Solve the equation $ x^3 \minus{} 3x \equal{} \sqrt {x \plus{} 2}$.

$ \cos 3\theta \equal{}\left|\cos \frac{\theta}{2}\right|.$

[quote="kunny"]$ \cos 3\theta = \left|\cos \frac {\theta}{2}\right|.$[/quote]

I think you are wrong, if I don't misunderstand.

Your proposal seems to consider $ x=\cos(\theta)$. but :

$ 1)$ some solutions have an absolute value $ > 1$ (for example $ x=2$) and can't be written as cosine of real numbers.

$ 2)$ $ |\cos(\frac{\theta}{2})|\neq \sqrt{cos{\theta+2}}$

[quote="kunny"]$ x \equal{} 2\cos \theta$[/quote]

  OK, I'm sorry  :blush:  :blush:

Let $x=2\cos\theta$.

Then $8\cos^3\theta-6\cos\theta=\sqrt{2+2\cos\theta}\implies \cos 3\theta=\left|\cos\frac{\theta}{2}\right|$ by just using the tiple angle and half angle formulae.

Either $3\theta=k\pi+\frac{\theta}{2}\implies\theta=\frac{2k\pi}{5}$ or $3\theta=k\pi-\frac{\theta}{2}\implies\theta=\frac{2k\pi}{7}$.

Then, we take all solutions in $0\leq\theta<2\pi$ where $\cos 3\theta>0$. These are $\theta\in\left\{0,\frac{4\pi}{7},\frac{4\pi}{5},\frac{6\pi}{5},\frac{10\pi}{7}\right\}$

Final solution:

$x=2$

$x=2\cos\frac{4\pi}{5}=\frac{-1-\sqrt{5}}{2}$

$x=2\cos\frac{4\pi}{7}=-\frac{1}{3}-\frac{1+i\sqrt{3}}{12}\sqrt[3]{28+84i\sqrt{3}}-\frac{1-i\sqrt{3}}{12}\sqrt[3]{28-84i\sqrt{3}}$

because $\sin{\theta}=\sin{\theta+2\pi}$, so no need to consider $\theta \ge 2\pi$


Find all functions $f,g:R \rightarrow R$ such that
$(i)$ $g(g(x)+y) = x-y+2g(y)$ and
$(ii)$ $f(g(x)+y) = g(x+f(y))$
for all real numbers $x,y$

Let's put $y=0$, so we have $g(g(x))=x+2g(0)$, so for every $t \in R$ there is only one $x$ such that
$g(x)=t$.
Because $f(g(x)+y)=g(f(x)+y)$ for every $x,y \in R$, we conclude that $f$ has  the same (as $g$) property mentioned.
In addition, there exists such $x_0$ that $g(x_0)=0$, so $f(g(x_0)+y)=g(x_0+f(y))$ so
$f(y)=g(x_0+f(y))$ but from the property of $f$ - $f(y)$ can be every value from $R$, so for every
$t \in R$ $g(t+x_0)=t$, so $g(t)=t-x_0$, so $g(g(x))=g(x-x_0)=x-2x_0$ so $x-2x_0=x+2g(0)$ so
$x_0=-g(0)$, so $g(x)=x+g(0)$ and $f(g(x)+y)=g(x+f(y))=x+f(y)+g(0)$. Putting $y=0$ we obtain
$f(g(x))=x+f(0)+g(0)$ but $f(g(x))=f(x+g(0))$ so $f(x)=x+f(0)$.
Thus, $f(x)=x+f(0)$ and $g(x)=x+g(0)$ are the only solutions.


Let $ a_1, a_2, ...$ be a sequence for which \[a_1 \equal{} 2\,\hspace{.2in}a_2 \equal{} 3\, \hspace{.2in}\text{and}\hspace{.2in}a_n \equal{} \frac {a_{n \minus{} 1}}{a_{n \minus{} 2}} \text{ for each positive integer } n \ge 3.\]What is $ a_{2006}$?

$\textbf{(A) } \frac 12 \qquad \textbf{(B) } \frac 23 \qquad \textbf{(C) } \frac 32 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } 3$

[hide]Playing around, we see that the first few terms are 2,3,3/2, 1/2, 1/3, 2/3, 2,3, and so on.  It repeats with period 6.  2006 is 2 greater than a multiple of 6, so the answer is 3. E.[/hide]


Let $s(n)$ denote the sum of the digits of $n$.  What is $s(s(s(2^{2008})))$

my ti89 said so

[url]http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_theorem[/url]

[quote="Tecnogram888"]why is $2^{2008}=7 \mod9$?[/quote]

More elementarily:

$2^{2008}\equiv 2 \cdot 2^{3 \cdot 669}\equiv 2 \cdot 8^{669}\equiv 2 \cdot (-1)^{669}\equiv-2 \equiv 7 \bmod 9$

[quote="junggi"][hide]$2^{2008}$ has $\lfloor2008 \log 2\rfloor+1<2009$ digits.
Thus an upper bound for $s(2^{2008})$ is $9\cdot2009=18081.$
So $s(s(2^{2008}))$ can be at most $36,$ which is when $s(2^{2008})=9999.$
Finally, $s(s(s(2^{2008})))$ is at most $11,$ when $s(s(2^{2008}))=29$
Now note that $2^{2008}=7 \mod9$ which means that $s(s(s(2^{2008})))$ is also $7\mod9.$
Since $s(s(s(2^{2008})))<11$ and $s(s(s(2^{2008})))=7 \mod9,$
$s(s(s(2^{2008})))=7$

[/hide][/quote]

How do you know that taking the sum of the digits preserves the congruency of the original number modulo 9?

[quote="NeverOddOrEven"]How do you know that taking the sum of the digits preserves the congruency of the original number modulo 9?[/quote]

[hide="One line"] $\sum_{k=0}^{n}d_{k}10^{k}\equiv \sum_{k=0}^{n}d_{k}(1)^{k}\bmod 9$ [/hide]
[hide="For example..."] A number such as $12345$ can be written $10000+2000+300+40+5 = 1(1+9999)+2(1+999)+3(1+99)+4(1+9)+5$. [/hide]

This problem is very similar to this: http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteWrite.php


Find all the reals x if we know that (x^n)+x^(-n) is integer for each integer n

Isn't this completely trivial?

Obviously $x + 1/x$ needs to be an integer and that is sufficient (it is well known that if $x + 1/x$ is an integer for some real number $x$, then so is $x^n + x^{-n}$ for all positive integers $n$ - by a simple induction). So we just look at all roots of the equations $x + 1/x = n$ or $x^2 - nx + 1 = 0$ where $n$ varies over the integers. These are then the only solutions.


modular arithmetic
quadratics
number theory proposed
number theory

Let $ p$ be a prime number. Find $ r\in \{0,1,\cdots, p\minus{}1\}$ such that $ \prod_{i\equal{}0}^p(i^2\plus{}1)\equiv r\pmod{p}$.

If $ p \equal{} 4k \plus{} 1$ then there is $ i$ such that $ i^2 \plus{} 1 \equiv 0 \pmod{p}$, so $ r \equal{} 0$ in this case. Suppose that $ p \equal{} 4k \plus{} 3$.

We claim that $ r \equal{} 4$. Indeed, it is equivalent to $ \prod_{i \equal{} 2}^{p \minus{} 2} (i^2 \plus{} 1) \equiv 1 \pmod{p}$. To see that it holds multiply both sides by $ \prod_{i \equal{} 2}^{p \minus{} 2} (i^2 \minus{} 1)$ which is clearly non-zero mod $ p$. So we see that that congruence is equivalent to
\[ \prod_{i \equal{} 2}^{p \minus{} 2} (i^4 \minus{} 1) \equal{} \prod_{i \equal{} 2}^{p \minus{} 2} (i^2 \minus{} 1) \pmod{p}
\]
but this holds, since $ p$ is of the form $ 4k \plus{} 3$ and every quadratic residue is also $ 4$-th power residue. Too see this just note that if $ a^4 \equiv b^4 \pmod{p}$ then $ (a \minus{} b)(a \plus{} b)(a^2 \plus{} b^2) \equiv 0 \pmod{p}$. So, we must have $ a \equiv b$ or $ a \equiv \minus{} b$ since $ p$ is of the form $ 4k \plus{} 3$ and $ a^2 \plus{} b^2 \not \equiv 0 \pmod{p}$. It means that both product's contain the same factors, so they are congruent.

If $ p\in\{2,3\}$ then $ r\equal{}\cdots$? :P

Oops, missed the case $ p\equal{}2$. But I will work on that :D.

See at $ f(x)\equal{}(x\plus{}1^2)\cdots (x\plus{}[(p\minus{}1)/2]^2)\minus{}x^{(p\minus{}1)/2}\plus{}(\minus{}1)^{(p\minus{}1)/2}$ and thanks to Lagrange's theorem.
ZetaX@: Maybe $ 4>p\minus{}1$.

If $ p\equal{}3$ then that is a wrong answer.

The answer should be in $ \{0, 1, 2, \ldots, p\minus{}1 \}$. For $ p\equal{}3$ it's not, since $ 4 > 2$ :P.

[quote="TomciO"]The answer should be in $ \{0, 1, 2, \ldots, p \minus{} 1 \}$. For $ p \equal{} 3$ it's not, since $ 4 > 2$ :P.[/quote]
You are right! English is my problem.  :blush:


The area of the smallest square that will contain a circle of radius
4 is what?

this is asking area of the smallest square that can hold a circle of diameter 8
this is one which is tangent on all 4 sides to the circle.
the side length would be 8.
so area is given by formula
$ A\equal{}s^2$
and plugging in 8 gives
$ A\equal{}8^2\equal{}64$
64

Yes
The diameter of the circle would be equal to the side length of the square.
This leaves us with (2*4)^2=64, the area.


How many positive integers less than 100 are divisible by neither 5 nor 2?

[hide]There are 50 numbers less than 100 divisible by 2. There are 20 divisible by 5. Adding 50 and 20 together we get 70 numbers that are divisible by those. But we counted some twice. Those divisible by 10. There are 10 of those. Now we only have 60 that are divisible by 2 or 5. Therefore, we have $40$ that aren't.[/hide]

[quote="Necrowarrio0"][hide] 50 positive intergers below 100 can be divided by 2. 20 can be divided by 5. 50+20=70. But, the answer is not 70 because there are some common multiples of 5 and 2. After counting, we find out that there are 10 numbers that can be divided by 5 and 2. 70-10=60. 100-60=40. Therefore, we have 40. :spam:[/hide][/quote]

?????? whoa why is this post the same as kstans?

Yes, please please please do not copy other people's solutions.  You can still post a solution, but make it different in some way.  If you're going to paraphrase (to make something simpler), please change more than a few words.

there are 50 numbers that are not divisible by 2. 80 numbers not divisible by 5. some of those are also in 2 one so that makes 90 not divisible by 5.

90-50=40

Another way, sometimes simpler is just use the fact that ($\frac{1}{2}$)   are not divisible by 2 and ($\frac{4}{5}$ ) (of those remaining)  are not divisible by 5

[hide]so all you have to do is $100 \cdot \frac{1}{2}\ \cdot \frac{4}{5}= 40$
[/hide]
What makes it simpler, is you don't have to worry about "double counting".

(For example, if I changed the problem,  to:
[quote]How many positive integers less than 120 are divisible by neither 5 nor 2 nor 3 ?[/quote]

[hide="the answer"]   $120 \cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{4}{5}=32$ will give the correct answer.  [/hide]

[quote="hairymonkey2"]heres the really long way i did this-counted them all from 1-100 and theres 40. id name them but i dont think thats necessary :D[/quote]

After doing this, you should have noticed a pattern: there are exactly $4$ numbers out of the first ten that aren't divisible by $5$ or $2$, exactly $4$ out of the next ten, and so forth.

So, since there are ten sets of $10$, there are $4 \cdot 10 = \boxed{40}$ total.  (Even if you were to count them out, if you had noticed this pattern you would only have needed to consider the first ten positive integers ;) )

[hide="And this is because..."] When we are considering divisibility by both $5$ and $2$ the remainders repeat every $10$ numbers (since $lcm(5, 2) = 10$).

Moreover, $\varphi(10) = 4$. [/hide]


Let $ \sin x\plus{}\sqrt{2\minus{}\sin ^ 2 x}\equal{}t$.

Thank you. We can done it by use Inequality.

Using kunny's hint:
[hide="Factorization of Given Equation"]
$ (\sin x \plus{} \sqrt{2 \minus{} \sin^2 x} \plus{} 4 ) ( \sin x \plus{} \sqrt{2\minus{}\sin^2 x} \minus{} 2 ) \equal{} 0$.[/hide]


probability
IMC
college contests
Probabilistic Method

200 students participated in a math contest. They had 6 problems to solve. Each problem was correctly solved by at least 120 participants. Prove that there must be 2 participants such that every problem was solved by at least one of these two students.

This looks too trivial to be true, so I must be missing somewhere...

[hide="My Solution"]Consider the best student:
If he solved 6 problems, this is trivially true.
If he solved 5 problems, there are still 120 students who solved the other problem
If he solved 3 or less problems, not each problem can be solved by 120 students.

So, say student A solves {1,2,3,4}. Now, if there is no student solving both 5 and 6, there's no way we can get those problems solved by 120 people. So we're done? :o [/hide]

how do you solve the problem by the "probabilistic" method? (in the "Downloads" sections it is said that it can be done in this way)

Pick two students at random.  We know that the probability both students failed to solve problem $1$ is strictly less than $(80/200)^2=4/25$.  This holds for any other problem, so the probability that a problem exists that was unsolved by both students is at most $24/25$.  Since this is less than 1, there must be a pair of students for which this isn't true, i.e. one who combined solved every problem.

This seems like the most trivial IMC question ever :?

[and the official solution is over half a page, using long combinations and putting an upper limit on the number of couples...]

With Peter' solution this is trivial for smaller values also. (until $101$)

Indeed, and a model counterexample can be found for $100$; see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=487528&hilit=solved+problem[/url] for another posting of this very problem.

