it so difficult to get in before junior year?  Are they biased against underclassmen?

Well, yes, they'd ask about it because it's relevant, but it's not [i]required[/i].  That is, people get into them who haven't done any research, but having done research is almost certainly a plus.  And yes, they discriminate very heavily against underclassmen because they get many many applications compared with the number of spots they have available.  (As my friend joked this afternoon, "I got a rejection letter that said, 'We're sorry, but there were 4,000 applicants for our 9 available spots, and we decided we would leave 7 spots unfilled, and you weren't one of the top two.'")

Somewhat related...Anybody know how hard it is to get a (pure) math UROP at MIT (before junior year, I guess)? I know one of the big things at MIT is its UROP program, but I've looked at the listings of UROPs on their website, and I've never seen anything with pure math. Not sure if that means you just have to directly e-mail professors to find them, instead of simply hoping one pops onto the website.

What about applied math UROPs? Applied stuff in general is easier to find than theoretical stuff (like theoretical computer science or physics), right?

And...what do you actually do in theoretical research?  :? I understand stuff like biology and chemistry where you can do experiments in labs, but do you just...sit around and think for more theoretical stuff?

You try and prove things.  I mean, that's kind of silly sounding -- the process by which you pick the things you try and prove is not arbitrary, but then you just work on it and try and crack it, same as any science except that you don't run the risk of doing your experiments for 2 months to discover that someone gave you the wrong chemical and so all that work is wasted.  (That actually just happened to two chem friends of mine.)

REUs vary widely in what qualifications they seek.  Most will want students to have had abstract algebra and real analysis (that's why it's much harder to qualify before junior year).  Whether prior research experience will help or not depends on the attitude of the REU.  The original purpose of the program (sponsored via NSF) was to give undergraduates exposure to original research so that they could (a) be more prepared for graduate school and (b) have a taste of what grad school would be like, so they'd have a better idea of whether they wanted to go to grad school or not.  So some REUs really want students who have no prior research experience and who aren't sure they want to attend grad school.  (One of my sophomores who is pre-law will be attending such an REU this summer, and hasn't had algebra or analysis.)  Others want the best students they can get, where the definiton of "best" varies of course, but often includes qualities such as persistence, ability to work with others, and evidence of deep understanding or creativity in standard college mathematics courses.

This question is a bit broad... It depends greatly on which college you want to apply to, and what field you are looking to do research in. I recommend doing some research during the school year as well as in the summer. For instance, in my freshman year I did research (as a research assistant) in Condensed Matter Physics starting in October for about 15 hours per week. I had a paper published in Phys Rev. B by the end of the year. I have prior summer engagement, but I'm betting that even as a Freshman I could get an REU pretty much anywhere I apply. It all depends on your drive, you can get an REU at any age.


limit
logarithms
integration
calculus
derivative
algebra
functional equation

$ \underset{y\rightarrow0}{\lim}\frac{\log\left(  1\plus{}y\right)  }{y}\equal{}1$

Don't use L'Hôpital rule to find the limit!

How to prove it?

We take it as the definition of the natural logarithm. That's the proof.

If not that, it's something very close. What are some reasonable ways to define the exponential and the logarithm?
One way: $ \ln x\equal{}\int_1^x \frac1t\,dt$. Then the derivative at $ 1$ is $ \frac11\equal{}1$ by the FTC, and we have this limit.
Another way: We define the base $ b$ exponential by defining it algebraically for rational exponents, and interpolate by continuity. $ e$ is the unique base so that the derivative of $ e^x$ at zero is $ 1$, and $ \ln$ is the inverse of $ e^x$. A simple calculation gives that $ \frac{d}{dx}e^x\equal{}e^x$, and $ 1\equal{}\frac{d}{dx}e^{\ln x}\equal{}x\frac{d}{dx}\ln x$, so the derivative of $ \ln$ at $ 1$ is $ 1$.

This limit [i]is[/i] the derivative of the logarithm. If that's not the natural logarithm- which is quite logical if you're just starting to develop their calculus- it's wrong and the limit is some number that depends on the base and doesn't seem to be nice for any of the usual values like $ 2$ or $ 10$.

You're not only first class mathematicianan also a good teacher.

we can also use the not-so-well-known (but useful) expansion of :
$ {\log\left( 1\plus{}y\right)} \equal{} y \minus{} \frac{y^2}{2} \plus{} \frac{y^3}{3} \minus{} \frac{y^4}{4}...$

:)

I would say that is well known. It makes the short list of power series to memorize.

If you're going to base the definition on power series, start with the exponential; that series converges everywhere. The derivative $ \frac{d}{dx}e^x$ is obvious, and the functional equation $ e^{x\plus{}y}\equal{}e^xe^y$ can be done by manipulating the double series and invoking the binomial theorem. Then $ \ln$ is the inverse, and the rest is as in the second definition I posted. You can easily derive power series for $ \ln$ once you know its derivative, but taking a series as the definition is not advised because of the limited radius of convergence.

$ \underset{y\rightarrow0}{lim}\frac{\ln(1+y)}{y} = \underset{y\rightarrow0}{lim}\ln(1+y)^{1/y}$

If $ \underset{x\rightarrow x_0}{lim}u(x)^{v(x)}$ is an indeterminate form of $ 1^\infty$ then $ \underset{x\rightarrow x_0}{lim}u(x)^{v(x)} = e^{\underset{x\rightarrow x_0}{lim}v(x)(u(x)-1)}$ provided that the latter limit exists. According to this $ \underset{y\rightarrow0}{lim}(1+y)^{1/y} = e^{\underset{y\rightarrow0}{lim}(1+y-1).\frac1y} = e$

$ \underset{y\rightarrow0}{lim}\ln(1+y)^{1/y} = {\ln e = 1}$

You effectively assumed this limit in making that step to exponential form- after all, $ u(x)^{v(x)}\equal{}\exp(v(x)\ln u(x))$, and you asserted that this is equal to $ \exp(v(x)(u(x)\minus{}1))$ in the limit. Your argument is circular, and therefore useless.

On notation: use \lim; it looks nicer. It also allows for much simpler code, like \lim_{y\to 0}.

I am not clear why my argumentation is circular. Please could you explain a bit more? The equation $ \lim_{y\to 0}\frac {\ln(1 \plus{} y)}{y} \equal{} \lim_{y\to 0}\ln(1 \plus{} y)^{1/y}$ is true all time or is true if the limit at the lefthandside of the equation exists ? If it is true all the time my argumentation is not circular, I think. :)

That's fine; it's the next equation that's the problem. Going from $ \ln u$ to $ u\minus{}1$ in the exponent assumes the conclusion.

[quote="category"]
If $ \lim_{x\to x_0}u(x)^{v(x)}$ is an indeterminate form of $ 1^\infty$ then $ \lim_{x\to x_0}u(x)^{v(x)} \equal{} e^{\lim_{x\to x_0}v(x)(u(x) \minus{} 1)}$ provided that the latter limit exists.
[/quote]

The phrase "provided that the latter limit exists" is important, without that we get in trouble. If the limit at righthandside of equation above exists, then the limit at the lefthandside also exists. Is this interpretation correct or mathematical nonsense ? I would just like to learn about it. Thank you.