I think the official solution [b]Peter[/b] describes can be shortened as follows

[hide=solution]
Represent each problem as a row of a matrix and each participant as a column of that matrix. Let the element in the $i$-th row and $j$-th column be $m_{i, j}$ where $m_{i, j}=1$ if student $j$ solved problem $i$ and $m_{i, j}=0$ otherwise. Let $X$ be the number of unordered pairs of elements $(m_{i, x}, m_{i, y})=(0, 0)$ where $x, y \in [1, 200]$.

Suppose the statement is false. Then each pair of students has at least one problem not solved between the two students so $X \ge \dbinom{200}{2} = 19900$. Also, for each row of the matrix there exists at most $80$ elements that are $0$ so $X \le 6 \cdot \dbinom{80}{2} = 18960$. But this means $18960 \ge 19900$, untrue.
[/hide]

A solution using probabilistic method (it's eventually the same as AMN300).
Pick two contestants randomly, the probability that those two can't so a problem is at most $\frac{\binom{80}{2}}{\binom{200}{2}}$. Hence, if $X$ is the number of problems that $2$ contestants can't solve then $\mathbb{E}[X]= 6 \cdot \frac{\binom{80}{2}}{\binom{200}{2}}<1.$
Hence, there exists a way to pick two contestants so that $\mathbb{E}[X]=0$, or each problem can be solved by at least one of them.

[hide = Solution]
Consider two random students. The probability none of them solved a particular problem is $\frac{2}{5} \cdot \frac{2}{5} = \frac{4}{25}$. Therefore, the expected number of problems that none of the two students solve is $\frac{24}{25}$. But since this number is less than one, there must be two students such that the number of problems that none of them solve is $0$, as desired. $\square$
[/hide]

Assume for contradiction that this is not the case, then consider the student with the highest score, $m$
[b][color=#f00]Case 1:[/color][/b] $m=6.$
It's clear that this is impossible.
[b][color=#f00]Case 2:[/color][/b] $m=5.$
This is also impossible as the other problem which is solved incorrectly/unsolved by the student must not be solved by any of the other students, contradicting the condition given.
[b][color=#f00]Case 3:[/color][/b] $m=4.$
In this case, consider the remaining $2$ problems solved incorrectly/unsolved by the the student, it's clear that there must be one problem which is not correctly solved by at least $120$ students. Thus, a contradiction.
[b][color=#f00]Case 4:[/color][/b] $m\leqslant 3.$
In this case, the total number of problem solved correctly by all the participants is less than or equal to $3\times 200=600$ which is less than $6\times 120=720$ correctly solved problems required by the problem statement.

[quote=brianzjk][hide=Solution]If we choose any 2 contestants, then the probability that they will both get a problem wrong is $\left(\frac{2}{5}\right)^2=\frac{4}{25}$. Then, the expected value of the number of problems both contestants get wrong is $6\cdot\frac{4}{25}=\frac{24}{25}$. However, this is less than 1, so there exist a pair of contestants such that each problem is solved by at least 1 contestant, as desired. [/hide][/quote]

Isn't this particular solution incorrect, because technically, once you've chosen the first contestant, you know that the second contestant must be distinct, so conditionally, the second probability should be 79/199, as in @shinichiman's solution. Because then a "pair" that exists that you are counting is a person paired with themselves, which is absurd, so although you used a weaker bound, you also a got a weaker statement.

[hide=Solution]
Consider the highest score.

Case 1: The highest score is $6$. Then this is obviously true.

Case 2: The highest score is $5$. This is also true because at least $120$ people solved the other problem.

Case 3: The highest score is $4$. Assume they solved problems #1, #2, #3, and #4. If no one solved both problems #5 and #6, the $\ge120$ people who solved #5 are all different from the $\ge120$ people who solved #6. So there are at least $240$ students, a contradiction. Therefore, at least one other person solved the other two problems so there are 2 participants such that every problem was solved by at least one of these two students.

Case 4: The highest score is 0, 1, 2, or 3. Assume each problem solved gives you one point. The sum of all students scores is at most $600$. But by the problem conditions, the sum of all students scores is at least 720, a contradiction. [/hide]

As $\frac{120 \cdot 6}{200} > 3$ there exists a student who solved at least four problems. If any student solved five or six problems, the conclusion is trivial.

Now consider the two problems that the student cannot solve. If no student solved both such problems, they would have received less than 199 solves total, an obvious contradiction. Thus, the aforementioned student and the student who solved both problems form the desired pair.

First, there are at least $120 \times 6=720$ correctly solved problems, and $\frac{720}{200} >3,$ so there is at least a student that solved $4$ problems.
[b]Case 1:[/b] The student with highest score solved $6$ problems.
This is obviously true.
[b]Case 2:[/b] The student with highest score solved $5$ problems.
There must be another $120$ students that solved the other problem, so it is true.
[b]Case 3:[/b] The student with highest score solved $4$ problems.
Fix the highest student to solve $4$ problems out of the $6,$ consider the other $2$ problems, if no students solved both problems. it is clearly contradicting the problem statement so we are done. $\blacksquare$

Note that the probability that a pair of students both didn't solve a problem is $\frac{4}{25}$, so the expected number of problems that neither has solved is $\frac{24}{25}$, implying that there exists a pair such that this number is $0$.


function parameterization trigonometry calculus algebra domain derivative

Is it true that
the function $ y(x)$ is differentiable iff $ x(t), y(t)$ are differentiable and $ x'(t)\neq0$,
where $ t$ is a parameter?

However, for a function $ y(x)$, it's $ x(t), y(t)$ may not be unique. For instance, $ x^2 \plus{} y^2 \equal{} 1$ can be rewritten as $ y \equal{} \sin t, x \equal{} \cos t$, or $ x \equal{} \cos t, y \equal{} \sin t$.

Because of that non-uniqueness, it can't be an "iff."

For instance, consider $ x\equal{}t^3,y\equal{}t^3.$ This is just $ y\equal{}x$ and is we certainly have $ \frac{dy}{dx}$ existing everywhere, even though the parameterization suggests that $ (0,0)$ could be a singular point.

but how about the case that $ x\equal{}x(t)$ satisfy $ x'(t)\neq0$?

Then you should state the claim differently:

Suppose that $ x(t)$ is differentiable at $ t\equal{}t_0$ and $ x'(t_0)\ne 0$. The function $ y(x)$ is differentiable at $ x(t_0)$ iff $ y(t)$ is differentiable at $ t_0$.

This follows from the inverse function theorem and the chain rule.

Is the inverse function theorem a theorem in multivariable calculus? Can it be stated in single variable calculus (maybe with less rigor)?

It's usually stated in multivariable terms, but the single variable version can be found in calculus textbooks, and is easier to prove. This is from Stewart's Calculus:
[quote]If $ f$ is a one-to-one differentiable function with inverse function $ g \equal{} f^{ \minus{} 1}$ and $ f'(g(a))\ne 0$, then the inverse function is differentiable at $ a$ and $ g'(a) \equal{} \frac {1}{f'(g(a))}$[/quote]
This statement is preceded by informal discussion ("we can think of a differentiable function as one whose graph has no corner or kink to it" :lol: ) and followed by an actual proof. :o
The assumption that $ f$ is one-to-one can be omitted if $ f'$ is continuous at $ g(a)$, provided that the domain of $ f$ is reduced to a small interval around $ g(a)$.

Or, I think it can be proved by chain rule: Let $ f$ be an invertible differentiable function with inverse $ g \equal{} f^{ \minus{} 1}$ and $ f'(g(x))\neq0$, then $ f(g(x)) \equal{} x$. Hence $ [f(g(x))]' \equal{} (x)' \equal{} 1$. Moreover, by the chain rule, we have

$ [f(g(x))]' \equal{} f'(g(x))g'(x).$

Thus, $ f'(g(x))g'(x) \equal{} 1$, i.e. $ g'(x) \equal{} \frac1{f'(g(x))}.$

[quote="ifai"]by the chain rule, we have $ [f(g(x))]' \equal{} f'(g(x))g'(x).$[/quote]
No, you don't - until you checked that the hypotheses of the chain rule are satisfied. Among them is: $ g'(x)$ exists.

Oops! Thank you. But can we use a way other than the general method to prove the existence of g'?

You can proceed in any way you want, as long as it's a correct proof. :)

A typical proof of the existence of $ g'(a)$ goes like this: $ \frac{g(x)\minus{}g(a)}{x\minus{}a}$ can be written as $ \frac{g(x)\minus{}g(a)}{f(g(x))\minus{}f(g(a))}$, which is the reciprocal of $ \frac{f(g(x))\minus{}f(g(a))}{g(x)\minus{}g(a)}$, which converges to $ f'(g(a))$ as $ x\to a$.

In any case, both here and for implicitly defined functions, the actual calculation of the derivative is a routine application of the chain rule. Existence of an appropriate differentiable function is a nontrivial theorem, and that theorem is best handled with multivariable techniques.

I beg to disagree. The single-variable inverse function theorem is actually easy: if $ f'(x_0)\ne 0$ and $ f'$ is continuous at $ x_0$, then (by the MVT) $ f$ is strictly monotone in a neighborhood of $ x_0$, which implies (with the IVT) that is has an inverse. The differentiability of the inverse can be proved as in my previous post.

Things get complicated in the multivariable case, where we don't have monotonicity (well we kind of do), and the derivative is no longer the limit of finite quotients.


inequalities
analytic geometry
algebra
linear equation

I have to solve problems like the following, and show all the work...Only problem is that this is an online class, and i totally forgot how to do this nonsense...

I would appericate if someone could do a few of these for me, while showing the work. So that this way ill kinda have an idea...

Thanx Fellas.....

1. Solve for x. Collect xs on the LEFT side. The solution should be presented with
x first, the correct inequality symbol, then the number.
3(4x – 5) < 9

2. Solve for x. Collect xs on the LEFT side. The solution should be presented with x first, the correct inequality symbol, then the number.
- 7x + 2 + x – 10 > -3(x – 1)

3. Name the coordinate described here:
From the origin, the point is four places to the left and six places down.

4. Find the midpoint between these points:
(1, -6) and (-5, -2)

5. The linear equation x = -4 graphs as a horizontal/vertical/diagonal line (choose the correct label and type it below).

[quote="theflash0216"]1. Solve for x. Collect xs on the LEFT side. The solution should be presented with
x first, the correct inequality symbol, then the number.
3(4x – 5) < 9
[/quote]

Instead of the inequality sign, imagine as an equal sign, so

$ 3(4x-5) = 9$

$ 12x-15=9$

$ 12x=24$

$ x=2$

Now remember to add your inquality sign so the answer is

$ x<2$

[quote="theflash0216"]2. Solve for x. Collect xs on the LEFT side. The solution should be presented with x first, the correct inequality symbol, then the number.
- 7x + 2 + x – 10 > -3(x – 1)
[/quote]

$ -7x+2+x-10 >-3(x-1)$

Simplifying gives

$ -6x-8 >-3x+3$

$ -3x > 11$

$ x <-\frac{11}{3}$

[quote="theflash0216"]3. Name the coordinate described here:
From the origin, the point is four places to the left and six places down.
[/quote]

The origin is located at $ (0,0)$ and so the coordinates will be $ (-4,-6)$.

[quote="theflash0216"]4. Find the midpoint between these points:
(1, -6) and (-5, -2)
[/quote]

Midpoint formula- $ \left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Plugging in, we get

$ \left(\frac{1-5}{2}, \frac{-6-2}{2}\right)$

$ (-2,-4)$

[quote="theflash0216"]5. The linear equation x = -4 graphs as a horizontal/vertical/diagonal line (choose the correct label and type it below).[/quote]

It is a vertical line.

I hope all those were right... :D


So, everybody, what are your favorite IRC channels?

I spend a little time on #tcpa now and then.

I'd use it for the USACO channels, but I can never find out how to actually get it running.  I tried using it from the internet, but that doesn't work.  And gaim doesn't work for me either.  So i guess it's just IRC-free for me  :(   Hey, that rhymes, IRC-Free


Remainder Theorem stuffs:
What is the remainder when $ x^{10}$ is divided by $ x^2\minus{}a$ ?

Denote $ y \equal{} x^2$. By Remainder Theorem, there exist polynomials $ Q$ and $ R$, with $ \deg R < 1$, such that

$ y^5 \equal{} Q(y)(y\minus{}a) \plus{} R(y)$. For $ y\equal{}a$ this yields $ R(a) \equal{} a^5$, so $ R$ is the constant polynomial equal to $ a^5$.

Please do not double-post the same problem in multiple forums. If it is urgent, AoPS is not the right place for it.

[url]http://www.mathlinks.ro/viewtopic.php?t=304749[/url]

[quote="greenhand"]$ \plus{} a^6$[/quote]This term is wrong. It's $ a^5$.


Is the sequence x_n=sin(n!) divergent? I am absolutely sure, but I don't know how to prove. What about the sequence x_n=sin(2004^n)?