Would this be circular as well? Or just plain wrong?

$ \lim_{y\to 0} \frac{\ln(1\plus{}y)}{y}\equal{}\lim_{y\to 0} \ln(1\plus{}y)^{\frac{1}{y}}\equal{}\lim_{y\to \infty}\ln (1\plus{}\frac{1}{y})^y\equal{}\ln \lim_{y\to \infty}(1\plus{}\frac{1}{y})^y\equal{}\ln e\equal{}1$

Isnt this by definition $ dy/dx log(1\plus{}x) at x \equal{} 0$. This equals 1

If you try to calculate the derivative of the logarithm by the definition, you inevitably come back to this exact limit. We absolutely need an independent way to evaluate it- and the best way to do that is to embed it into the definitions. What is $ e$? $ e$ is the unique base that makes this limit $ 1$. bike_me's post effectively defines $ e$ as $ \lim_{y\to\infty}(1\plus{}\frac1y)^y$, which is basically equivalent.


Let real sequence x0=0,xn=x(n-1)/2004  +  (-1)^n
  Calculate an follow n. and find lim xn^2


"What is the maximum value of $n$ for which there is a set of distinct positive integers $k_{1}, k_{2},... ,k_{n}$ for which

$k_{1}^{2}+k_{1}^{2}+...+k_{n}^{2}= 2002.$"

I can't find an answer, yet its placement among the other AMC questions indicates only moderate difficulty. What am I missing?

[quote="DiscreetFourierTransform"]I think that sum is for consecutive integers, rather than distinct, not necessarily ordered integers.
We have to show that the maximum occurs over consecutive integers for that to be the case.

I don't know how to go about showing that, though.[/quote]
Well keep in mind that the numbers are distinct, lower numbers are more efficient.

However, we have no guarantee that using smaller numbers works.

[hide]Using $S_{k}=k(k+1)(2k+1)/6$, $S_{17}<2002$ and $S_{18}>2002$
Clearly, then, $n\leq 17$
The question is, [i]can[/i] we find a case of $n=17$?

Such a solution would exist if we can find a solution to $S_{17+j}-2002=k_{1}^{2}+k_{2}^{2}+...k_{j}^{2}$
In other words, take an integer above $17$ and pull out a few squares.

For $j=1$, at least 3 squares need to be taken out: $S_{18}-2002=1^{2}+5^{2}+9^{2}$
For $j=2$, $S_{19}-2002=468=12^{2}+18^{2}$, which satisfies the condition, so the minimum is in fact $17$[/hide]

[i]Follow-Up Questions:[/i]
$(1)$ What is the minimum $n$ for which $\sum_{i=1}^{n}k_{i}^{2}= 2002$ is true?
$(2)$ How many different ways are there to write $2002$ as the sum of this minimum number of squares?

(1) Attack problem mod 7 to show that two squares will not work. So we need at least three squares. Since we have
\[2002=43^{2}+12^{2}+3^{2}\]
3 is indeed the minimum.

(2) I'm not brute-forcing this one... :roll:

According to http://www.math.niu.edu/~rusin/known-math/95/three.sq , the number of ways that $n$ can be written as a sum of three squares, if it can be written at all, is $12h$ if $n \equiv 3 \bmod 8$ and $24h$ otherwise, where $h$ is the class number of $\mathbb{Q}[ \sqrt{-n}]$.


Prove that if x is irrational, then we can find infinitely many n such that
 (-1)^[x]+(-1)^[2x]+...+(-1)^[nx]=0.

[i]Mathematics Magazine[/i]

Due to Hurvits's theorem we can find infinitely many irreducable fractions m/n such that |x-m/n|<1/2n2. Let m/n be such fraction.
We have |kx-km/n|<k/2n2<=1/n for all k=1,2,3,...,2n. Hence [kx]=[km/n] except k=n,2n.

1) Suppose m is even number ==> n is odd.
Consider [kx] and [(n-k)x] where k=1,2,...,[n/2]. Since n is odd we conclude kx<>(n-k)x. We know [kx]=[km/n], [(n-k)x]=[(n-k)m/n]=[m-km/n]. It is clear [km/n]+[m-km/n]=1(mod 2).
Thus (-1)[x]+(-1)[2x]+...+(-1)[(n-1)x]=0.

2) Suppose m is odd number.
Consider [kx] and [(2n-k)x] where k=1,2,...,n-1. We have [kx]=[km/n] and [(2n-k)x]=[2m-km/n] ==> [kx]+[(2n-k)x]=1(mod 2).
   A) Suppose nx<m ==> [nx] is even and [2nx] is odd.
   B) Suppose nx>m ==> [nx] is odd and [2nx] is even.
Thus (-1)[x]+(-1)[2x]+...+(-1)[2nx]=0.


inequalities
calculus
derivative
trigonometry
inequalities unsolved

x<=1
x+y<=5
x+y+z<=14
x+y+z+t<=30
Find max của $\sum \sqrt{x}$(with $x,y,z,t \in R+$)

Find max and min of:
$P=Sin^{3}_{x}-Cos^{3}_{x}$

This problem is quite easy. First notice that t is not depend on x, y, z so that we have
$t \leq 30-x-y-z$
then
$\sum{\sqrt{x}} \leq \sqrt{x} + \sqrt{y} + \sqrt{z}+\sqrt{30-x-y-z}$
Apply the Karamata inequalities with {x,y,z,t} and {1,4,9,16} we have
$max \sum{\sqrt{x}} = 10$

OK,you have true answer ,but problem 2:
Find max and min of Sin^3x-Cos^3x you dont finish it

Well, the second question I think that we can use derivative to solve this.
First let call $f(x)=sin^3x-cos^3x$ then
$f'(x)=3sin^2xcosx+3cos^2xsinx=3sinxcosx(sinx+cosx)$
The equation $f'(x)=0$ is equivalent to $sinx=0$ or $cosx=0$ or $sinx+cosx=0$.
You easily solve these equation and find the min, max of $f(x)$

use Cauchy Ineq to solve the  fisrt problem.(1+2+3+4)(x+y/2+Z/3+t/4)>(x^0.5+y^0.5+z^0.5+t^0.5)^2[/hide][/list][/code]

[quote="imathsvn"]Find the maximum and the minimum of $P(x)=\sin^{3}{x}-\cos^{3}{x}$.[/quote]
$|P(x)|\leq|\sin^{3}{x}|+|\cos^{3}{x}|\leq\sin^{2}{x}+\cos^{2}{x}=1\Longrightarrow\{\begin{array}{c}P(\frac{\pi}{2})=1=P_{\max},\\\\P(0)=-1=P_{\min}.\end{array}$


graph theory
Ramsey Theory
combinatorics proposed
combinatorics

I don't where is the suitable place to post, so I post it here:

Between 10 people, show that there exist 3 people who are acquainted with each other, or 4 people who are not acquainted with each other.

Let's consider the complete graph whose vertices are the persons, and color the edge $ AB$ in red if $ A$ knows $ B$, and blue otherwise.
Then, it is asked to prove that there is a red triangle or a blue $ K_4$. Note that it means that the Ramsey number $ R(3,4)$ is not greater than $ 10$.

It is well-known that in any bicoloration of a $ K_6$ there exists a monochromatic triangle.

Now, let's consider the vertex $ P_1$.
Among the $ 9$ edges between $ P_1$ and the other vertices there are at least $ 6$ blue edges or at least $ 4$ red edges.

- If there are blue edges between $ P_1$ and, say, $ P_2, \cdots, P_7$ :
If among these $ 6$ vertices, there exists a blue triangle, then adding $ P_1$ leads to a blue $ K_4$. Otherwise, from above, there exists a red triangle and we are done.

- If there are red edges between $ P_1$ and, say, $ P_2,P_3,P_4,P_5$ :
If there is any red edges between these four points, say $ P_2P_3$ is red, then $ P_1P_2P_3$ is a red triangle. Otherwise the four points form a blue $ K_4$.

Pierre.


Can you please give example on how to apply cauchy criterion in Series? Thanks!

i dont quite understand your question. can you clarify a bit? but there are lots of places where cauchy criterion is used. for instance in the Weirstrass test for determining uniform convergence of a series of functions.

What is the negation of Cauchy criterion?

$\exists \epsilon>0\;\forall n_{0}\in\mathbb{N}\;\exists m,n>n_{0}: |x_{m}-x_{n}|\geq \epsilon$

Curious.

Using the Negation of Cauchy criterion.
If $\sum(a_{n})$ is divergent. Can we claim that for any $n$ there is an $m>n$ such that $|s_{m}-s_{n}|\ge e$?
Where $s_{n}$ is a partial sum.

$\exists \epsilon>0\;\forall n_{0}\in\mathbb{N}\;\exists m,n>n_{0}: |s_{m}-s_{n}|\geq \epsilon$ is true. Is it equivalent to your assertion? You can check.

So there is an $m,n >n_{0}$. so this implies only a few blocks of the entire series are actually greater than epsilon.

I was wondering whether by tweaking $n_{0}$ we can make it so that for any $n$ there is a block $|s_{m}-s_{n}|\ge e$.

So that means the series can be composed of blocks one after the other such that they are all greater than epsilon.


college
polynomial
algebra
college contests

Prove that $f(x) = x^4 + x^3 +x^2 + 6x + 1 $ is irreducible over Q.

(Berkeley 96 - For Training 04)

Perhaps some high algebra would solve it quickly (something in the same vein as Eisenstein), I don't know.

It's enough to prove f(x) is irreductible in Z[X]. Since f has no integer roots (which should be +-1), if f is reductible it must be written (x + b*x + c)(x + d*x + e), with b,c,d,e integer.

Expanding, this gives b + d = 1, bd+c+e = 1, be+cd=6, ce = 1.

So c=e=+1 or c=e=-1. which leads to a contradiction in either case.

Yes, Eisenstein's criterion can be used.
Write $f(x) = x^4 + x^3 +x^2 + x + 1 + 5x=\frac{x^5-1}{x-1}+5x$
$f(y+1) = y^4+5y^3+10y^2 +10y +5+5y+5 = y^4+5y^3+10y^2+15y+10$, which is irreducible by Eisenstein with p=5.

Sorry ...I don't know about  Eisenstein's criterion
can someone tell me ?
Thank you very much..

Eisenstein's criterion:
A polynomial $p(x)=a_nx^n+...+a_1x+a_0$ with integer coefficients is irreducible in the rational numbers if there exists a prime $p$ so that $p \nmid a_n$, $p|a_i$  $0\leq i<n$ and $p^2 \nmid a_0$.

May I learn what do you mean by Berkeley, training 2004
which contest was this question asked? May anyone give me the explicit name of it.


logarithms
function
limit
calculus
calculus computations

Hey guys,

Im currently studying for an exam and just need some help.  The professor gave us some similar problems to what we might expect on the exam but no solutions.  I know how to solve the problems but just need to verify that I am getting the correct answer.  So if anybody could just solve the following questions and post only the final answer, it would be very helpful.

4. Let f(x) = log(7 − x) and g(x) = x2 − 9.
(a) [4 points] Find and simplify (g (f(X))(x) + 3g(x).
(b) [5 points] Solve f(x) is greater than or equal to 2.

5. Let P(t) = C(1 + be−t/3). Suppose P(0) = 3e and P(3) = 4.
(a) [10 points] Find b and C. Write your answers as expressions involving e, and
not as decimals.
(b) [3 points] Find P(6) rounded to two decimals.

7. (a) [7 points] In what year will an investment of $2455.65 made today, grow to$4000.00 if the interest rate is 10% compounded semi-annually.
(b) [10 points] A debt of $12, 000, which is due 10 years from now, is instead to be paid off by four payments:$3000 now, $2000 in 3 years,$2000 in 6 years, and
a final payment at the end of 8 years. What would this payment be if an interest
rate if 5.5% compunded semi-annually is assumed?

Please, edit your post: it is quite hard to read. Besides, it is not clear what some expressions are: is x2 in 4 supposed to be $x^2$? Is log in 4 the natural or the decimal logarithm (or logarithm to some other base)? Is  C(1 + be−t/3) in 5 actually $C(1+be^{-t/3})$? And so on. Of course, we can try to guess what is meant here and there but it would make it much easier if you write your questions correctly from the start. :) Also, I'm moving it to Computations and Tutorials section, which is a more appropriate place for questions of that type..

Yeah, sorry, I can see its not very clear.  Im not too good with math symbol posting but ill try to explain it
better.  Hopefully this is better:

4. Let f(x) = log(7 − x) and g(x) = x^2 − 9.
(a) [4 points] Find and simplify (g (f(X))(x) + 3g(x).
(b) [5 points] Solve f(x) is greater than or equal to 2.

The log is a base 10 log.

5. Let P(t) = C(1+be^{-t/3}). Suppose P(0) = 3e and P(3) = 4.
(a) [10 points] Find b and C. Write your answers as expressions involving e, and
not as decimals.

(b) [3 points] Find P(6) rounded to two decimals.

7.(a) [7 points] In what year will an investment of 2455.65 made today, grow to 4000.00 if the interest rate is 10% compounded semi-annually.

(b) [10 points] A debt of 12, 000, which is due 10 years from now, is instead to be paid off by four payments:3000 now, 2000 in 3 years,2000 in 6 years, and a final payment at the end of 8 years. What would this payment be if an interest rate if 5.5% compunded semi-annually is assumed?

Thank you.

5.
(a)

$b = 3e - 4$

$C = \frac{e}{e-1}$

(b)

$P(6) = 2.47$

7.

(a)

5 years from today, so I guess February 5, 2011....

(b)

I'm not sure if I have done this correctly, I'm notorious for messing up algebra, but I came to the conclusion that the final payment should be 9039.04

4a) $(\log(7-x))^2+3x^2-36$ (I assumed that mysterious g(f(X))(x) is just $g(f(x))$)
4b) $x\le -93$

I got the same answers as ankur87 for 5ab and 7a. As to 7b, I was too lazy to attempt it without a calculator and I haven't got one at hand at the moment.

thanx for all those solutions.  All of mine match.  I just had a few more questions.  thank you.