[quote="harazi"]Is the sequence x_n=sin(n!) divergent?[/quote]

I wonder if the Stirling's formula will help us here.. :)

yes i think it works...i must write the problem nicely however in order to give a coherent solution :D


[i]Σε ένα ξένο  μαθηματικό περιοδικό βρήκα την παρακάτω άσκηση.Αναφέρεται ως 2η ολυμπιάδα ZAUTIKOV (KAZAKHSTAN 2006). Νομίζω ότι έχει ενδιαφέρον και αξίζει να την έχουμε στο ελληνικό φόρουμ.[/i]

Δίνεται τρίγωνο $ ABC$ και τα σημεία $ K,L$ των πλευρών $ AB,AC$, ώστε  $ BK\equal{}CL$ . Οι $ BL, CK$ τέμνονται στο $ P$. Η παράλληλη από το $ P$  προς τη διχοτόμο $ AD$   του τριγώνου $ ABC$   τέμνει την $ AC$  στο $ M$. Να αποδειχθεί ότι $ CM\equal{}AB$

 Μπάμπης

Αρχικά είχα κάνει μια πατατιά . Μετά πήγα είδα τον κωλοαγώνα πίκρα και από εκεί και είπα να πιάσω να λύσω καμιά geom
Ευτυχώς βγήκε κάτι πολύ όμορφο . :D

Έστω $ R,P$ τα μέσα των $ BC,\ KL$ και $ N$ το μέσο του $ AP$ . Από την ευθεία Gauss έχουμε ότι $ R,P,N$ συνευθειακά και
εύκολα ή με παράλληλη μεταφορά ή με δύο μενελάους μπορούμε να δείξουμε ότι $ AT\equal{}AE$ επομένως RE//AD//PM και επειδή
$ N$ το μέσον του $ AP$ θα είναι $ E$ το μέσον του $ AS$ άρα $ RD\equal{}RQ$ άρα $ BD\equal{}QC$ .Τώρα τα τρίγωνα $ ABD,\ MQC$
έχουν μια γωνία ίση και μια παραπληρωματική επομένως $ \frac{AB}{BD}\equal{}\frac{MC}{QC}$  οπότε $ AB\equal{}MC$ and QED  :)

einai oraia klasiki askisi. uparxei mia omada askiseon pou exoun auto to sxhma toulaxiston etsi ta exo taktopoihsei ego sto mualo mou. gia paradeigma entelos omoia borei kaneis na apodeiksei oti to symmetriko tou p os pros to meso ths bc anhkei sth dixotomo ad. (btw auth einai mia apo tis askiseis pou exo panta sto mualo mou-dld kai plysh egefalou na mou kanoune, auth tha meinei mazi me merikes alles)

[quote="silouan"]Όντως Nick αυτό το σχήμα έχει πάρα πολλές ιδιότητες . ένα ωραίο πολύ εύκολο όμως συμπέρασμα είναι ότι η AD περνά από το δεύτερο κοινό σημείο των περιφερειών των τριγώνων $ PKB,\ PLC$  :)[/quote]

 [color=red]Πολύ ωραία ! [/color]

 Σιλουανέ - Νίκο, μάζέψτε την καλά με τις παρατηρήσεις σας και τις αποδείξεις , θα στείλλω και γω μια μετρική λύση(αυτή του περιοδικού) με ένα Μενέλαο και λίγη βασική ομοιότητα , οπότε την πάμε καρφωτή για το βιβλίο.

 Μπάμπης


When I tried to use \fcolorbox{black}{gray}{long text}, it doesn't appear correctly, I mean exceeds the PDF width.

See:

[img]http://img237.imageshack.us/img237/7204/asdfasdwa1.png[/img]

Is there a way to fix this so that full text fits in the box?

Either
1) reduce the size of the text so that it fits;
or
2) reduce the margins (using the geometry package);
or
3) put the text on more than one line using \\ inside a parbox
\fcolorbox{black}{gray}{\parbox{[i]width[/i]}{long text \\ new line}}
adjusting [i]width[/i] to whatever is suitable.


αν a,b,c,d,x,y πραγματικοί αριθμοί να δείξετε ότι

[img]http://img401.imageshack.us/img401/4558/newwwwuz9.png[/img]

Οι πρώτες δύο με Cauchy ή διανύσματα (τις έχει και το σχολικό βιβλίο) .Η τελευταία Μinkowski ή διανύσμτα είναι απόρια μια πάλι του σχολικού βιβλίου . :)

Ωράια siloune

Ακόμη μία τότε :

[img]http://img365.imageshack.us/img365/9938/rrrrrrrrrrrwj0.png[/img]

Kalispera kai xronia polla Gianni,

Tha ithela na se rwtisw ean oles oi rizes stin askisi pou edwses einai $2n$-ostes. Mou fainetai parakseno apo tin apopsi oti ean ipswsw eis tin $2n$ feugoun oles oi rizes... Mipws kapoies rizes einai $n$-ostes? Mporeis na to elenkseis se parakalw?

Alexandros

Ναι σωστά χωρίς ν είναι η άσκηση

giannis ise siguros..?an oles i rizes itan xoris n diladi tetragonikes... afu ta a,b,c ine pleyres trigonu..bgeni (me xrisi tu irona +tu 1ou diameson..) oti $m_{a}\frac{\sqrt{a}}{2}\geq E$ to opioi den isxii panta..(p.x nomizo gia a=3,b=4,c=5 vgazi atopo..)
eno an itan oles 2n ostes ektos apo ekini pu ine mono tu to a na itan n osti..
tote 8a evgene $m_{a}\frac{a}{2}\geq E$ pu isxii panta afu $m_{a}\geq h_{a}$...

 :D

Akrivws RAKAR... To mono pou apomenei twra einai i orthi diatipwsi tis askisis wste apla na dwthei i lisi h opoia profanws tha dwthei me thewrima diameswn kai ton tipo tou Hrwna gia ton Emvadon ('h gia to ipsos isws).

Ean exeis tin ekfwnisi Gianni kane tis diorthwseis pou xreiazontai gia na mi xreiastei na dokimasoume oles tis dinates epiloges tou 2n kai tou n.

Alexandros

Ναι σωστά , η βασική ιδέα της άσκησης ήταν αυτή..


\[ \begin{array}{l}
 In\_{\rm{a}}ny\_\Delta ABC,show\_th{\rm{a}}t \\
  \minus{} 2 \le \sin 3A \plus{} \sin 3B \plus{} \sin 3C \le \frac{{3\sqrt 3 }}{2} \\
 \end{array}
\]

[quote="anorithdialga"]
\[ \begin{array}{l} In\_{\rm{a}}ny\_\Delta ABC,show\_th{\rm{a}}t \\
- 2 \le \sin 3A + \sin 3B + \sin 3C \le \frac {{3\sqrt 3 }}{2} \\
\end{array}
\]
[/quote]
[hide="right side"]
$ \sin3A + \sin3B + \sin3C = \sin3A + \sin3B + \sin3(A + B) =$
$ = (1 + \cos3A)\sin3B + \sin3A\cos3B + \sin3A\leq$
$ \leq\sqrt {(1 + \cos3A)^2 + \sin^23A} + \sin3A =$
$ = 2|\cos{\frac {3A}{2}| + \sin3A\leq \frac {{3\sqrt 3 }}{2} ,}$ which is easy to check.
For example, if $ \sin3A=\frac{\sqrt3}{2}x$ then
$ 2|\cos{\frac {3A}{2}| + \sin3A\leq \frac {{3\sqrt 3 }}{2}\Leftrightarrow9(x-1)^2(x^2-10x+33)\geq0.}$  :wink: [/hide]

[quote="apratimdefermat"]what about the left side :?:[/quote]
[hide="obvious"]
Let $ A \equal{} \min\{A,B,C\}.$ Hence, $ A\leq60^{\circ}.$
Thus, $ \sin3A\geq0.$
Id est, $ \sin3A \plus{} 1 \plus{} \sin3B \plus{} 1 \plus{} \sin3C\geq0.$  :wink:
[/hide]

[b] Ji Chen:[/b]
$ \sum sin3A\le \sqrt{3 \sum sin^2 3A}$
$=\frac{1}{2}\sqrt{27-3(1+cos6B+cos6C)^2-3(sin 6B-sin6C)^2}\le\frac{3\sqrt{3}}{2},$
Equality holds for $\{A,B,C\}=\{\frac{7\pi}{9},\frac{\pi}{9},\frac{\pi}{9}\}.$
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=544999&hilit=sin3A+sin3B+sin3C]here[/url]

 http://www.artofproblemsolving.com/community/c6h1130425p5248480:
\[\frac{3}{2}\geq \cos{3A}+\cos{3B}+\cos{3C} \geq  -3\]

In $ \triangle ABC  $  . Prove that
$$cos3A+2cos 2B+3cos C>-\frac{25}{16}$$
$$cos3A+3cos B+3cos C>1$$

In $ \triangle ABC  $  . Prove that
$$6>cos3A+2cos 2B+3cos C>-\frac{25}{16}$$
$$7>cos3A+3cos B+3cos C>1$$
$$3>cosA+cos2 B+cos C>-\frac{1}{8}$$

In acute triangle $ABC.$ Prove that
$$(sin^2A+sin^2B+sin^2C)(cos^2A+cos^2B+cos^2C)\geq \frac{27}{16}$$

[quote=sqing]In $ \triangle ABC  $  . Prove that
$$6>cos3A+2cos 2B+3cos C>-\frac{25}{16}$$[/quote]
$$A=\pi -arccos\frac{3}{8},B=arccos\frac{3}{8}, C=0$$


function
logarithms
calculus
derivative
real analysis
real analysis solved

Hi, I'm having trouble solving the following 2 problems, could someone help me out? Thanks  :)

1.) Let L be a differentiable function on (0, infinity) such that L'(x)=1/x and L(1)=0. Prove that for any two positive numbers a and b, L(ab)=L(a)+L(b). The hint given: to show that for each x>0, L(ax) and L(x) have the same derivative.

2.) Prove that cosx<1-(1/2)x^2+1/24x^4 for all x in (0, infinity).
    The hint was to prove that x-(1/6)x^3<sinx for all x in (0, infinity).

For #1, I think the game is to pretend that we never heard of the logarithm function.

For fixed $y,$ define $f(x)=L(xy).$ Differentiate. By the chain rule,

$f'(x)=yL'(xy)=\frac y{xy}=\frac1x.$

Since $f(x)$ has the same derivative as $L(x)$ (for $x>0$), we have:

$f(x)=L(x)+C.$

Now, how can we find $C$? Try setting $x=1: \,L(y)=f(1)=L(1)+C=C$ so $C=L(y).$

Put it together and we have $L(xy)=L(x)+L(y).$

For #2, can you prove the following results? (Hint: Mean Value Theorem.)

If $f'(x)< g'(x)$ for $x>0$ and $f(0)=g(0),$ then $f(x)<g(x)$ for $x>0.$

If $f'(x)< g'(x)$ for $x<0$ and $f(0)=g(0),$ then $f(x)>g(x)$ for $x>0.$

Then iterate this theorem on top of itself.


how do i graph y=sec(x)tan(x)? i think i we need some calc knowledge. like derivatives...... not sure if this is inthe right forum...

Find the function at known values, such $ 0, \pi/6, \pi/4, \pi/3$, etc. Some of these points will be undefined and have a vertical asymptote; try to determine where the limit is from each side of the asymptote. Do this for as many rational multiples of $ \pi$ as you can in $ [0,2\pi)$, and you'll get a pretty good picture of what the graph is like.

If it needs to involve some kind of derivate $ \sec x \tan x$ is the derivate of $ \sec x$.  From the $ \sec x$ graph, graph the slope. Not in anyway simple.

thanks i get it now
we just graph secx and then just draw a lot of tangent lines then connect them together..


algorithm
floor function
symmetry
quadratics
logarithms
continued fraction
Diophantine equation

Lagrange proved long ago that if $ N$ is a squarefree integer, then it has a [b]periodic[/b] continued fraction expansion of the form
\[ [a_0;\overline{a_1,a_2,a_3,...,a_2,a_1,2a_0}].
\]
Now I need an algorithm (Maple or other program) to compute $ a_0,a_1,...,a_n$, as well as the period of repetition.

How to find the exact period? I guess we must end in a situation where back at $ N \minus{} \lfloor\sqrt {N}\rfloor$, but I don't see how to check this efficiently. Also to compute the actual coefficients, I assume there are faster ways than just recursion? I found some documents on the internet but they weren't really helpful (one featured this as an exercice, but without solution).

Any help is appreciated!

is [url=http://www.research.att.com/~njas/sequences/A003285]this[/url] or [url=http://www.u.arizona.edu/~miller/webreport/node8.html]this[/url] useful ? :maybe:

How do you mean exactly? In neither reference I can find an algorithm to determine the period?
The first 5000 periods and the fact that it is periodic do not help in finding the period?

There is a reasonably fast algorithm available to calculate the coefficients: see http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion .

I don't know of a fast way to find the period of a continued fraction without explicitly calculating its coefficients.

Nice. I came up with that same algorithm myself, and didn't know how well-known it was. The integers involved are no larger than $ 2\sqrt{S}$ in magnitude, and there's a lot more symmetry than is obvious. Most periods are pretty short, so this is an efficient algorithm.
A slightly less obfuscated version: we can define $ d_{n\plus{}1}\equal{}\minus{}d_nq_n^2\plus{}2m_nq_n\plus{}d_{n\minus{}1}$ instead; the discriminant of this quadratic is $ 4(m_n^2\plus{}d_nd_{n\minus{}1})\equal{}4S$, and the quotients $ a_n$ come from approximating the zeros- they're the last integers before we hit the wrong side. This version of the iteration also doesn't directly involve $ S$ except as an initial condition $ d_{\minus{}1}\equal{}S$.
If the $ n$th convergent $ a_0\plus{}\frac1{a_1\plus{}\frac1{\cdots\plus{}\frac1{a_{n\minus{}1}}}}$ is $ \frac{r_n}{s_n}$, $ d_n\equal{}(\minus{}1)^n(r_n^2\minus{}Ds_n^2)$, and $ m_n\equal{}(\minus{}1)^{n\minus{}1}(r_nr_{n\minus{}1}\minus{}Ds_ns_{n\minus{}1})$. Calculating the exponentially growing $ r_n$ and $ s_n$ is of course expensive; you need to work with very large integers.

You know you've reached a period when you see a quotient of $ 2\lfloor\sqrt{S}\rfloor$ or a $ d_n\equal{}1$; you don't need to watch the whole triple.

An example with a long period, so you can see some of the patterns:
($ S\equal{}61$)
$ \begin{array}{cccc}n&a&d&m\\0&7&1&0\\1&1&12&7\\2&4&3&5\\3&3&4&7 \\4&1&9&5\\5&2&5&4\\6&2&5&6\\7&1&9&4\\8&3&4&5\\9&4&3&7\\10&1&12&5 \\11&14&1&7\end{array}$
That only took a few minutes to do by hand, and it would be immediate in a spreadsheet.

Since that one has an odd period length, you have to double the period to get a nontrivial solution to the Pell equation $ x^2\minus{}61y^2\equal{}1$. Incidentally, [i]all[/i] positive integer solutions to $ x^2\minus{}Sy^2\equal{}k$ with $ k<\sqrt{N}$ are generated by the continued fraction.

[Notation copies wiki version; I used different letters for everything)

Ok, that does it for small numbers. But when the numbers get big, this seems to get in an infinite loop. Which is of course impossible, so I assume the problem is machine/software-based.