The ICALLU cell phone company has a monthly plan that charges a basic,
flat rate of 10dollars a month. There is an extra charge of 20c/ for each minute of call-time used up to and including two hours. There is a further 10c/ a minute for all call-timebeyond two hours. Find the total monthly charge B(t) in dollars as a function of thetotal number of minutes, t, of call-time used per month.

In this question let P(t) =N/(1 + ae^−.2t) where N, a > 0 are constants.

(a) [8 points] If P(0) = 12 and P(25) = 90, find N and a rounded to two decimals.
(b) [4 points] Show mathematically why P(t) < N for all t is greater than or equal to 0.

I don't know if you can use piecewise defined functions but here is a nice way to show $B(t)$

$B(t)=\{\begin{array}{cc}10 + 0.20t,&\mbox{ if } t\leq 120\\10 + 0.20(120) + 0.10(t - 120), & \mbox{ if } t>120\end{array}$

For the next one, I calculated that $N = 94.15$ and $a = 6.85$

and I don't know exactly how to show it mathematically, but if you are allowed to use limits, it shouldn't be that hard to show that

$\lim_{t\rightarrow\infty} 1+ 6.85e^{-.2t} = 1$

I'm pretty bad at showing/proving things...perhaps someone could provide something concrete...


geometry
trigonometry
integration
calculus
calculus computations

Find the area of the region that lies inside the closed polar curve $r=3\cos \theta, 0 \leq \theta \leq 2\pi$ and outside the closed polar curve $r=1+cos\theta, 0\leq \theta \leq 2\pi$

I think I must first find the are of the region inside $r=3\cos \theta$, then find the area of the region lying inside the cross-section of $r_{1}$ and $r_{2}$, then subtract the second area from the first.  However, I just can't seem to grasp how to find the area of the cross section.

It's not just subtract!!

The variation of angle need to be equal!!

see

http://tutorial.math.lamar.edu/AllBrowsers/2414/PolarArea.asp

The two curves $r=3\cos \theta$ and $r=1+cos\theta$ intersect at $\pm\frac{\pi}{3}$, so therefore the area inside the former and outside the latter is equal to
\[\frac{1}{2}\int_{-\pi/3}^{\pi/3}((3\cos\theta)^{2}-(1+\cos\theta)^{2})\,d\theta\]
\[\frac{1}{2}\int_{-\pi/3}^{\pi/3}(8\cos^{2}\theta-2\cos\theta-1)\,d\theta\]
\[3t-2\sin\theta+2\sin 2\theta\Big]_{-\pi/3}^{\pi/3}=\boxed{\pi}\]


A certain lottery has tickets labeled with numbers 1, 2, ....., 1000. The lottery is run as follows:
First, a ticket is drawn at random. If the number on the ticket is odd, the drawing ends; if it is even, another ticket is randomly drawn (without replacement). If this new ticket has an odd number, the drawing ends; if it is even, another ticket is drawn (again without replacement) until an odd number is drawn. Then every person ticket number was drawn (at any point of the process) wins a prize. You have ticket number 1000. What is the probability that you get a prize?

First time is $ \frac {1}{1000}$. There's a $ \frac {1}{2}$ chance that it'll be an even number, so second time is $ \frac {1}{2}\cdot\frac {1}{999}$. Third time is $ \frac {1}{2}\cdot\frac {1}{2}\cdot\frac {1}{998}$.
So overall, it'll be $ \frac {1}{1000} \plus{} \frac {1}{2}\cdot\frac {1}{999} \plus{} \left(\frac {1}{2}\right)^2\cdot\frac {1}{998} \plus{} ... \plus{} \left(\frac {1}{2}\right)^n\cdot\frac {1}{1000 \minus{} n} \plus{} ... \plus{} \left(\frac {1}{2}\right)^999 \equal{} \sum_{n \equal{} 0}^{999}\left(\frac{1}{2}\right)^n\cdot\frac{1}{1000 \minus{} n}$. But I feel like I'm missing something... And something went awry with my LaTeX.

It won't be $ \frac12$ that the second number is even. Also, you have to check for even numbers not including 1000 for your $ "\frac12"$'s.

Oh yeah, I just realized your second point after posting, although it took me a while to understand the first one.

Ok, first time is $ \frac{1}{1000}$.
Second time around, there's a $ \frac{999}{1000}$ chance that it didn't pick $ 1000$ the first time around, $ \frac{1}{2}$ that the 1st number was even, and $ \frac{1}{999}$ chance that it'll pick $ 1000$. Total, $ \frac{999}{1000}\cdot\frac{1}{2}\cdot\frac{1}{999}$
Third time around, there's a $ \frac{998}{999}$ that it didn't pick $ 1000$, $ \frac{449}{999}$ that the 2nd number is even and $ \frac{1}{2}$ that the first number was even, and $ \frac{1}{998}$ chance that it'll pick $ 1000$. Total, $ \frac{998}{999}\cdot\frac{449}{999}\cdot\frac{1}{2}\cdot\frac{1}{998}$
Continuing in this fashion, the general term is... I have no idea. I'm stuck.

[hide="Solution"]Whenever any even number (other than 1000) is drawn, it is as if nothing happened, since another number is drawn, and since the ratio of odd numbers to the number 1000 stays at 500 : 1, so we can ignore all even numbers other than 1000.

Thus, we have 500 odd numbers, and the number 1000. If an odd number is chosen, the game ends, so 1000 must be chosen first. Thus, the probability is $ \boxed{\frac{1}{501}}$.[/hide]

EDIT: Fixed $ \text{\LaTeX}$ error.

[hide="Solution"]The even numbers are almost irrelevant.  The answer is just the probability that if you take a random order of your ticket and the odd tickets, your ticket is before them.  There are $ 500$ odd tickets plus your ticket, and because each arrangement is equally likely, the answer is just $ \dfrac{1}{501}$.[/hide]


Let $f,g: [0,1]\to [0,1]$ be continuous funtions over $[0,1]$. Prove that there is some $x_{0}\in [0,1]$ such that $g(f(x_{0}))=f(g(x_{0}))$.

Counterexample
$g(x)=1$ and $f(x)=0$ for all $x$.

I am sorry, I edited, and here is problem

[b]Problem[/b]. Let $f,g: [0,1]\to [0,1]$ be continuous funtions over $[0,1]$ satisfying $f(f(x))=g(g(x))$ forall $x\in [0,1]$. Prove that there is some $x_{0}\in [0,1]$ such that $g(f(x_{0}))=f(g(x_{0}))$.


if $x_i > 0$ and $x_i y_i - {z_i}^2 > 0$ then prove that
\[ \frac{n^3}{( \displaystyle\sum_{i=1}^{n} x_i ) (\displaystyle\sum_{i=1}^{n} y_i ) - ( \displaystyle\sum_{i=1}^{n} z_i )^2 } \leq \displaystyle\sum_{i=1}^{n} \frac{1}{x_i y_i - z_i ^2}  \]

Maybe $\displaystyle\sum_{i=1}^{n} \frac{1}{x_i y_i - z_i^2}$? :|

Yes(again I;m misstypo),  I still can't solve this inequality  :P , any Idea??

that's really nice :) I've try a subtitution, and I thinks I've found it (also using Cauchy).

[quote="DPopov"]Can you show this solution? I used a substitution also and am getting close with Cauchy but I keep hitting bumps.[/quote]
 Since, $\frac{\sum_{i=1}^{n}x_{i}\cdot\sum_{i=1}^{n}y_{i}-(\sum_{i=1}^{n}z_{i})^{2}}{\sum_{i=1}^{n}x_{i}}\geq\sum_{i=1}^{n}\frac{x_{i}y_{i}-z_{i}^{2}}{x_{i}}$  then
$\left( \sum_{i=1}^{n}x_{i}\cdot \sum_{i=1}^{n}y_{i}-\left( \sum_{i=1}^{n}z_{i}\right)^{2}\right)\cdot\sum_{i=1}^{n}\frac{1}{x_{i}y_{i}-z_{i}^{2}}\geq$
$\geq\sum_{i=1}^{n}x_{i}\cdot\sum_{i=1}^{n}\frac{x_{i}y_{i}-z_{i}^{2}}{x_{i}}\cdot\sum_{i=1}^{n}\frac{1}{x_{i}y_{i}-z_{i}^{2}}\geq n^{3}.$ :wink:

[quote="arqady"]let, $ f_{i}(x) \equal{} x_{i}x^2 \minus{} 2z_{i}x \plus{} y_{i}.$ Hence, $ \min_{x\in\mathbb R}f_{i}(x) \equal{} \frac {x_{i}y_{i} \minus{} z_{i}^2}{x_{i}}.$
$ \min_{x\in\mathbb R}\sum_{i \equal{} 1}^{n}f_{i}(x)\geq\sum_{i \equal{} 1}^{n}\min_{x\in\mathbb R}f_{i}(x).$
Hence, $ \frac {\sum_{i \equal{} 1}^nx_{i}\cdot\sum_{i \equal{} 1}^{n}y_{i} \minus{} (\sum_{i \equal{} 1}^{n}z_{i})^2}{\sum_{i \equal{} 1}^{n}x_{i}}\geq\sum_{i \equal{} 1}^{n}\frac {x_{i}y_{i} \minus{} z_{i}^2}{x_{i}}$ and Cauchy.  :)[/quote]
Well $ \min_{x\in\mathbb R}\sum_{i \equal{} 1}^{n}f_{i}(x)\equal{}\sum_{i \equal{} 1}^{n}x_i.x^2\minus{}2\sum_{i \equal{} 1}^{n}z_i.x\plus{}\sum_{i \equal{} 1}^{n}y_i$ and what's next?


ax+by+cz+ \sqrt ((a^2+b^2+c^2)(x^2+y^2+z^2))  >=  2/3 * (a+b+c)(x+y+z) for arbitrary reals a,b,c,x,y,z

The official solution doesn't use any force :)

[quote="harazi"]And from this problem we find an immediate solution to a problem from Mathlinks contest! And the solution to this one is quite natural: Cauchy. We have 2/3*(a+b+c)(x+y+z)-ax-by-cz=sum a(2/3z+2/3y-1/3x)<=sqrt(sum a^2)*sqrt(sum (2/3y+2/3z-1/3x)^2)<=sqrt(sum a^2)*sqrt(sum x^2).[/quote]

aH Harazi!! Maybe you are too quick.
How can you be sure that
sqrt(sum (2/3y+2/3z-1/3x)^2)<=sqrt(sum x^2).
can you explain it more carefully!!

Dear SSB, Sum (2x+2y-z)^2 = 9Sum x^2!!!

BTW, this inequality can be generalized: For 2n reals a1, ..., an, x1, ..., xn we have
    Sum a_i.x_i + sqrt((Sum a_i^2)*(Sum x_i^2)) >= (2/n)*(Sum a_i)*(Sumx_i).

Namdung


trigonometry
limit
real analysis
real analysis unsolved

Let [tex]f ( x ) = \sin ( \cos ( x ) )[/tex]. Is it true that [tex]\lim_{n \rightarrow
\infty} f^{( n )} ( \pi ) = \frac{\sqrt{2}}{2}[/tex] ? (By [tex]f^{( n )} ( x )[/tex] I mean [tex]f^{( n )} ( x ) = f ( f^{( n - 1 )} ( x ) ) )[/tex])
(I guess the answer is no : but (1) I didn't gave it much tought (2) who knows ?  :lol: )

It is not hard to see that $f$ is a contraction on $[-1,1]$, so $f$ has a unique fixed point $x_*$ that is a limit of the sequence $f^{(n)} (x_0)$ for any $x_0\in \mathbb R$. Using a computer, we can find $x_*=0.69481969\dots$, which is slightly less than $\frac{\sqrt 2}2=0.7\dots$, so the answer is, indeed, "no".


I think you meant 324 for (6).

(1) edit, stupid mistake

(2) is simply 9*8*7=504.

(5) Look at say, 991. You have 991, 919, 199. Choosing the first and second digits differently gives 9*8, then 3 permutations. 9*8*3=216.

(4) edit did this one wrong too

[quote="igor12"]Let S = {1,2,3,4,5,6,7,8,9} be the set of nonzero digits. Let D denote the set of all 3 digit numbers that can be built using the elements of S as digits and allowing repetition of digits.

1) What is |D|?
2) How many elements of D have three different digits?
3) How many elements of D are multiples of 99?
4) How many elements of D are multiples of 3?
5) How many elements of D have exactly 2 different digits?
6) How many even numbers belong to D?[/quote]

[hide]
1) 9*9*9 = 729

2) 9*8*7 = 504

3) They are: 198, 297, 396, 495, 594, 693, 892, 991. That's 8.

4) For every possible choice of the first two digits, the ones digit can be 3 different values to yield a multiple of 3. So that's 9*9*3 = 243.

5) First choose the two digits. There are 9*8 = 72 ways. Then choose whether the first, second, or third digit will be the unique one. So there are 72*3 = 216 ways.

6) The first two digits can be whatever, the third digit must be 2, 4, 6, or 8. So that's 9*9*4 = 324.
[/hide]

[hide="#1"]$9\times9\times9=729$[/hide]
[hide="#2"]$9\times8\times7=504$[/hide]
[hide="#3"]Only $198$, $297$, $396$,... $891$ work, and there are $8$.[/hide]
[hide="#4"]Once you pick any two for the first two digits, there are $3$ digits left that can be the last one, so there are $9\times9\times3=243$.[/hide]
[hide="#5"]There are $9$ ways to pick the first digit, $8$ ways to pick the second digit, and $3$ ways to pick where the second digit will go, so there are $9\times8\times3=216$ numbers.[/hide]
[hide="#6"]There are $9\times9=81$ ways for the first two digits, and the last one can be $2$, $4$, $6$, or $8$, so there are a total of $81\times4=324$ ways.[/hide]

[hide]1. $9^3=729$
2. $9(8)(7)=504$
3. There are $99(2)$, $99(3)$, ... $99(9)$ . So there ar $8$ .
4. It must be $\frac{1}{3}$ of the total. Therefore, there are $243$ possibilities.
5. First choose the digits. That is $9*8=72$ . Then there are 3 ways to order the digits. Therefore $72*3=216$ .
6. The first 2 digits dont matter. The last one must be even so
$9*9*4=324$[/hide]

EDIT: forgot to hide it.