I did this in Maple. For $ N \equal{} 2^{2^5} \plus{} 1$, it works perfect to compute $ \sqrt {2N},\sqrt {3N},...$. But for $ N \equal{} 2^{2^6} \plus{} 1$, the computation of the continued fraction of $ \sqrt {2N}$ seems neverending (computed for hours, $ d \equal{} 1$ never triggered (and I computed over 500 000 terms), versus some seconds on the previous one). Same for $ \sqrt {3N}$.

Here's my code. Can someone tell me what I do wrong? (or provide me a working alternative)

[code]restart: k:=2: N:=2^(2^6)+1; Digits:=20; m:=0; d:=1; a(0):=floor(sqrt(k*N));

i:=0:
do
  i:=i+1:
  m:=d*a(i-1)-m:
  d:=(k*N-m^2)/d:
  a(i):=floor((sqrt(k*N)+m)/d):
  if d=1 then break; end if;
end do:[/code]

Ran up to 20 000 000 terms in the continued fraction tonight, still no period...

Something is wrong here. :P

Maple handles arbitrary-precision integers, right?  It's possible that the continued fraction's period is simply too long.  There are asymptotics available; check what the expected length is.

I'm pretty sure that the expected length for $ \sqrt{N}$ is of order $ \sqrt{N}$. For multiples of $ 2^{64}\plus{}1$, you should expect the periods to be close to $ 10^9$ in length, and you've only run it out to $ 10^7$.

Some empirical data: I ran this for fifteen random integers between $ 10^6$ and $ 2\cdot 10^6$. The periods ranged from $ 32$ to $ 738$, averaging $ 335$. If I restrict to the nine squarefree integers, that's $ 48$ to $ 714$, averaging $ 323$.
All fifteen periods have even length; odd periods are rare.

According to http://mathworld.wolfram.com/PeriodicContinuedFraction.html , an upper bound for the period is $ O(\ln D \sqrt{D})$.  See also http://projecteuclid.org/handle/euclid.pjm/1102811631 and http://www.math.princeton.edu/mathlab/jr02fall/Periodicity/mariusjp.pdf .

Hmm... it seems my teacher is a little bit sadistic then, since for application purposes I don't need $ \mathcal O(10^9)$ terms at all, only a few thousand, and the application works perfectly when not enforcing to find the period first. :mad:

Ah well, I guess I'll just ignore this part then. Thanks for your help!

(By the way, I knew that bound but I thought it was a high overestimate... for exponent $ 5$ the period was around $ 3\ 000$ only, so I didn't expect it to be over $ 20\ 000\ 000$ for exponen $ 6$... but since the application works, I assume the computed terms arre correct indeed.)


trigonometry
complex numbers
calculus
calculus computations

a. $ a_n\equal{}\frac{\sin{n}}{n}$

b. $ a_n\equal{}\frac{\cos{\pi}n}{n^2}$

If it converges, find its limit.

Both series converge.  Just consider series $ \frac{1}{n}$ and $ \frac{1}{n^2}$, which clearly converge to $ 0$.  Clearly, in the first case $ |a_n|\leq\frac{1}{n}$ and in the second case $ |a_n|\leq\frac{1}{n^2}$.  Hence both $ \{a_n\}$ series converge to $ 0$.

sin n/n is a converging sereis. This can be proved as folows.

you can find sin 1 +sin 2 +sin 3 +......................+sin n using euler's formula and sum of a Geometric progression. This sum turns out to be bounded by (2-sin 1)/(1-cos 1).(May be, someone can get a better estimate.)

Now, the secquence, 1/n is monotonically decreasing and, approaches to zero.

Hence, by Dirichlet's test, the sum sin n /n converges.

[quote="bharathhm"]you can find sin 1 +sin 2 +sin 3 +......................+sin n using euler's formula and sum of a Geometric progression. This sum turns out to be bounded by (2-sin 1)/(1-cos 1).[/quote]
Can you tell how did you find the bound?

the sum sin 1 + sin 2..............+sin n is the imaginary part of the sum e^(0) + e^(i) + e^(2i)..............+
e^(ni) This is a geometric sereis and can be summed up. You get an exression  {1-e^i(n+1)}/{1-e^1}

Subtract the conjugate of the above expression from itself. you get the imaginary part of it.

The sum turns out to be

  1/2 {sin (n+1)-sin 1 -sin n }/{1-cos 1}  (I think I forgot the factor 1/2 in my previous post)

Now, mod( sin(n+1) - sin n)  <= mod(sin (n+1)) + mod(sin n) <= 2. This is rather, a rough estimate, however, we dont need any better for our purpose.

use sandwich principle to get limit for both as 0


a)prove that if $ a$ ,$ b$, $ c$ are the sides of a triangle such that :
$ 2(ab^2\plus{}bc^2\plus{}ca^2)\equal{}a^2b\plus{}b^2c\plus{}c^2a\plus{}3abc$
then it is equilateral triangle.
b)prove that this equality may satisfated by reals which aren't the sides of  a triangle

i think it is more algebric than geometric


Laura and Pam decide to meet at the library. They both leave the house on their bikes at 8 a.m. Laura pedals 5phm faster that Pam and waits 12 minutes for her friend who pedals at a rate of 10mph. After spending 2 hours at the library they return home pedaling at the same rate they did in the morning. Laura arrives home at 11:48 a.m. What is the distance between Pam's house and the library?

I dont think you need the distance formula

[hide]Answer:  10 miles

Solution:

The first thing that i did was find out how far Laura lives from the library.  If she spent $ 2$ hours and $ 12$ minutes at the library (off of her bike) then that means that she spent $ 96$ minutes on her bike to travel to the library and back.  So if we divide $ 96$ by $ 2$ we get one trip to the library traveling at $ 15$ mph took her $ 48$ minutes.  Plugging these values into $ D \equal{} RT$ gives us $ D \equal{} 12$.  But this doesnt actually matter because we want to know how far Pam lives from the library.  Whe know that while Laura was biking for $ 48$ minutes and waiting for $ 12$ minutes at the library Pam was riding her bike.  Which means that Pam's time was $ 1$ hour and the rate was $ 10$ mph so plugging this into $ D \equal{} RT$ gives us that Pam lives $ 10$ miles away from the library.[/hide]


calculus
function
trigonometry
derivative
algebra
binomial theorem

This is my first topic and my first time using LATEX. I hope I did everything right. I'm not at all sure if this belongs in "High School Basics," but I'm in high school, and I'm basic, so...

Today I was teaching someone how to solve:

$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}$

When I started wondering how one would solve something like this:

$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\sqrt[5]{2+...}}}}$

I'm not sure whether it requires calculus, or whether it is even possible, but if you have a (preferably non-calculus) solution, that would be helpful.

I tried it on my calculator and it seemed to diverge at about [hide]1.869597307
This number seems familiar, but I don't know from where.[/hide]
I would appreciate any help. Thanks!

That number shouldn't be familiar.
I believe the Answer will basically just be
${2^\frac{1}[2}+2^\frac{1}{6}+2^\frac{1}{8}$ and so on.

Maybe I messed up on something but to me, they don't look particularly special. I mean if it was something like the first one it would give a close value at at least when the radical is 10 but nothing really came up.

$1.869597307$ is not special, nor is $1.87$.. (according to Google)
maybe i'm wrong.

[quote="8parks11"]That number shouldn't be familiar.
I believe the Answer will basically just be
${2^\frac{1}[2}+2^\frac{1}{6}+2^\frac{1}{8}$ and so on. [/quote]
You can't distribute a exponents / radicals over addition / subtraction. Further more, all of the radicals cover all of the subsequent terms, not only the first one.

I don't think that it's extremely special, just maybe something I've typed in to my calculator before.

Maybe e-1 could've been it even thought it is 1.71...

Most "arbitrary problems" don't have nice solutions.  I doubt this is related to any familiar constants.

It's related to a certain value of a certain type of hypergoemetric function :wink: I solved it with calculus, it's not too difficult actually (but WAY beyond this forum level!!!!)
PS t0r: I think you could do it also[/img]

can you please show and solve it for me?

I am borrowing this solution from the Algebra Trig book by Paul Foerester.
Let x = the infinite radical in question. Then you can see that, since the radical goes on infinitely, it contains itself. This means it reduces to x=rad(2+x). From there on out its easy to solve.

That's true for the first one, but not for the second, which is the one I am wondering about.

Can some super-calculus person give me an just an exact answer, even if the method is too painful for my eyes (I do have very basic experience with calculus).

[quote="me@home"][quote="t0rajir0u"][quote="me@home"]it's not too difficult actually[/quote]

 :huh:

I certainly can't think of any neat way to attack this.  I hope you're not talking about [hide="noodles"] the rational binomial theorem [/hide] or anything like that.[/quote]
No not at all.
You end up solving a differntial equation which is in fact an approximation of the function. I won't give the specifics, but there are many ways to do this actually.

First replace 2 by x and define a sequence of functions $\S_{n}$ (yes that's the variable I used in my type up) defined creatively (I'll let you find that...!) then there is a way of expressing the recursion in terms of x in an elementary method. Then, compare the derivatives. Take the derivative of an $\S_{n}$ and nice things drop out and you get a rather nice function in terms of $\S_{n}$ and $S_{n+1}'$ which is a recursion that can be used to find the Taylor series. I didn't do it that way... I approximated $\S_{n+1}$ by $\S_{n}$ and solving the resulting differential equation in Mathematica Integrator which has n nice closed form but it is a rather simple hypergeometric function, whereupon values can be plugged in.

Thats about it... :D[/quote]
If you guys want to do calculus don't you think this should be moved?

is this a known equation? I couldn't find anything at all.

[quote="me@home"]First replace 2 by x and define a sequence of functions $\S_{n}$ (yes that's the variable I used in my type up) defined creatively (I'll let you find that...!) then there is a way of expressing the recursion in terms of x in an elementary method. Then, compare the derivatives. Take the derivative of an $\S_{n}$ and nice things drop out and you get a rather nice function in terms of $\S_{n}$ and $S_{n+1}'$ which is a recursion that can be used to find the Taylor series. I didn't do it that way... I approximated $\S_{n+1}$ by $\S_{n}$ and solving the resulting differential equation in Mathematica Integrator which has n nice closed form but it is a rather simple hypergeometric function, whereupon values can be plugged in.[/quote]

Eugh, that was something like the other way I was considering...   :|

Its actually not bad... :lol:
About the calculus, well, that's why I summarized my solution.
And yes, this problem should be moved

Can I see the specifics of the entire solution, either t0r or me@home? If you could pm me with more details, that'd be cool.

can someone please post a solution to this?
i'm really curious :S

bump :S
trying but I don't get much
someone please clear this up1

[b]me@home[/b], I'm not precisely sure what definition of $\S_{n}(x)$ you're using, but neither of the ones I've tried pan out nicely.  One of them doesn't have a nice recursion, and your assumption isn't valid in the other one.

I don't seem to get anything nice..
please clear it for me :(


[quote="GameBot"]A shake from Jake's Shakes uses 2 tsp vanilla, 2 scoops of ice cream, 4 oz fresh banana and $ \frac 12$ cup milk.  A banana (without the peel) weighs 6 oz on average.  How many whole bananas must Jake order so that he can make 500 shakes?[/quote]

To make 500 shakes, he needs $ 2000$ oz of fresh banana.

Thus, $ \frac{2000}{6} \approx 334$.


Square $ABCD$ has side length 2. A semicircle with diameter $AB$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $AD$ at $E$. What is the length of $CE$?

[asy]defaultpen(linewidth(0.8));
pair A=origin, B=(1,0), C=(1,1), D=(0,1), X=tangent(C, (0.5,0), 0.5, 1), F=C+2*dir(C--X), E=intersectionpoint(C--F, A--D);
draw(C--D--A--B--C--E);
draw(Arc((0.5,0), 0.5, 0, 180));
pair point=(0.5,0.5);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));[/asy]

$ \textbf{(A)}\frac{2+\sqrt5}2\qquad
\textbf{(B)}\sqrt5\qquad
\textbf{(C)}\sqrt6\qquad
\textbf{(D)}\frac52\qquad
\textbf{(E)}5-\sqrt5 $

[hide] Say where CE hits the semicircle is X.  Say that AE=x.  This means that EX=x since EX and AE are tangents from a common point.  Also, BC=2=CX for the same reason.  Therefore, CE=2+x.  We also have that DE=2-x.  Finally, CD=2.  By the Pythogorean Theorem, we have that $2^{2}+(2-x)^{2}=(2+x)^{2}$.  Simplifying, we get $4+4-4x+x^{2}=x^{2}+4x+4$.  This means that 8x=4, so x=1/2.  This implies that CE=2+x=5/2.  Therefore, the answer is $D$. [/hide]

[hide]
[img]7077[/img]
Clearly, $EA=EF=BG$
Thus, the sides of right triangle $DEC$ are in arithmetic progression. Thus is is similar to the triangle $3-4-5$ and since $DC=2$ $CE=5/2$
[/hide]

[quote="#H34N1"][img]http://i96.photobucket.com/albums/l167/adasarathy/P22.gif[/img][/quote]

[hide="solution"]Let $AE=x$.  Since $BC=2$, $CE=2+x$.

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

$4+4x+x^{2}=4-4x+x^{2}+4$

$x=\frac{1}{2}$

Therefore, $CE=\frac{5}{2}$[/hide]

Call the point where CE intersects the circle F. Then x=AE=EF and 2=CB=CF, so EC=EF+FC=2+x.

[hide=dk why i did trigg] call $\angle{CED}=\theta$ ,  and point of contact of radius to $CE$ as $O$ we get $\angle{OCB}=\theta$ , then we get $\tan{\frac{\theta}{2}}=\frac{1}{2}$\\
so we get $\tan{\theta}=\frac{4}{3}$\\
so we get $EC=\sqrt{\frac{9}{4}+4}=\boxed{\frac{5}{2}}$[/hide]


If

$ \frac{\sin^4x}a \plus{} \frac{\cos^4x}b \equal{} \frac1{a\plus{}b}$

show that

$ \frac{\sin^6x}{a^2} \plus{} \frac{\cos^6x}{b^2} \equal{} \frac1{(a\plus{}b)^2}$

WLOG , $ a$ is positive. Also, it's easy to see that $ sin^4 x$ and $ cos^4 x$ cannot be zero.