Find all positive integers $n$ such that $21$ divides $2^{2^n}+2^n+1$.

I think $n$ should be $n\equiv 2\ \ \text{or}\ \ 4\ (mod6)$

Is it ok??
 :)

Try again, I corrected a typo in the problem statement.

$mod \ 3$ : $n$ even
then $mod \ 7$ : $n \in 2+6 \cdot \mathbb{N}$

$2^k=-1,1$ mod $3$,so $n$ must be even.$2^k=2,4,1$ mod $7$,but $2^{2^n}=2$ mod $7$ if $n$ is even.but $2^2=4,2^4=2,2^6=1,2^8=4,...$ mod $7$,so $2^n=4$ if $n$ is in this form $2+6s$,where $s$ is natural.so finally $n$ is in the form $2+6s$ to divisible by $21$.


calculus
real analysis
function
limit
calculus computations

[quote="debanik2"]could anyone recommend  a good calculus book where i can find a substantial treatment of Squeeze Theorem???[/quote] what does squeeze theorem stands for? Anyhow, I think the book principles of mathematical analysis by rudin is a good one indeed...

"Squeeze theorem" stands for this result, and for a number of closely related results.

Suppose that on some punctured neighborhood of $ a,$ $ f,g,$ and $ h$ are real-valued functions satisfying $ f(x)\le g(x)\le h(x),$ and suppose that $ \lim_{x\to a}f(x)\equal{}\lim_{x\to a}h(x)\equal{}L.$ Then $ \lim_{x\to a}g(x)\equal{}L$ (with part of the conclusion being that that limit exists.)

When I say "closely related results," I mean mostly other kinds of limits - one-sided limits, limit at $ \infty,$ limits of sequences ...

To prove it, use the raw $ \epsilon$-$ \delta$ definition of limit and be careful about applying inequalities. There's not much more to it than that.

[quote="Kent Merryfield"]"Squeeze theorem" stands for this result, and for a number of closely related results.

Suppose that on some punctured neighborhood of $ a,$ $ f,g,$ and $ h$ are real-valued functions satisfying $ f(x)\le g(x)\le h(x),$ and suppose that $ \lim_{x\to a}f(x) \equal{} \lim_{x\to a}h(x) \equal{} L.$ Then $ \lim_{x\to a}g(x) \equal{} L$ (with part of the conclusion being that that limit exists.)

When I say "closely related results," I mean mostly other kinds of limits - one-sided limits, limit at $ \infty,$ limits of sequences ...

To prove it, use the raw $ \epsilon$-$ \delta$ definition of limit and be careful about applying inequalities. There's not much more to it than that.[/quote]

O I C, :blush:  thank you!  :lol:


geometry
AMC
USA(J)MO
USAMO
Gauss
Math Kangaroo

Hey if you live in the chicago area you can participate in this competiton.

Aryth, are you going to naperville center,north, or south.

[quote="saszs"]
Aryth, are you going to naperville center,north, or south.[/quote]
That's a tad bit creepy. But to answer your question, he obviously goes to Naperville South High School.

Fermi, you're a jerk  :roll:

I'm going to Naperville Central, which is the "underdog" school in terms of math team, that will hopefully rise to the occasion tomorrow at ICTM state. We had 3 USAMO qualifiers, unlike North, which has a long tradition and history of doing well in everything math team related. This year they had 0 USAMO qualifiers.

Naperville South does not (yet) exist. There are plans for a "South" to be built in District 204 (NCHS and NNHS are in 203), since Waubonsie Valley and Neuqua Valley are hugely crowded.

Why do you ask? If you have any questions about schools or personal things just PM me.

Rofl.

Okay I forgot to say this but I was at state science olympiad.  Our team went into the awards ceremony at around 4:30.  I was all for sitting up at the top which was the only space left.  But then we thought it would be too hard to go up to get our awards.  So we sat on the floor.  I was thinking, "Shoot!  Oh well."  Then suddenly a name on a t-shirt caught my attention.  Kevin?  Kevin Hu?  Hmmm... OH.  *looks up*  Naperville Central High School.

New Trier Winnteka owned everyone.  Hmm I noticed that IMSA didn't do that good :( I also noticed that North was winning more than Central..

both richard rhoad and robert whipple teach math at new trier.

I have their geometry book.

it's also a fact that my local high school failed...

Does anyone here think that NSML gives too much time for each problem set...? I'm not really sure how it goes, but each question is worth 5 points, right? Any partial credit?

The point of contests is not to see who's the top like 3, it's to give all students experience with (fun and sportsmanlike competition). If NSML contests had a time limit of 10 minutes, some of us would be fine, but it'd have to be renamed No Such Math League.

Questions are weighted 4, 4, 5, 6, 6.  No partial credit.

It's more a test on consistency (not makign stupid mistakes) than on skill.  However, it is pretty difficult to not make any mistakes throughout the whole thing.

[quote="gauss1181"]bleh aren't there some beastly people like Michael Tarczon and Arjun Puranik who probably would never make stupid mistakes?[/quote]

False. Yesterday, Arjun['s team] was 4th out of 4 teams going into the last question of jr - sr 2 person demonstration (only to move to first thanks to a quick balls and urns problem). I'm sure Arjun was only 4th at that point due to stupid mistakes and not other reasons... cause most of the teams beating him at that point weren't all that amazing.

dude i was pretty disappointed in that demo. the last problem was like... let's do this in my head. oh, yay, 10 choose 3 - 6 choose 3 = 100.

hey kevin weren't you at the math kangaroo for the last few years

and USAMO.

day one
http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/09USAMOday1.pdf

day two
http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/09USAMOday2.pdf

Well, I all I know is that he did USAMO.

[quote="gauss1181"]huh what???

saszs, you attended NSML?[/quote]

lol i don't think he's in high school

I was talking about the math kangaroo.

I'm only in 6th.

so I saw gauss and suma in the mcstate.

Lol wow.
I see you live in Lisle...me too. o_o

Math Kangaroo is so bleh. There's an awards ceremony for that on Saturday May 9, but I think I'm gonna ditch that this year. I went last year and I won an..... alarm clock... without batteries included. One more question would have gotten me 100 dollars.   :(


AMC
AIME
geometry
geometric transformation
rotation
USA(J)MO
USAMO

Just curious, did anyone take the AIME II?

"Yeah, I did. I'm not a big fan of my score."

This is how most people feel after any competition, but you are Eric Morphis, right?

Yes, that's my name; what's yours (Boru Wang, perhaps)? I've actually felt good after many competitions this year - I guess I was due for a bad one.

Not that anyone has ever heard this from me before (yeah, I know. If you know me you've heard this lots before. Just tune out now!) Math Team is like golf. Tiger Woods is clearly the best golfer in the world. It's not even close. But he doesn't win every tournament. We have at least 7 students in Georgia that could win any tournament. Sometimes it just depends on who's having a good day or if the test suits them.

Tom

Yes, my name is Boru Wang. I rarely feel good after any competition, but I think I can largely remedy this situation by studying more.  :)
Eric, what kind of study materials do you use?