Case 1:$ b$ is negative , then
$ max\{|\frac{1}{b}|, |\frac{1}{a}|\} \ge |\frac{sin^4x}{a} \plus{} \frac{cos^4x}{b}| \equal{}|\frac{1}{a\plus{}b}|>max\{|\frac{1}{b}|, |\frac{1}{a}|\}$, impossible.

Case 2:$ b$ is positive , then by Cauchy Schwarz inequality,
$ (a\plus{}b)(\frac{sin^4x}{a} \plus{} \frac{\cos^4x}{b}) \ge (sin^2\plus{}cos^2)^2\equal{}1$

As equality holds , $ \frac{sin^4 x}{a^2}\equal{}\frac{cos^4 x}{b^2}\equal{}k$, where $ k$ is a constant.
So $ \frac{\sin^4x}a \plus{} \frac{\cos^4x}b \equal{} \frac{1}{a\plus{}b} \Leftrightarrow k(a\plus{}b)\equal{}\frac{1}{a\plus{}b}$,
$ \frac{\sin^6x}{a^2} \plus{} \frac{\cos^6x}{b^2}\equal{}k(sin^2 x\plus{}cos^2 x)\equal{}\frac{1}{(a\plus{}b)^2}$

[b][u]Here is an well-known extension[/u].[/b] Let $ n\in\mathbb N$ , $ n\ge 2$ and $ ab\ne 0$ . Then $ \boxed {\frac {\sin^4x}a \plus{} \frac {\cos^4x}b \equal{} \frac {1}{a \plus{} b}}\ \implies$

$ \cos 2x \equal{}$ $ \frac {b \minus{} a}{b \plus{} a}\ \implies\ \frac {\sin^2x}{a} \equal{}$ $ \frac {\cos^2x}{b} \equal{}$ $ \frac {1}{a \plus{} b}\ \implies\ \boxed {\frac {\sin^{2n}x}{a^{n \minus{} 1}} \plus{} \frac {\cos^{2n}x}{b^{n \minus{} 1}} \equal{} \frac {1}{(a \plus{} b)^{n \minus{} 1}}}$ .

[b][u]Two similar exercises[/u].[/b]

$ 1\blacktriangleright$ $ a\ne b$ , $ n\in\mathbb N^*$ , $ \frac {\sin^2x}{a}\plus{}\frac {\cos^2x}{b}\equal{}\frac {2}{a\plus{}b}\ \implies\ \frac {\sin^{2n}x}{a^n}\plus{}\frac {\cos^{2n}x}{b^n}\equal{}\frac {2}{(a\plus{}b)^n}$ .

$ 1\blacktriangleright$ Solve the equation $ \sin^{2n}x\plus{}\cos^{2n}x\equal{}\frac {1}{2^{n\minus{}1}}$ , where $ n\in\mathbb N^*$ .

[quote="Virgil Nicula"][b][u]Two similar exercises[/u].[/b]

$ 1\blacktriangleright$ $ a\ne b$ , $ n\in\mathbb N^*$ , $ \frac {\sin^2x}{a} \plus{} \frac {\cos^2x}{b} \equal{} \frac {2}{a \plus{} b}\ \implies\ \frac {\sin^{2n}x}{a^n} \plus{} \frac {\cos^{2n}x}{b^n} \equal{} \frac {2}{(a \plus{} b)^n}$ .

$ 2\blacktriangleright$ Solve the equation $ \sin^{2n}x \plus{} \cos^{2n}x \equal{} \frac {1}{2^{n \minus{} 1}}$ , where $ n\in\mathbb N^*$ .[/quote]

[b]1)[/b] Similar to my solution above.

[b]2)[/b]For $ sin x, cos x \neq 0$, $ \frac {(sin^2 x)^n \plus{} (cos^2x)^n}{2} \ge \frac {(sin^2 x \plus{} cos^2 x)^n}{2^n}$
$ \Leftrightarrow \sin^{2n}x \plus{} \cos^{2n}x \ge \frac {1}{2^{n \minus{} 1}}$, equality holds iff $ sin^2 x \equal{} cos^2 x,$
So $ x \equal{} k\pi \plus{} \frac {\pi}{4},k\pi \plus{} \frac {3\pi}{4}$ are the solutions.

For one of the $ sin x$, $ cos x \equal{} 0$, we have for $ n \equal{} 1$, $ x \equal{} \frac {k\pi}{2}$ are solutions also.


geometry
3D geometry
inequalities unsolved
inequalities

prove that for all $ a,b,c>0$ we have:
\[ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)>2(a^3\plus{}b^3\plus{}c^3\plus{}abc)\]

[quote="knoppix"]prove that for all $ a,b,c > 0$ we have:
\[ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) > 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} abc)
\]
[/quote]
Are you sure?

[quote="knoppix"]prove that for all $ a,b,c > 0$ we have:
\[ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) > 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} abc)
\]
[/quote]
Try $ a,b \rightarrow 0$

sorry, it's false ... :blush:

you can delete this topic :oops:


let $ H$ represent the set of natural numbers whose prime factors are less than or equal to 3.prove that the sum of the reciprocals of all the elements of $ H$ is less than 3

Hmm..
I'm not sure if this will work but I'll give it a try,

If I understood correctly then, $ H = \{2,3,4,6,9,\cdots\}$

So the sum of the reciprocals is,

\begin{eqnarray*} S & = & \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{2^2} + \dfrac{1}{2\cdot 3} + \cdots \\
S & = & \left(1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots\right)\left(1 + \dfrac{1}{3} + \dfrac{1}{3^2} + \cdots\right)-1 \\
S & = & \left(\dfrac{1}{1 - \dfrac{1}{2}}\right)\left(\dfrac{1}{1 - \dfrac{1}{3}}\right)-1 \\
S & = & (2)\left(\dfrac{3}{2}\right)-1 \\
S & = & \boxed{2} \end{eqnarray*}

Well I seem to have proven that the sum equals $ 3$, did I go wrong anywhere?

Edit: Thanks anderson, it's changed now.

All prime factors of 1 are at most 3.  (Similarly, all prime factors of 1 are yellow pigs.)


algorithm
complex numbers
real analysis
real analysis unsolved

The general idea:  the sum of a conditionally convergent series can be partitioned into a sum of positive and negative terms, both of which diverge independently.  Hence, an algorithm to determine $ \pi$ (we will assume WLOG that $ x > 0$):

Add positive terms until the partial sums are greater than $ x$.  Then add negative terms until the partial sums are less than $ x$.  Rinse, repeat.

$ \text{(Riemann Rearrangement Theorem)}$ Let $ \sum a_n$ be a conditionally convergent series. Then for $ \minus{} \infty \leq c \leq b \leq \infty$, there exists a permutation $ \sigma$ such that

$ \liminf_{n} \sum_{k \equal{} 1}^{n} a_{\sigma(k)} \equal{} c, \quad \limsup_{n} \sum_{k \equal{} 1}^{n} a_{\sigma(k)} \equal{} b$

[b]Idea of the proof:[/b] Rearrange the given sequence so that the partial sum zigzags between desired supremum and infimum.

Let $ a_{n}^{ \plus{} }$ be the subsequence of $ a_n$ consisting of positive terms of $ a_n$, and define $ a_{n}^{ \minus{} }$ similarly.
Also, let $ b_n \equal{} b$ if $ b$ is finite, $ b_n \equal{} n$ if $ b \equal{} \infty$, $ b_n \equal{} \minus{} n$ if $ b \equal{} \minus{} \infty$ and define $ c_n$ in the same way.

Choose a permutation $ \sigma$ so that

$ \sum a_{\sigma(n)}$ $ \equal{} (a_{1}^{ \plus{} } \plus{} \cdots \plus{} a_{p_1}^{ \plus{} }) \plus{} (a_{1}^{ \minus{} } \plus{} \cdots \plus{} a_{q_1}^{ \minus{} }) \plus{} (a_{p_1 \plus{} 1}^{ \plus{} } \plus{} \cdots \plus{} a_{p_2}^{ \plus{} }) \plus{} (a_{q_1 \plus{} 1}^{ \minus{} } \plus{} \cdots \plus{} a_{q_2}^{ \minus{} }) \plus{} \cdots,$

where $ p_1 < p_2 < \cdots, q_1 < q_2 < \cdots$ are chosen so that if we let $ s(n) \equal{} \sum_{k \equal{} 1}^{n} a_{\sigma(k)}$, then

$ s(p_1 \minus{} 1) < b_1, \ s(p_1) \geq b_1$
$ s(p_1 \plus{} q_1 \minus{} 1) \geq c_1, \ s(p_1 \plus{} q_1) < c_1$
$ s(p_2 \plus{} q_1 \minus{} 1) < b_2, \ s(p_2 \plus{} q_1) \geq b_2$
$ s(p_2 \plus{} q_2 \minus{} 1) \geq c_2, \ s(p_2 \plus{} q_2) < c_2$

and so on. (This is always possible because $ \sum a_n^ \plus{} \equal{} \infty$ and $ \sum a_n^ \minus{} \equal{} \minus{} \infty$).

Then $ \sigma$ gives the desired result.

[quote="amcavoy"]Does anyone know of a nice proof of the following?:

Let $ \sum a_{n}$ be a conditionally convergent series in $ \mathbb{R}$.  Then for each $ x\in\mathbb{R}$ including $ x \equal{} \pm\infty$, there exists a permutation $ \pi$ such that
\[ \sum a_{\pi\left(n\right)} \equal{} x.
\]
[/quote]
Here is what [b]t0rajir0u[/b] says in more detail.
1)There have to be infinite many positive terms and negative terms of $ \{ a_n \}$ otherwise the convergence can't be conditional.
2)Let $ b_1,b_2,...$ be the positive terms and $ c_1,c_2,...$ be the negative terms.
3)Since the series is conditionally convergent it means $ \sum_{n \equal{} 1}^{\infty}b_1 \equal{} \infty$ and $ \sum_{n \equal{} 1}^{\infty} c_1 \equal{} \minus{} \infty$.
4)Suppose $ x \equal{} 1$ for simplicity.
5)Let $ K_1$ be so large that $ \sum_{n \equal{} 1}^{K_1} b_n > 1$ which is possible by #3.
4)Let $ M_1$ be so large so that $ \sum_{n \equal{} 1}^{K_1}b_n \plus{} \sum_{n \equal{} 1}^{M_1}c_n < 1$
5)Let $ K_2$ be so large that $ b_1 \plus{} ... \plus{} b_{K_1} \plus{} ... \plus{} b_{K_2} \plus{} c_1 \plus{} ... \plus{} c_{M_1} > 1$.
6)Repeat. The new rearranged sequence converges to $ 1$ (okay I didn't actually prove it but that is the algorithm that was described).

By the way, I think Riemann is the guy to acredit this nice theorem to. :)

EDIT: Ah, I got beat by [b]sos440[/b].

This problem is well known as the Riemann rearrangement theorem. Now we have a new problem, that if  \sum a_{n} is a conditionally convergent complex series, then the set of all complex numbers to which our rearranged series might converge is a straight line in a complex plane or this set coincide with all complex plane. Does anyone know how to prove it using theorem above?


algebra
polynomial
number theory proposed
number theory

Let $f(x)$ be the polynominal with the coefficient of integer.
Prove that if $f(n)$ is prime number for all positive integers $n$, then $f(x)$ is constant number.

Let $p = f(1)$ be a prime.
Then, for each $n$, we have $f(np+1) = f(1) \mod [p]$ thus $f(np) = p$. It follows that the polynomial $f(x)-p$ has an infinite number of roots, and therefore is identically zero.
Thus $f(x) = p$ for all $x$, and we are done.

Pierre.


How many of the 200 smallest natural numbers are congruent to $1\mod 9$?

[hide="My Solution which contradicts the solution manuals solution"]
I got 22, because it basically asks for all multiples of 9, which I named $a$, where $0\le a\le 199$. So, there are $\lfloor 22 \frac{1}{9}\rfloor$ multiples, which is $22$. However, the solution manual says $23$. Its problem $12.2.6$ on page 224 of Intro to Number Theory.[/hide]

[hide]The numbers that you are looking for are $1,10,19,...,199$ which is basically the same as $9,18,...,207$ if you add 8 to each term.

Then divide each term by 9 and youg et $1,2,3,...,23$ and obviously there are $\boxed{23}$ numbers.

You can't assume that there are the same number of $0\pmod9$ terms as $1\pmod 9$ terms. [/hide]

Well, it's not really asking for multiples of 9, because it's asking for numbers that are 1 more than multiples of 9.  But the two numbers are the same, since $200\equiv 2\bmod{9}$.

You counted 199, 190, 181, etc. down to 10. But when you divided by 9, you missed the $9\times 0+1=1$, which is your extra solution.


number theory
greatest common divisor
modular arithmetic
number theory unsolved

Find all positive primes $p$ such that $\frac{2^{p-1}-1}{p}$ is a perfect square.

$p = 2$ obviously doesn't work, so assume $p \neq 2$. we want $pk^2 = 2^{p-1}-1 = (2^{\frac{p-1}{2}} - 1)(2^{\frac{p-1}{2}} + 1)$. notice that these two factors are relatively prime. (they differ by 2, so their gcd is $\leq 2$, but they're both odd, so their gcd must be 1.) this means that one of the factors is of the form $pa^2$ and the other of the form $b^2$. we have two cases.

1. $2^{\frac{p-1}{2}} + 1 = b^2$. then $2^{\frac{p-1}{2}} = b^2 - 1 = (b-1)(b+1)$. then $b-1$ and $b+1$ must both be powers of 2, which means $b - 1 = 2$, $b+1 = 4$, $b = 3$, $p = 7$. and $p = 7$ does indeed work.

2. $2^{\frac{p-1}{2}} - 1 = b^2$. if $\frac{p-1}{2} \geq 2$, that is, if $p \geq 5$, then $b^2 \equiv -1 \pmod{4}$, which is impossible. so in this case we must have $p \leq 3$. an easy check shows that $p = 3$ works.


Hi,

Are there packages in Latex that cannot be used together? Like mathabx and amsmath cannot seem to used together. Is there a way to get around that?

Thanks. :)

[quote="limac"]Like mathabx and amsmath cannot seem to used together.[/quote]Who says they can't? This document compiles without a problem:
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{mathabx}
\begin{document}
This is an amsmath symbol $\dfrac{1}{2}$, an amssymb symbol $\mathfrak{F}$ and this is a mathabx symbol $\varEarth$
\end{document}[/code]
Please give a [b]complete[/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url].


geometry
algebra
function
domain
topology
Functional Analysis

How can I discuss an area of math that I am not familiar with, like formulate the statements clearly and well posed?

My friend keeps saying I am talking math vaguely and formulate the question imprecisely, when I am totally unaware of the nontriviality of my question (due to many unexpected assumptions and conditions needed in the statement).

I am feeling so bad wasting my friend's time figuring out the precise meaning of my question.

there's nothing to feel bad kid - it takes time to get used to formal mathematics.
if you keep at it, you'll develop the intuition eventually, and quicker than you expect.
but could you post an example here?

say this:
http://www.mathlinks.ro/Forum/viewtopic.php?t=193462

First, I didn't make clear the domain $ D$
Then I didn't make clear what is the invertibility condition of my operator $ T$
Then I didn't make clear on the measure of the space
and probably more...

At first I am just expecting a simple answer, say a subset of C, but I didn't expect so many things I am missing  :(

My friend is feeling mad at me now...

As far as I can see, you just need to be a bit careful.
First write the question as you think of it.
Next go through it carefully looking for possible shadowy places.
See if you have defined everything carefully.
You should have started by mentioning what C is I think.
Try it next time.
I'm sure your friend will have the patience to endure this - mathematicians always do. :)

C is the complex numbers...
I tried to type the math boldface, but seems I forget the command  :maybe:

You are looking for \mathbb{C}.  This yields $ \mathbb{C}$.

$ \mathbb{THX}$

so, how can I be aware of those "shadowy" places if I am not too familiar with the subject?

You've got to see what is well established and what isn't.
For example, simply calling something a number may result in a raised eyebrow or two, while there is nothing shadowy about the real numbers.
You must be clear about the problem, that's all.
If it's a metric space only, call it an arbitrary metric space.
If it's a Banach space, you have to say so.
If the norm is arbitrary, it's enough to say it's a Banach space, but otherwise you must mention the norm.
Something that helps is hunting down common generalizations, because in maths you'll very frequently find things which mean something specific to you, but which has a generalized version too, which may get you into the kind of trouble you're speaking of. (For example, we often take norm and inner product to mean the euclidean norm and inner product, but you should always mention this in your problem.)
Trust me, it's not as big a problem as it looks - things will get better with time.


group theory
abstract algebra
number theory
greatest common divisor
superior algebra
superior algebra unsolved

Is $\mathbb{Q},+= grp\{\frac1{p}\ |\ p \ prime \}$?

[hide="what I found myself"]
I only managed to prove that $\mathbb{Q}\setminus \{0\},. = grp\{\frac1{p}\ |\ p\ prime \}$ That's all :(
[/hide]

Why is $\mathbb{Q}$ not finitely generated?

[hide="what I found myself"]
I said, take some $q \in \mathbb{Q}$, then $q=a/b$, with $a,b \in \mathbb{Z}$. If you factor $b$ *uniquely* in prime factors such that $b=p_{1}^{z_{1}}...p_{n}^{z_{n}}$ we can split $q=a/b$ *uniquely* by doing partial fraction: $q=a/b = \frac{x_{1}}{p_{1}^{z_{1}}}+...+\frac{x_{n}}{p_{n}^{z_{n}}}$ with $x_{i}\in \mathbb{Z}$ for $i=1,...,n$. Because there are infinitely many elements in $\mathbb{Z}$, there are also infinitely many *unique* prime factorizations and hence you need infinitely many generators to generate $\mathbb{Q}$. This is not really rigorous and it does not feel right at all.  Corrections are always welcome.
[/hide]

[quote="qleak"][quote="yannick"]

I only managed to prove that $\mathbb{Q}\setminus \{0\},. = grp\{\frac1{p}\ |\ p\ prime \}$ That's all :(

[/quote]

Did you know that you get zero for free?
I cannot comment on the correctness of your argument as you haven't shown it.

If you have $\mathbb{Q}\setminus \{0\}\subseteq \left< \frac{1}{p}\mid p \text{ is prime }\right>$

Then you have 1,-1 in the above mentioned set and hence you have 0=1+(-1) in the set by closure of addition.

cheers

-Q[/quote]
I succeeded to do it for the multiplicative group. But not for the additive group of $\mathbb{Q}$ (so I did not make a typo).

[quote="ZetaX"]For the first: how can $\frac 1{p^{2}}$ be in the set¿[/quote]

Thanks, I did not think about that.

For why the additive group of $\mathbb{Q}$ isn't finitely generated:
Show that every finitely generated subgroup is generated by a single element (the gcd of the generators). A singly generated subgroup can't be everything.


geometry
geometric transformation
reflection
incenter
homothety
trigonometry
geometry proposed

Let $ABC$ be an acute triangle with incenter $I$. The incircle touches $BC$ at $X$ and the median $AD$ is drawn. Let $F$ be the reflection of $X$ with respect to $I$ and $E$ be the reflection of $X$ with respect to $D$. Show that $A, E, F$ are collinear.

Hint: consider a homothety with center $A$ which turns incircle into $A$-excircle.

P.S. This was posted before at least few times.

[b]The first (Megus') method.[/b] $\frac{AI}{AI_a}=\frac{p-a}{p}=\frac{r}{r_a}\Longrightarrow \frac{AI}{AI_a}=\frac{r}{r_a}\Longrightarrow E\in AF\ .$

 [b]The second method[/b]. Suppose $b\ne c$ and define $L\in (BC),\ AL\perp BC$.

You observe that $\frac{FX}{AL}=\frac{2r}{h_a}$ and you can prove easily that $\frac{FX}{AL}=\frac{EX}{EL}$

because $EX=|CX-CE|=\left| (p-c)-(p-b\right|=\left|b-c\right|$

and $EL=\left| BE-BL\right|=\left| (p-c)-c\cdot \cos B\right|=$

$\left|\frac{a(a+b-c)-(a^2+c^2-b^2)}{2a}\right|=\frac pa\cdot \left|b-c\right|.$ But  $2S=2pr=ah_a.$

could  you explain the first  method more clearly  please??  thanks

Here $E$ is the touch point with the excircle opposite to $A$.
Also, consider $M$ midpoint of $AX$, we know $MI$ meet $BC$ at $D$ (by old theorem). Then $AF$ meet $BC$ at $E$ ($XD=DE$).

[img]http://i43.photobucket.com/albums/e387/phamduyhiep/Mathlinks/Thanghang.jpg[/img]
This is an old problem.
We have $\frac{AP}{AC}$ = $\frac{AQ}{AB}$ = $\frac{PQ}{BC}$ = $\frac{p-a}{p}$ = $\frac{FP}{p-b}$...


[quote="Inspired By Nature"]What is the image of $\left( \begin{array}{c}3 \\ 1 \\ 2 \end{array}\right)$ under the mapping $\left( \begin{array}{ccc}1 & 4 & 1 \\-2 & 0 & 0 \\ 3 & 2 &-3\ \end{array}\right)$?[/quote]

Don't you multiply them or something...

[hide]$\left( \begin{array}{ccc}1 & 4 & 1 \\-2 & 0 & 0 \\ 3 & 2 &-3\ \end{array}\right) \cdot \left( \begin{array}{c}3 \\ 1 \\ 2 \end{array}\right)=\left( \begin{array}{c}9 \\-6 \\ 5 \end{array}\right)$
[/hide]

oh, ofcourse, I posted the wrong question. :blush:
Thanks ch1n353ch3s54a1l
wasted all that time... :(


A boy plays 10 games.  The probability of him winning any game is (independent) and .25.  Let X = the number of games he wins.  Find E(X) (the expected value of X) and P( X <= 3) (probability that X is less than or equal to 3).

[hide]E(X) = 2.5 since 0.25*10 = 2.5

P(X :le: 3) can be done using the binomcdf function on a TI-83+ or by calculating each individual probability:

[tex] P(X=m)={n\choose m}*0.25^m*0.75^{n-m}[/tex]

You can also perhaps do (1 - case X = 5)/2 - case X = 4.[/hide]


function
integration
calculus
limit
analytic geometry
graphing lines
slope

Question 1) Let $ f$ and $ g$ be functions such that $ f''$ and $ g''$ exist everywhere on $ \Re$. For $ a<b$, suppose that $ f(a)\equal{}f(b)\equal{}g(a)\equal{}g(b)\equal{}0$, and $ g''(x)\neq0$ for every $ x\in(a,b)$.
	(i) Prove that $ g(x)\neq0$ for every $ x\in(a,b)$
	(ii) Show that there exists a number $ c\in(a,b)$ for which $ \frac{f(c)}{g(c)}\equal{}\frac{f''(c)}{g''(c)}$

	Question 2) Let $ f$ be a continuous function on $ \Re$ such that $ \stackrel{lim}{x\rightarrow0}\frac{f(x)}{x}$ exists. Define the function $ g(x)\equal{}\int^{1}_{0}f(xt)dt$, $ x\in\Re$. Determine whether $ g'$ is continuous at $ x\equal{}0$. Justify your answer.

[quote="DineshManohar"]Question-2

Given $ g(x) \equal{} \int_{0}^{1}f(xt).dt$
 Assume $ u \equal{} xt$ thus $ dt \equal{} \frac {du}{x}.$Limits vary from $ 0 \to x$ thus

$ g(x) \equal{} \frac {1}{x}\int_{0}^{x}f(u).du$

Thus by first fundamental theorem of Calculus,

$ g'(x) \equal{} \frac {f(x)}{x}$ :maybe:

Since $ \stackrel{lim}{x\rightarrow0}\frac {f(x)}{x}$ exists $ g'(x)$ is continuous. :roll:[/quote]

No, the product rule gives $ g'(x) \equal{} \frac{\minus{}1}{x^2}\int_{0}^{x}f(u) du\plus{}\frac {f(x)}{x}$.  If the last term $ \rightarrow L$, then the first term $ \rightarrow \frac{\minus{}L}{2}$ (that needs proof). So $ g'(x) \rightarrow L/2$.

Hints for question 1:

i) Consider the concavity of $ g.$

ii) What happens if you apply Rolle's Theorem to $ h(x) \equal{} f(x)g'(x) \minus{} f'(x)g(x)?$

[quote="Kent Merryfield"]Hints for question 1:

i) Consider the concavity of $ g.$

ii) What happens if you apply Rolle's Theorem to $ h(x) \equal{} f(x)g'(x) \minus{} f'(x)g''(x)?$[/quote]

For ii) I think you meant $ g(x)$ in that last term. [color=green][Agreed, and edited - K.M.][/color]

For i) we could also use Rolle: If $ g(c) \equal{} 0$ for some $ c \in (a,b)$, then the vanishing of $ g$ at the end points shows $ g' \equal{} 0$ at two points of $ (a,b)$, hence $ g''$ ...

ii)

Applying Rolle's theorem there exists $ c \in (a,b)$ such that $ h'(x)|_{x \equal{} c} \equal{} 0$ thus,

$ 0 \equal{} f(x)g''(x) \minus{} f''(x)g(x)|_{x \equal{} c}$

$ \frac {f(c)}{g(c)} \equal{} \frac {f''(c)}{g''(c)}$

i) Suppose there exists a $ c \in (a,b)$ such that $ g(c) \equal{} 0$ then $ g'(c)$ makes clear that $ f$ vanishes at two points in $ (a,b)$, thus $ g''(x) \equal{} 0$ is a contradiction as $ g''(x) \neq 0 , \forall x \in (a,b)$. Is this what you meant?  :blush:

hmm.. the other one. As $ g''(x) \neq 0 , \forall x \in (a,b)$ $ g$ happens to be either concave or convex. Thus $ g(x) \equal{} 0$ for some $ c \in(a,b)$ contradicts  $ g''(x)$ condition.
But here too Rolle's theorem lies foremost in shift of curve from cocave to convex or viceversa.. :| .

[quote]the product rule gives $ g'(x) \equal{} \frac { \minus{} 1}{x^2}\int_{0}^{x}f(u) du \plus{} \frac {f(x)}{x}$. If the last term $ \rightarrow L$, then the first term$ \rightarrow \frac { \minus{} L}{2}$(that needs proof). So $ g'(x) \rightarrow L/2$.[/quote]

Proof:

Assume $ \frac {f(x)}{x} \equal{} L$

Consider
 $ \lim_{x \to 0} g'(x) \equal{} \lim_{x \to 0} \frac {f(x)}{x} \minus{} \lim_{x \to 0} \frac {\int_{0}^{x}f(u).du}{x^{2}}.$

Applying L-Hopital's rule we get ,

$ \equal{} L \minus{} \lim_{x \to 0} \frac {f(x)}{2x}$

$ \lim_{x \to 0} g'(x) \equal{} L \minus{} \frac {L}{2} \equal{} \frac {L}{2}$ :roll:

[quote]How do you solve part (i), does it have something to do with concavity.[/quote]

Sorry, I did a mistake in the numbering in my posts  :maybe: . :roll:
If that was not to convince you , then here is my explanation,

i) $ g''(x) \neq 0$ says geometrically that the rate of change of slope cannot be $ 0$ thus the function $ g(x)$  cannot be having a constant slope for $ x \in (a,b)$. It's  slope varying throughout $ (a,b)$ . Thus a condition $ g(x)\equal{}0$ for some $ c \in (a,b)$ contradicts the above argument.

[quote="WWW"]

No, the product rule gives $ g'(x) \equal{} \frac { \minus{} 1}{x^2}\int_{0}^{x}f(u) du \plus{} \frac {f(x)}{x}$.  If the last term $ \rightarrow L$, then the first term $ \rightarrow \frac { \minus{} L}{2}$ (that needs proof). So $ g'(x) \rightarrow L/2$.[/quote]

Sorry WWW, I don't quite see how the product rule gives this. Could you explain this a little? I'm a little confused.

And Kent how did you come up with the idea to consider $ h(x)\equal{}F(x)g'(x)\plus{}g(x)f'(x)$? I'm at a loss to comprehend how people come up with such constructions under time pressure.

Thanks in advance!

[quote]Sorry WWW, I don't quite see how the product rule gives this. Could you explain this a little? I'm a little confused. [/quote]

Its the usual product rule for the differentiating product of functions.( In this case product of two functions).

$ ( u.v)' \equal{} u'v \plus{} uv'$. Applying this we get,

$ g'(x)$ = $ \int_{0}^{x}\frac {f(u)}{u}.du$ $ \left [ \frac {1}{x} \right ] '$ +  $ \frac {1}{x} \left[ \int_{0}^{x}\frac {f(u)}{u}.du \right ]' : )$

$ \int_{0}^{x}  f(u).du \equal{} f(x).$ . Fact that derivatives and integrals are inverse operations. :)

I hope this will be clear??  :maybe:

[quote] Kent how did you come up with the idea to consider h(x)=F(x)g'(x)+g(x)f'(x)? I'm at a loss to comprehend how people come up with such constructions under time pressure.

Thanks in advance![/quote]

Hmmm.. Experience !! Am I right Proff.Kent ?  :)

Thanks Dinesh, I didn't read your post carefully the last time round  :blush:  Sorry about that


Find out all the squares wich are written with only two distinct digits.
(For example 1221)

Hmmm, maybe you put too difficult questions. Only the problem of finding the squares that have k digits, k-1 of which are equal reprezents an extremely difficult problem solved by Laurentiu Panaitopol (Journal of Integer Sequences). So, forget about it. Maybe when you will be very old and famous you will be able to solve it( though I doubt).

I was almost sure it cant be solved.Do you have the solution of that problem of Panaitopol?I saw a particular case of this before(with k digits from which k-1 are 6 and one is 5)

You must be kiding. It is 12 pages long and I don't think you want me to do such an effort to write it.


Edward starts in his house, which is at (0,0) and needs to go point (6,4), which is coordinate for his school. However, there is a park that shaped as a square and has coordinates (2,1),(2,3),(4,1), and (4,3). There is no road for him to walk inside the park but there is a road for him to walk around the perimeter of the square. How many different shortest road routes are there from Edward's house to his school?

Your answer is correct. Mind on showing "how" you counted them?

1   5  15 25 40 66 110
1   4  [u]10 10 [/u]15 26 44
1   3  |6 ....| 11 18
1   2  |[u]3   4   5[/u]|  6   7
S   1  .1  .1  .1  .1  .1

As you may or may not be able to tell by this terrible diagram, there is a bounded region (the park) and a starting point (S). the number of routes to that number is found by adding the two numbers numbers leading to it (kind of like a Pascals triangle, except with a top of S, a bounded region, and opening to the right and up.)
At the destination, there are 110 unique shortest paths.

Yeah, same thing I did. It's kind of hard to explain without a diagram, but you basically draw a 6x4 grid with a 2x2 square rather than a 1x1 at the specified spot. From there, it's the standard counting thing.

Yah... I was trying to draw a grid with autocad, but it was screwing up, so I gave up and drew that crappy ASCII drawing =p


linear algebra
matrix
Putnam
linear algebra unsolved

If $ A^{ \minus{} } \plus{} B^{ \minus{} } \equal{} A^{ \minus{} }(A \plus{} B)B^{ \minus{} }$, then for this fixed $ (A^{ \minus{} },B^{ \minus{} })$,
we have $ A^{ \minus{} } \plus{} B^{ \minus{} } \equal{} B^{ \minus{} }(A \plus{} B)A^{ \minus{} }$. Here $ A^{ \minus{} }$, $ B^{ \minus{} }$ are any matrix wtih the same size satisfying $ AA^{ \minus{} }A \equal{} A$, $ BB^{ \minus{} }B \equal{} B$.

Forget the pseudoinverses!

[color=blue][b]Theorem 1.[/b] Let $R$ be a commutative ring with unity. Let $n$ be a positive integer. Let $A$, $B$, $P$, $Q$ be $n\times n$ matrices over $R$ such that $P+Q=P\left(A+B\right)Q$. Then, $P+Q=Q\left(A+B\right)P$.[/color]

Before we start proving this, an easy lemma:

[color=blue][b]Theorem 2.[/b] Let $R$ be a commutative ring with unity. Let $n$ be a positive integer. Let $U$ and $V$ be $n\times n$ matrices over $R$ such that $UV=I_n$. Then, $VU=I_n$.[/color]

[i]Proof of Theorem 2.[/i] We have $\det U\det V=\det\left(UV\right)=\det I_n=1$. Now,

$VU=\underbrace{1}_{=\det U\det V}\cdot VU=\det U\det V\cdot VU$
$=VU\cdot\det U\cdot\det V\cdot I_n=VU\cdot\det U\cdot V\mathrm{adj}V$ (since $\det V\cdot I_n=V\mathrm{adj}V$)
$=\det U\cdot V\underbrace{UV}_{=I_n}\mathrm{adj}V=\det U\cdot \underbrace{V\mathrm{adj}V}_{=\det V\cdot I_n}=\underbrace{\det U\cdot\det V}_{=1}\cdot I_n$
$=1\cdot I_n=I_n$,

and Theorem 2 is proven.

Now a more serious lemma, generalizing an old Putnam problem:

[color=blue][b]Theorem 3.[/b] If $X$, $Y$ and $T$ are three $n\times n$ matrices over a commutative ring $R$ such that $X+Y=XTY$, then $X+Y=YTX$.[/color]

[i]Proof of Theorem 3.[/i] We have

$\left(TX-I_n\right)\left(TY-I_n\right)=T\underbrace{XTY}_{=X+Y}-TX-TY+I_n=T\left(X+Y\right)-TX-TY+I_n$
$=I_n$.

Thus, Lemma 2 (applied to $U=TX-I_n$ and $V=TY-I_n$) yields $\left(TY-I_n\right)\left(TX-I_n\right)=I_n$.

Now, $X+Y=XTY$ yields $Y=XTY-X=X\left(TY-I_n\right)$ and thus

$Y\left(TX-I_n\right)=X\underbrace{\left(TY-I_n\right)\left(TX-I_n\right)}_{=I_n}=X$,

so that $X=Y\left(TX-I_n\right)=YTX-Y$. In other words, $X+Y=YTX$, and Theorem 3 is proven.

[i]Proof of Theorem 1.[/i] We apply Theorem 3 to $X=P$, $Y=Q$, $T=A+B$ and immediately obtain Theorem 1.

  Darij


Suelyn counts up from 1 to 9, and then immediately counts down
again to 1, and then back up to 9, and so on, alternately counting
up and down
\[ (1, 2, 3,4,5,6,7,8,9,8,7,6,5,4,3,2,1,2,3,4,\ldots ).\] What is
the $ 1000^{\text{th}}$ integer in her list?

The number blocks are 16 numbers long, then they repeat. 1000/16= 62.5, or mod 8. So, this is the 8th number, or $ \boxed{8}$


modular arithmetic
function
combinatorics unsolved
combinatorics

We have been given a playing card deck. We deal it to the four players, each of whom gets 13 cards. Assume that the number values of cards are $1,2,\cdots,13$. How many ways there are to deal the deck such that there is an $i$ such that when dealing a round number $i$ some player gets a card which value is $i \pmod{13}+1$?

(I'm not sure if "round" is a correct term. I mean that in first round we deal four cards, one for each player. On the secound round we deal four more cards and so on.)

Suppose that we have a deck with $l$ suits and $n$ cards in each suit (so, the total number of cards in the deck is $nl$). Suppose that values of the cards within each suit are simply numbers $1, \dots, n$
Dealing the deck among $l$ players takes $n$ rounds, during each round $l$ cards are dealt, one to each player.

First, for each $1\leq k\leq n,$ we will count the number of shuffled decks such that dealing such decks does not give card $i$ to any player during round $i$ for all $i=1,2,\dots,k$ (in other words, there is no card $1$ within first $l$ cards in the deck, no card $2$ within second set of $l$ cards, etc., no card $k$ within $k$-th set of $l$ cards in the deck). The inclusion-exclusion principle implies that there are
\[F(n,l,k) = \sum_{j_{1},\dots,j_{k}=0}^{l}(-1)^{j_{1}+\dots+j_{k}}(nl-j_{1}-\dots-j_{k})! \prod_{i=1}^{k}{l\choose j_{i}}^{2}j_{i}! \]
such decks.

Now, again according to the inclusion-exclusion principle, the number of shuffled decks such that at each round $i$ at least one card $i$ is dealt is
\[G(n,l)=\sum_{k=0}^{n}(-1)^{k}{n\choose k}F(n,l,k). \]
Computing $F(n,l,k)$ for large $n,l,k$ using the definition above may be time-consuming. There is an alternative approach. Let
\[f(m,l,k) = \sum_{j_{1},\dots,j_{k}=0}^{l}(-1)^{j_{1}+\dots+j_{k}}(m-j_{1}-\dots-j_{k})! \prod_{i=1}^{k}{l\choose j_{i}}^{2}j_{i}!. \]
Then $F(n,l,k) = f(nl,l,k)$ and
\[f(m,l,k) = \sum_{j=0}^{l}(-1)^{j}{l\choose j}^{2}j! f(m-j,l,k-1) \]
with $f(m,l,0)=m!.$ With this formula, the function $f(m,l,k)$ can be computed using dynamic programming (assuming $l$ fixed).

In particular, we have
[code]G(13,4) = 20366826385239437910994563581940510272786822469946975293800448[/code]
and that's the answer to the original question.

P.S. This is a straightforward implementation of functions $F$ and $G$ in [url=http://pari.math.u-bordeaux.fr/]PARI/GP[/url]:
[code]{ F(n,l,k) = local(r,s); r=0; forvec(j=vector(k,i,[0,l]), s=sum(i=1,k,j[i]); r+=(-1)^s * (n*l-s)! * prod(i=1,k, binomial(l,j[i])^2*j[i]!) ); r }
{ G(n,l) = sum(k=0,n, (-1)^k * binomial(n,k) * F(n,l,k) ) }[/code]

[quote="puuhikki"]Are you sure? I think there are only $\binom{52}{13}=635013559600$ possible ways to deal the deck among four players such that each players gets 13 cards. Am I missing something?[/quote]
First, there are
\[{52\choose 13\ 13\ 13\ 13}= \frac{52!}{13!^{4}}= 53644737765488792839237440000 \]
ways to distribute 52 cards among 4 players.
Second, I did not count such distributions. I counted the number of shuffled  decks (out of total number of decks $52!$) that satisfy the specified condition.


quadratics
geometry
3D geometry
number theory proposed
number theory

Find all positive integers $k$ for which number $3^k+5^k$ is a power of some integer with exponent greater than $1$.

Checking residues mod 16 we can see that if $3^k+5^k=n^t$ then $2\mid n$ and $4\nmid n$. Since $16\nmid n^t$ we find that $t\le 3$.
Hence we have two cases: $3^k+5^k=(2m)^3$ and  $3^k+5^k=(2m)^2$.
The second can hold only for odd $k$ because of the remainder mod 4. But then, $3^k+5^k\equiv -1 (\mod 3)$ which is not a quadratic residue.
Hence only one case remained. Even in this case $k$ must be odd, because of the remainder mod 4. If $k\ge 2$ we have that $3^k+5^k\equiv 5^k\equiv (10)^i\cdot 5 \equiv 5 (\mod 9)$. But $5$  is not a cubi residue mod 9. The only posible situation remained is $k=1$ which works.

[quote="xirti"] $3^k+5^k\equiv 5^k\equiv (10)^i\cdot 5 \equiv 5 (\mod 9)$[/quote]
Hmm, what do you mean by that? E.g $k=3$ we get $5^3\equiv -1 \mod 9$  :?

Ok,I will explain more this.We have $x^3 \equiv 0,1,-1 (mod 9)$ for all $x \in Z$
Because $9 \nmid 3^k+5^k$ (if k \geq 3) so that $3^k+5^k \equiv 1,-1 (mod 9)$.But $5^k \equiv 5^1,5^3,5^5 (mod 9)$ (because $k$ is odd number and $k \geq 3$)
Check these case we see that $k \equiv 3 (mod 6)$.Put $k=6k_{1}+3 (k_{1} \in N^{*})$
We have $3^{(2k_{1}+1)3}+5^{(2k_{1}+1)3}=a^3$ (this is Fecmat's equation) so we don't have the integer solution of this equation.Hence $k=1$ is the only solution of our problem. :)  :)

just a funny way to finish it.

You can get pretty quickly that it boils down to

3^k + 5^k = a^3 for some a.

Now, assume k≥2, then examine it mod 9. The perfect cubes mod 9 are 1, -1, and 0. Thus 5^k is congruent to 1, -1, or 0. Using the fact that phi(9) = 6, we can see that k must be divisible by 3. But then we have that

3^k + 5^k = a^3. Where 3 | k. This contradicts fermat's last theorem. And hence k < 2.

oh sorry, i didn't read the solution above O_O

$k=1$ is a trivial solution.
when $k>1$ if $k$ is even,then$2||3^k+5^k$ contradiction
hence $k$ is odd.
it's trivial that $8||3^k+5^k$
so if it is a power,it's a cube!
let$3^k+5^k=t^3$
mod 9,we get $3|k$
let $k=3l$then
$3^{3l}=(t-5^l)(t^2+5^l+5^{2l}$
since $gcd(t-5^l,t^2+5^lt+5^{2l})|5^{2l}*3$
so $t-5^l=3$
yielding $5^{2l}+5^l+3=3^{3l-2}$
mod 3 and we get a contradiction.

[hide=Solution (This is also 2016 MOP Homework problem N1)]
I claim $k$ is odd.
Proof: Assume not. Then $3^k\equiv5^k\equiv1\pmod4$, so the sum is $2\pmod4$, which can never be a power of some integer with exponent greater than $1$ (it will either have too many or too little powers of $2$). Thus, $k$ is odd.

Taking $\pmod{16}$ gives that $3^k+5^k\equiv8\pmod{16}$. So the exponent must be equal to $3$.

Since $\boxed{k=1}$ works, assume $k>1$.

Now, taking $\pmod9$ gives that $k\equiv0,3\pmod6$ because a cube must be $-1,0,1\pmod9$. But since $k$ is odd, $k\equiv3\pmod6$. Now taking $\pmod7$ gives that $3^k\equiv6\pmod7$ and $5^k\equiv6\pmod7$. These add up to $5\pmod7$, which cannot be a perfect cube.

So no $k>1$ works. [/hide]


[quote="every"]Let $ (u_n)$ be sequence given recursively as follows :
$ u_1 \equal{} \minus{} 1,u_2 \equal{} \minus{} 2$
$ n.u_{n \plus{} 2} \minus{} (3n \minus{} 1)u_{n \plus{} 1} \plus{} 2(n \plus{} 1)u_n \equal{} 3$
Let $ S \equal{} u_1 \plus{} u_2 \plus{} ... \plus{} u_2009 \plus{} 2(2^{2009} \minus{} 1)$
Prove that S is divisible by 2009[/quote]

Are you sure of the exact definition of the sequence ?

At first glance, $ u_n$ is not an integer for $ n>6$ and, very likely, $ S$ is not an integer, so .... .

[quote="mathVNpro"][quote="pco"][quote="every"]Let $ (u_n)$ be sequence given recursively as follows :
$ u_1 \equal{} \minus{} 1,u_2 \equal{} \minus{} 2$
$ n.u_{n \plus{} 2} \minus{} (3n \minus{} 1)u_{n \plus{} 1} \plus{} 2(n \plus{} 1)u_n \equal{} 3$
Let $ S \equal{} u_1 \plus{} u_2 \plus{} ... \plus{} u_{2009} \plus{} 2(2^{2009} \minus{} 1)$
Prove that S is divisible by 2009[/quote]

Are you sure of the exact definition of the sequence ?

At first glance, $ u_n$ is not an integer for $ n > 6$ and, very likely, $ S$ is not an integer, so .... .[/quote]
I am sorry but the problem statement is true. By the way, this is exactly the problem in the [b]Mathematics and Youth magazine[/b] in [b]VN[/b]. So please wait until the $ 25th$ of May, then we can discuss it.[/quote]

I'm sorry, but I proved that this S (sum up to 2009) is not an integer.
So, i'm quite sure there is a mistake in this problem and I suggest  nobody tries to solve it.

I can give the method I used to prove S is not an integer (since direct calculus up to 2009 is nearly impossible). I think this would not be forbiden. Just ask if you want I post it here.

[quote="pco"]We are now beyond the limit date for this problem.

Could the OP (or somebody who can access the VN magazine) confirm or not the problem?[/quote]
Hey pco you are right, The statement above is wrong ( it is the typo of MYM ). The right version here:

[quote]Let $ (u_n)$ be sequence given recursively as follows :
$ u_{1} \equal{} \minus{} 1,u_{2} \equal{} \minus{} 2$
$ n.u_{n \plus{} 2} \minus{} (3n \plus{} 1)u_{n \plus{} 1} \plus{} 2(n \plus{} 1)u_n \equal{} 3$
Let $ S \equal{} u_1 \plus{} u_2 \plus{} ... \plus{} u_{2009} \plus{} 2(2^{2009} \minus{} 1)$
Prove that S is divisible by $ 2009$.
[/quote]
It is time for us to discuss this nice problem!


Janelle averages 40 kilometers per hour biking on level ground. She averages 60% of her level-ground speed riding uphill, and she averages 120% of her level-ground speed riding downhill. The course is level for 5 kilometers, uphill for 6 kilometers, and then downhill for 6 kilometers. How many kilometers per hour is her average speed foe her entire course?

[hide]$40$ kph level
$24$ kph up
$48$ kph down

$\frac{1}{8}$ hr. level
$\frac{1}{4}$ hr. up
$\frac{1}{8}$ hr. down

TOTAL: $\frac{1}{2}$ hours iin $17$ kilometers.

$34$ kph.[/hide]


Let $d(n)$ be the number of divisors n has. find all n such that $d(n)=\frac{n}{4}$

You posted this before, why are you reposting it?

$d(n)<=2\sqrt{n}$ because if $ab=n$ then $min(a,b)<=\sqrt{n}$. so we have $n<=8\sqrt{n}$ so $n<=64$. on the other hand if $4d(n)=n$ we must have that $n$ is of the form $4k$. this leaves $16$ possible numbers! So we only have to check 4k for $k=1,2,3...,16$ Now iff k is not a square, then $d(n)$ is even and then $k$ is even. so we only need to prove it for $k=1,2,4,6,8,9,10,12,14,16$ we can now finish manually...

Arne, i was sure someone was going to attack it in this forum  :D


Let $S$ be the set of positive integers with the largest possible product which add up to $2005$. What is the product?

theres an interesting solution to this in the art and craft of problem solving's section on eulerian mathematics, i can post it later if you want.

[hide="Guess"]I think that just like when the product of 2 numbers (which add to a given sum) is maximized when they are equal, then if you have x^y where xy = 2005, it is maximized when x=y or in this case, x is closest to y.

So we can take the square root of 2005 to get approx 45.
The cube root = approx 13
The 4th root = approx 6.6
The 5th root = approx 4.5

So the answer MIGHT be something like, S is a set containing 401 5's, and the product is 5^401.[/hide]

[hide="another guess/idea"]Well here's how i would see it...
Look at each of the following...
$\sqrt{2}$
$\sqrt[3]{3}$
$\sqrt[4]{4}$
and so on.  We see that the greatest out of these is $\sqrt[3]{3}$.
Thus, we will get the largest product if we use 3s.  Note that $\sqrt{2}=\sqrt[4]{4}$, and past 4, it decreases.  Therefore, the largest product will be $3^{667}2^2$ , with 3*667+2*2=2001.  I think...[/hide]

The answer is correct but your justification needs a bit more work. Consider any integer $n>4$. We can replace $n$ by $3$ and $n-3$ without changing the sum while increasing the product.

\[3(n-3)>n\]
\[2n>9\]

The above is clearly true since $n>4$. Also, $n=4$ can be broken down into $2 \cdot 2$ without altering the sum or product (1 does nothing).

i remember seeing somewhere that the best way to do these types of problem is to break it down into as many groups of 2 and 3 as possible, with 3 having higher precedence.  this has something to do with the fact that it's closer to $e$ although i really couldnt give anyone a good explanation as to why $e$ maximizes this problem.

therefore, i'm guessing the answer is $3^{667}2^2$

(edit)

upon reading k81o7's post, when you plot the function $\sqrt[x]{x}$, one see's $(e, \sqrt[e]{e})$ is the maximum value.

[hide]$S=\{ a_1, a_2, a_3, \ldots, a_n \}$, and $a_1+\ldots+a_n=2005$. To maximize $a_1a_2 \cdots a_n$, we want all the numbers to be as close to each other as possible. The problem is, we don't know the value of $n$. $2005=5 \cdot 401$, so if all the numbers are equal, then we have $S_1=\{ 5,\cdots,5 \}$ with 401 $5$s, or $S_2=\{ 401,401,401,401,401 \}$. The product of $S_1=5^{401}, S_2=401^5.$

$S_2 \approx 400^5$, and $S_1 \approx 5^{5 \cdot 80} = (5^{80})^{5}$, and $5^{80} > 400$, so $S_1$ is bigger. The answer is then $5^{401}.$

EDIT: Nevermind, the previous answers look right...[/hide]

So I did a little bit of work today:

[hide]My guess is that for any number $k$, if $a_1+a_2+...+a_n$ = $k$, the maximal product of the $a_i$ is:

for $3|k$, $3^{k/3}$
for $3|(k+1)$, $(3^{(k-4)/3})(4)$
for $3|(k+2)$, $(3^{(k-2)/3})(2)$

In other words, put as many threes as you can. If you have a $1$ left over, make a $4$. If you have two left over, multiply by $2$.

I don't have a proof yet, but it's clear that if you start with as many $2$s as possible, replacing three $2$s with two $3$s will increase the product. I'm trying to see a way to prove that this is maximal as well.[/hide]

Hey man, doesn't Am-GM give anything useful???

No, except that we want to make all the numbers as close as possible. It doesn't give much for deciding just what number we should make all of 'em close to.

You want as many 6's as possible and I have shown why in my post above.

You mean you want as many 3s as possible?

No, as many 6's, because I said for $n>4$. Notice that if you have $n=4$ left you can still put a three and a one but two two's would be better.

But for $n=7$ you would want $2,2,3$, not $1,6$, right?
I think your above proof is using $n, n-3$...

Bleh, yeah, I meant you have to watch out, for if you just divide your integer by $3$ then if the remainder is $1$ you should use one less $3$ and then two $2$'s.

Yeah, if the remainder is 1, you make a 4, if it is 2, you multiply by 2.

And when I said as many 6's I meant divide $n$ by 6 and then multiply that number to give you the number of $3$'s since you have to play with the remainder upon division by six as we have previously said.

whoever mentioned $e$ was probably right, it seems the closer the numbers get to e, the greater the product, of course i cannot prove it but it seems right: consider the case for 15
15/e = 5.518...
e^5.518 = 249...
whereas if we did the 2's and 3's, 3^5 = 243
that shows that the closer the numbers get to e, the higher the product, so we are restricted by integers so 3 would be closer to e than 2 so we would want to maximize the number of 3's

here is a step to showing that $e$ would be the maximum
consider a number $a$ where we would like to find the set of numbers whose sum is a and the set with the maximum product (we will consider nonintegers as well)
let us consider using just 1 number ($x$) in our set so $a/x=$ the number of x's in the set so the product would be $x^{a/x}$
to maximize this, we would use calculus ( i am not sure that i can post this here but it definately helps so...)
maximize $f(x) = x^{a/x}$
$ln(y) = \frac{a}{x}ln(x)$
$\frac{y'}{y} = a(-x^{-2}ln(x) + x^{-2})$
$y' = a(\frac{1}{x^2})(1-ln(x))(x^{a/x})$
to find the inflection points, set $y' = 0$
$0 = a(\frac{1}{x^2})(1-ln(x))(x^{a/x})$
the only factor that yields 0 is $(1-ln(x))$
so $0 = (1-ln(x))$
$1 = ln(x)$
$x=e$
i don't feel like writing the test for it being a maximum instead of a minimum, but it is obvious from the graph

Yes indeed, That's the continous proof I was thinking... thanks  :)


Please post solutions to each of the questions in Round 1 here. Post questions also here.

EDIT: Also, I have recently received 2 requests for problems on #2.

The other root---well, it doesn't mean that the 2 roots are different. :wink:

What is the answer to #5 on Mock MathCounts 1?

[quote="anirudh"]Oh...its 64[/quote]

How do you get that?

I dunno. :maybe: It said that was the answer to the question on one of the school handouts we got. :maybe:

That's what I thought but SplashD thinks otherwise

[quote="mathgeniuse^ln(x)"][quote="sonny"][quote="mathgeniuse^ln(x)"]Well, then does anyone else have a good solution?

A friend told me that is just 4 possibilites for each of the three letters, so

$4^{3}$[/quote]

Ya I dont believe there is a faster way. I added the possibilites instead of mulitplying so i got 12  :furious:   :wallbash:[/quote]

Oh, sorry about that.

And if you wanted to do the reverse it would be 3^{4} right.[/quote]

Yes I think mapping B into A would be $3^{4}=81$.

[quote="anirudh"]But its mapping A into B...

So its $4^{3}$. :)[/quote]

No i was refering to mathgeniuse^ln(x)' question about being the other way around.


algebra polynomial calculus derivative algorithm

If $ax^4 +4bx +c$ has a double root at $x=\alpha$ show that $ac^3=27b^4$

if there is a double root at $\alpha$, then there will be a critical point there ($f'(\alpha)=0$), also $f(\alpha)=0$, the result follows

can you please explain it please.  :blush: sorry i am new to polynomials and I still dont see the trivial things.

here is a proof with calclulus,

if there is a double root at $x=r$, then $P'(r)=0$

say there is a double root at $r$, then the polynomial can be written as

$P(x)=(x-r)^2*Q(x)$ where $Q(x)$ is either a constant, or a polynomial, then

$P'(x)=(x-r)^2Q'(x)+2(x-r)(-r)Q(x)$
then obviously, $P'(r)=0$

[quote="amcavoy"]$4a\alpha^{3}+4b=a\alpha^{4}+4b\alpha+c=0$
[/quote] How did we get this?

oh calculus.. :(  have no idea wot derivatives are yet. but i found a way to do the question with polynomials and sum of the roots and stuff.

you can just do the division algoritm to factor out $(x-\alpha)^2$, but this way is faster


A friend of mine is a math teacher at a small high school in the Boston area.  She's put together a small math team there, and is looking for some contests to join for the remainder of the year.  I think she's looking for something that won't be too discouraging to her first-time math team students (so e.g. HMMT is probably out of the question, though they may take the AMC).  Anyone have any suggestions?

well in the fall, her team can join the Greater Boston Math League (GBML)
but currently, the only competition i can think of is AMC and ARML(?), the one done at your school

NEML is fairly unintimidating.  Whether or not NEML accepts mid-year registrations, she could purchase a few books of old NEML questions ($ \$$12.95 a book or somesuch) and work problems with her students.

For next year she should join a local math league.  If she's looking for "gentle," she should avoid GBML and probably MML.  SMML might be a good choice.  SMML doesn't have a website but some students here could give you good information.

Thanks!  The school already does AMC (and has a handful of AIME qualifiers each year), but I think they're looking for something to do throughout the year.  Anyone want to give an assist on SMML (= Southern Massachusetts Math League?)?

GBML is not so gentle indeed, especially with all those tricky errors in the answer key.

there's the Purple COmet Math COmpetition, that already happened, you could do next year. IT's online and it's for around the world, You could look up the specifics.