[quote="llolloll"]Eric, what kind of study materials do you use?[/quote]
I practice almost exclusively by doing problems from old competitions. I have a lot of trouble reading math books because I have an extremely short attention span. :maybe: Naturally, two of the three tournaments that I have won this year (UGA and Furman) have been the two tournaments that post the most past tournaments on their websites because I have so much experience with their test-type.

[quote="Boy Soprano II"]Well, variability does have something to do with it, but also, from what I hear, this AIME was particularly hard. I would not be too surprised if the highest score in the state were eight or so.[/quote]
Actually I might be the top score given that no other ARMLers in the state took the AIME II (edit: Boru forgot to mention that he got an eight  :wink: ). I don't think this AIME was all that challenging - it just had quite a few concepts that I am unfamiliar with (inversion, Pell equations, rotations, etc.). Also, my short attention span doesn't help me during a three hour competition...

Boru, don't be that modest, you should feel good this time. You know you get in USAMO with an 8 on AIME II anyways, not to mention your AMC 10 score. In my opinion, this AIME was more trick-oriented than the first one. I don't know if this is fortunate or unfortunate, but I personally got a much higher score on this one (unofficially): 1-9, 11. But this doesn't mean anything, since I didn't take it...

interesting....i did really bad on both amc and aime...oh well

tj only had 6 haha >.<

[quote="gtf"]We have at least 7 students in Georgia that could win any tournament.[/quote]
Well... in my opinion, there are only two students in Georgia who could potentially be the USAMO state winner. :wink:

[quote="cognos599"][quote="Math Geek"]Yay!  I made USAMO by 1.5 points!  ZOMGs!  There was a 5th grader who made it, again.  idk if it was the same person who made it in the 4th grade last year.

EDIT: The preliminary list of people who made it in GA:

Harrison Brown 12 Centennial HS Alpharetta GA
Howard Tong 12 Chamblee HS Atlanta GA
Miles Edwards 11 Lassiter HS Roswell GA
Sitan Chen 8 Northview HS Johnscreek GA  :lol:
Benjamin Hu 10 Northview HS Johns Creek GA
Oliver Huang 10 Pace Academy Atlanta GA
Boru Wang 10 Parkview HS Lilburn GA
Tianqi Wu 9 Parkview HS Lilburn GA
Eddy Ferreira 12 Roswell HS Roswell GA
John Dorminy 11 Soli Deo Gloria Home Ed Mcdonough GA
Santhosh Karnik 11 Wheeler HS Marietta GA

PS: I didn't know Billy was a nickname for John. :huh:[/quote]

who the heck is Sitan Chen??????????? wow he must be really smart.  :wink:[/quote]

lol Sitan Chen is Math Geek.

[quote="Math Geek"]I didn't know Billy was a nickname for John. :huh:[/quote]

Billy is not a nickname for John, actually. $ \pi$ is a nickname for John; brotherdear Johnny the youngest goes by $ \pi$. Otherwise Billy ... approximates random. I'm stubborn, I'm not going to say what my middle name is (NOT william).

[quote="Twiz"][quote="academicsrule"]lol Sitan Chen is Math Geek.[/quote]
I think cognos was being sarcastic, but I can't be sure. :maybe:[/quote]

you are right, i was being sarcastic that is why i had a  :wink: at the end. lol  :)

[hide]I didn't I don't even know what that is[/hide]

[quote="llolloll"]who is cognos599?[/quote]

Not a Georgian, I think.

i don't know him very well, but i think cognos599 is a hoosier.


Is it possible to write some ten numbers from the set $ \{ 0,1,2,...,14 \}$ in the circles in the diagram in such a way that the absolute differences of two numbers joined by a segment are all different? Justify your answer.

Assume that this task is possible. The differences must be 1, 2, 3, ..., 14. The sum of this differences is 1+2+3+...+14=105. But each number is counted twice (maybe added or subtracted) because each circle is connected to two other circles, so this sum must be even. We get a contradiction, so the task is impossible.


integration
calculus
function
complex analysis
complex analysis theorems

[b]Cauchy's Theorem for a disk.[/b] Let $f$ be analytic on a closed disk and let $\gamma$ be a path arund the boundary of the disk. Then $\int_{\gamma}f(z)dz=0.$
I must be misunderstanding something, but i'm just interested in question: does the theorem conclusion remains true if there,say,one singularity on the bopundary of the disk. For example does it follows from the theorem that $\int_{|z-1|=1}\frac{1}{z}dz=0$ ???

???
But this integral diverges! :?
Anyway, Cauchy's theorem requires that the function is continuous in $\overline{D}$ and analytic inside of $D$, $D$ is an open region with good boundary.

If there are poles inside the region, the residue theorem applies, allowing you to calculate the value of the integral based on the residues at the poles.
If there is a simple pole on the boundary of the region (at a smooth point), the integral diverges but we can identify a "principal value" by taking a limit of integrals that omit a small length centered at the pole. This principal value is halfway between the value if the pole were inside and the value if the pole were outside.

Don't even ask about other kinds of singularities.

There was the reason i'was asking about this questioon regardless it's divegres,because at the calculus lesson we were calculating the integral $\displaystyle\int_{|z-1|=1}\frac{e^{z^2}}{z(z-6)}dz$ and my teacher said that it follows from Cauchy theorem that this integral equal to $0$ regardless it has singularity(simple pole $z=0$) on the integration boundary.
It seems to me that if even it were true it would be for example $\displaystyle\int_{|z-1|=1}\frac{1}{z}dz=0$ in the same way, nevertheless it diverges.

I think that it is not true and you words:Myth,jmerry confirmed it(as far as i understand)

Are you sure your teacher was right?  :?
Thanks for PM ;)

it seems to me that he wasn't right. But like a proverb says the teachers shouldn't be argued with...


Find the best constant $k$ such that the inequality true for all $a,b,c,d$ positive :

$\frac{abc}{ab+bc+ca}+\frac{bcd}{bc+cd+da}+\frac{cda}{cd+da+ab}+\frac{dab}{da+ab+ba} \ge k\sqrt[3]{\frac{abcd}{a+b+c+d}}$

The same question for three veriable has result $k=\frac{5}{2}$
For 4 variable I believe that $k=\frac{11}{6}$

For $a=b=c=d$, the best constant $k$ is $0$. :P

It's $(\frac{abcd}{a+b+c+d})^{1/3}$
I don't use latex well.

are you sure about the denominator of the l.h.s.?

[quote="hungkhtn"]Find the best constant $k$ such that the inequality true for all $a,b,c,d$ positive :

$\frac{abc}{ab+bc+ca}+\frac{bcd}{bc+cd+da}+\frac{cda}{cd+da+ab}+\frac{dab}{da+ab+ba} \ge k\sqrt_{3}{\frac{abcd}{a+b+c+d}}$

The same question for three veriable has result $k=\frac{5}{2}$
For 4 variable I believe that $k=\frac{11}{6}$[/quote]

I mean, should it be
$\frac{abc}{ab+bc+ca}+\frac{bcd}{bc+cd+da}+\frac{cda}{cd+da+ab}+\frac{dab}{da+ab+ba} \ge k\sqrt_{3}{\frac{abcd}{a+b+c+d}}$
or
$\frac{abc}{ab+bc+ca}+\frac{bcd}{bc+cd+db}+\frac{cda}{cd+da+ac}+\frac{dab}{da+ab+ba} \ge k\sqrt_{3}{\frac{abcd}{a+b+c+d}}$

Just a remark about $\LaTeX$: to get $\sqrt[3]{\text{whatever}}$ type (in dollar signs) \sqrt[3]{whatever}. So your right hand side is $k\sqrt[3]{\frac{abcd}{a+b+c+d}}$, isn't it?

The inequality can be write clearly:

Prove that if $a,b,c,d$ are positive reals st $abc+bcd+cda+dab=1$ then:

$\sum\frac{1}{a+b+c} \ge \frac{11}{6}$

[quote="hungkhtn"]Prove that if $a,b,c,d$ are positive reals st $abc+bcd+cda+dab=1$ then:
$\sum\frac{1}{a+b+c} \ge \frac{11}{6}$[/quote]

Dear hungkhtn! I don't think it's true!
Let c,d-->0; a,b--> + infinity then:
1/(a+b+c)+1/(b+c+d)+1/(c+d+a)+1/(a+d+b) -->0


Give triangle ABC, $ \widehat{BAC}\equal{}90^{o}$. Let M is midpoint of BC. a,b is straight lines, which perpendicular with BC at B and C. D is a point on AM straight lines. BD, CD interects b and c at E and F. Prove that: $ \widehat{EAC}\equal{}\widehat{FAB}$

See the figure below.
We will prove that if $ \angle FAC\equal{}\angle EAB$ then $ EC, BF, AM$ are concurrent
iff $ \frac{AL}{LB}\equal{}\frac{AP}{PC}\Leftrightarrow \frac{S_{BAF}}{S_{BCF}}\equal{}\frac{S_{EAC}}{S_{EBC}}$
$ \Leftrightarrow \frac{AB.AF}{BC.CF}\equal{}\frac{AC.AE}{BC.BE}$
$ \Leftrightarrow \frac{AB}{AC}\equal{}\frac{FC}{AF}.\frac{AE}{BE}$
$ \Leftrightarrow \frac{AB}{AC}\equal{}\frac{sin \angle FAC}{sin \angle ACF}.\frac{sin \angle EBA}{sin \angle EAB}$
$ \equal{}\frac{sin \angle ACB}{sin \angle ABC}$
We are done.


1. Select $p$ and $q$ at random from the interval $[\frac{1}{4}, 1]$. Draw two arcs in Quadrant I as follows: draw an arc of $90^\circ$ of radius $p$ with center $(0,0)$ and an arc of $180^\circ$ of radius $q$ with center $(1,0)$. What is the probability that the arcs intersect?

2. A line segment of length 6 is divided at random into 3 parts with the restriction that the longest part is at least 0.5 more than the next longest part and the next longest part is twice the shortest part. Find the probability that the three parts can serve as the 3 sides of a triangle.

(Source: ICTM State Math Contest)

[quote="frt"]1. Select $p$ and $q$ at random from the interval $[\frac{1}{4}, 1]$. Draw two arcs in Quadrant I as follows: draw an arc of $90^\circ$ of radius $p$ with center $(0,0)$ and an arc of $180^\circ$ of radius $q$ with center $(1,0)$. What is the probability that the arcs intersect?[/quote]

1)Since they are in the first quadrant we have a quarter-circle in the "corner" of the first quadrant with radius p and a half-circle sitting on the x-axis centered at (1,0) with radius q. The two circles intersect if the sum $p+q$ is more than 1. So, to solve, make a graph with p being vertical axis and q being horizontal axis. So we have a few lines. By the problem we have $\frac{1}{4} \leq p,q \leq 1$ and to make it intersect we have $p + q > 1 \Rightarrow p > -q + 1$. The lines intersect at the points $(\frac{1}{4},\frac{3}{4})$ and $(\frac{3}{4},\frac{1}{4})$. So, we have a square with sidelength $\frac{3}{4}$ and an isosceles right triangle with legs $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ in the bottom left corner of the square. Since the line $p > -q + 1$ is greater than, the entire square EXCEPT the triangle is acceptable area, so the probability is $\frac{Area_{square}-Area_{triangle}}{Area_{square}} = 1 - \frac{\frac{1}{2}\frac{1}{2}\frac{1}{2}}{\frac{3}{4}\frac{3}{4}} = 1 - \frac{2}{9} = \frac{7}{9}$.

2) has been solved before. It really bothered me and I posted it. This is the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=254604&highlight=#254604]link.[/url]

Edit: I did not read carefully. #2 is not the one behind the link. But you can still go look if you want.

Your answer is correct, Pork_Chop8.

[quote="Pork_Chop8"]2) has been solved before. It really bothered me and I posted it. This is the link.[/quote]
Sorry, I didn't realize that :(

The interesting thing is, though, my problem is from 2004 ICTM State Geometry competition, and yours was from 2002 ICTM Regional Algebra II competition.

Just so you know, the answer for #2 is
[hide]$\frac{1}{11}$[/hide] :)

If the shortest one is length x, then the next one is 2x, and the longest is 2x+0.5+a, where a is some number between 0 and 5.5.  We know that the maximum for x is 1.1.  So, assume that each possible length for x is equally likely.  The three segments can serve as the three sides of a triangle if and only if (2x+x)>6(1/2), because the perimeter will be 6.  This occurs when x>1.  Therefore, the probability that the three segments can be the sides of a triangle is (0.1)/(1.1)=$\frac{1}{11}$


Prove that the only solution in rational numbers of the equation:
$ x^3 \plus{} 3y^3 \plus{} 9z^3 \minus{} 9xyz \equal{} 0$ is $ x \equal{} y \equal{} z \equal{} 0$.
http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf at page 13
I didn't understand this part: "[i]Hence, we may divide powers of 3 out of our integer solution an arbitrary
number of times, contradiction[/i]"
Could someone explain it to me?

The argument goes as follows:
-A solution in rationals corresponds to a solution in integers, and so if there are no solutions in integers, there aren't in rationals.
-If there was a solution in integers, $ (x_0,y_0,z_0)$ then it was proven that $ (x_0/3, y_0/3,z_0/3) \equal{} (x_1,y_1,z_1)$ is also a solution in integers. But then $ (x_0/9, y_0/9,z_0/9)$ is also a solution, and you can divide by powers of 3 and still get a solution. But all the numbers here are integers, and this process could not be carried out indefinitely and still get integers each time. So there are no solutions in integers.

Perhaps it might make more sense to you to pick the "smallest" solution then apply the same reduction method to get a "smaller" one, which is a contradiction. The two argument types are really the same here.

EDIT: Explained it better.


Ok, here's a little fun problem my freind asked me. I don't actually have an answer myself so I kinda wanna hear other people's opinions and ideas  :P

You have 100 loonies, 20 of them are heads up, 80 tails up. You are blindfolded and have gloves on. How do you split up those 100 loonies into 5 piles with an equal number of heads up loonies in each pile?

I don't think you can 'feel' which face is up up since you have golves...so..hmmm? :D
Btw... loonies or one dollar coins in Canada... just saying :blush:

This might seem a bit awkward...but if they are really really clean coins, use you're tongue?

What...and you don't say you *have* to keep the blindfold on.

Just saying...that's what you could do. :oops: