probability
geometry
calculus
integration
linear algebra

Let x and y be uniformly distributed, independent random variables on [0,1]. What is the probability that the distance between x and y is less than 1/2 ?

Extra: generalize it to... "is less than a, where 0 < a < 1/2" .

Find the area of the region bounded by $0\le x\le 1$, $0\le y \le1$, and $|x-y|\le \frac{1}{2}$, and that's your area of good outcome, then divide by total area of outcome.

Correct, Elemennop.

I was looking for a "unidimension" (only one real line) argument. Do you (or someone else) have an idea of how could this problem be done without the "plane trick" ?  (Even more restricted... real line and without Calculus!?)

Cheers,
fskilnik

I don't think that could be solved "unidimensionally", since the number of dimensions always depends on the number of independent variables.

Hi, Farenhajt.

Thanks for the reply.  You´re right if "dimension" is considered in the sense of Linear Algebra, but my idea was not related to it directly.

Let me explain what I have in mind: consider $x_0$ chosen (and therefore fixed) , say in the first half of the interval. What is the probability that, "afterwards",  y is chosen such that $\left| {y - x_0 } \right| \le \frac{1}{2}$ ?   The answer is $x_0 + \frac{1}{2}$ , which is geometrically evident.

Now Calculus enters (necessarily?  That´s my question!) the "arena"...
\[ \int\limits_0^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-\nulldelimiterspace} 2}} {\left( {x + \frac{1}{2}} \right)\,\,dx = \frac{3}{8}}  \]
and the answer is twice this value... got it?   :)

Regards,
fskilnik

Is that "not calculus"? :) Moreover, try solving three-variable problem that way... nasty and complicated, while using 3 dimensions you get neat volume computing.

Hi there, Farenhajt.

I know I used Calculus (I mean Differential Calculus, for sure) ...  that´s why I asked if it is possible to solve the problem "unidimensionally" (in a "certain sense" I did, because I used one dimension "after" the other...) AND without Calculus (I don´t see how).

Let us wait. Perhaps someone else may surprise both of us!   :)

fskilnik

Dear fskilnik,

That might happen (in the sense that the corresponding probability is not zero :)), but I strongly doubt it, because if you have continuous random variables, calculus is unavoidable. This particular case is easy since the resulting area is easy to compute by the elementary methods, but the general case must involve integrals - I've never heard of any other possible method.

It is hard to say when a tool is unavoidable, Farenhajt. But your argument "for Calculus" is good, no doubt. I love Calculus, by the way. I just think that if we can do it without it (and without to many trickeries) , it is also fun. With trickeries may also be fun, I must say.   :blush:   Math is great, that´s a fact!   :)

Just a small adjustment to your good-humour: events with probability ZERO may still occur. They are called "almost impossible" but... they are not impossible if they can occur!

An event is only called impossible if (by definition) it is not included in the universe of events that may occur.

Example:

What is the probability that, in a throw of dice, you get 7 ?
Answer: zero and it is an impossible event.

What is the probability that, in successive throwings of a (fair) dice, you get "1, 2, 1, 2, ..." (ad infinitum) ?
Answer: ZERO but it is an almost impossible event, not an impossible event.

Tricky, isn´t it?!   You see why this happens?  If you try to associate a probability to this event (and you CAN) it is of course greater or equal to 0 (and less or equal to 1, sure) but it is NOT any value in ]0,1] therefore... it is zero!

In other words, you may PROVE that, pre-fixed an epsilon positive, the probability of this event is LESS than epsilon, therefore since this epsilon is arbitrary... got it?!   ;)

See you in other problems!
fskilnik


Find all solutions to the following equation \[ \frac{dx}{dt}=\frac{x}{t}+\frac{t}{x^{2}}. \]

se face substitutia x=ty si se obtine $\ y^2dy=\frac{1}{t^2}dt$ . de aici prin integrare si extragerea radicalului avem solutia \[ x=(3ct^3-3t^2)^\frac{1}{3} \].

As a moderator, I feel obliged to remind sweetonect that one should write in [b]English[/b] outside national subforums  :).


function
limit
integration
real analysis
floor function
calculus
irrational number

today is the first time i've seen weyl's equidistribution theorem ( in a fourier analysis book!). it seems like i dont understand the definition of equi-distributed or as some ppl know it, uniformly distributed. the definition given is the one given in mathworld

[url]http://planetmath.org/encyclopedia/UniformlyDistributed.html[/url]

can someone shed light on the definition , why it makes sense ? i cant understand where we account for the uniformity in   " uniformly distributed" . sorry if its a stupid question  :(

A uniform distribution is one for which the likelyhood of landing in a certain set is dependent only on the size of that set (for some appropriate meaning of the word size), not the particular location of that set.  In this case, it says that the fraction of values which fall into a given interval is strictly a function of the length of that interval, not the particular values of the endpoints.

Say, we have a sequence of numbers $(x_{n})$ in $[0,1]$. It turns out that out of the first $1000$ elements, $56$ are in the interval $[0,0.1]$. Does this look like a uniform distribution? No, with a uniform distribution you'd expect about $10\%$ of the elements to be in $[0,0.1]$. And $56/1000=0.056$ is only $5.6\%$. Of course, this may be only a fluctuation in the beginning of the sequence, so we look further, at the limit $\lim_{N\to\infty}\frac{\#\{n\le N\colon x_{n}\in [0,0.1]\}}{N}$. If this limit is $0.056$, then the sequence is not uniformly distributed. If the limit is $0.1$, it [u]may[/u] be uniformly distributed, but we must also compute such limits for $[0.1,0.2]$, $[0.003,0.004]$... and all other subintervals of $[0,1]$ before we know for sure.

Another way to look at it: $x_{n}$ is equidistributed in $[0,1]$ if for all continuous functions $f$ we have

$\lim_{n\to\infty}\frac1k\sum_{k=1}^{n}f(x_{k})=\int_{0}^{1}f(x)\,dx.$

I think I'd want continuity in that definition - mere Lebesgue integrability for $f$ wouldn't give the right definition.

For a famous example, let $\alpha$ be an irrational number and let $x_{n}=\{n\alpha\}=n\alpha-\lfloor n\alpha\rfloor.$ Then $x_{n}$ is equidistributed. But the function $f(x)=1$ for all $x$ in the sequence $x_{n}$ and $f(x)=0$ otherwise is zero almost everywhere and has integral zero. You can't do that with a continuous $f.$


$ A \equal{} \begin{pmatrix} 5 & 3\cos\left(\frac{7\pi}{6}\right ) &3\cos\left(\frac{14\pi}{6}\right )&3\cos\left(\frac{21\pi}{6}\right )&3\cos\left(\frac{28\pi}{6}\right ))&3\cos\left(\frac{35\pi}{6}\right) \\

3\cos\left(\frac{7\pi}{6}\right )&5 &3\cos\left(\frac{7\pi}{6}\right )&3\cos\left(\frac{14\pi}{6}\right ))&3\cos\left(\frac{21\pi}{6}\right )&3\cos\left(\frac{28\pi}{6}\right) \\

3\cos\left(\frac{14\pi}{6}\right ) &3\cos\left(\frac{7\pi}{6}\right )&5&3\cos\left(\frac{7\pi}{6}\right )&3\cos\left(\frac{14\pi}{6}\right )&3\cos\left(\frac{21\pi}{6}\right) \\

3\cos\left(\frac{21\pi}{6}\right )&3\cos\left(\frac{14\pi}{6}\right )&3\cos\left(\frac{7\pi}{6}\right )&5&3\cos\left(\frac{7\pi}{6}\right )&3\cos\left(\frac{14\pi}{6}\right) \\

3\cos\left(\frac{28\pi}{6}\right )&3\cos\left(\frac{21\pi}{6}\right )&3\cos\left(\frac{14\pi}{6}\right )&3\cos\left(\frac{7\pi}{6}\right )&5&3\cos\left(\frac{7\pi}{6}\right) \\

3\cos\left(\frac{35\pi}{6}\right )&3\cos\left(\frac{28\pi}{6}\right )&3\cos\left(\frac{21\pi}{6}\right ))&3\cos\left(\frac{14\pi}{6}\right )&3\cos\left(\frac{7\pi}{6}\right) &5\\
\end{pmatrix}$

$ 11,11,2,2,2,2$ because:
let $ B\equal{}[b_{ij}]$ with $ b_{ij}\equal{}cos((i\minus{}j)7\pi/6)$.
Then $ B^2\equal{}3B$ with $ rank(B)\equal{}2$.


Prove or disprove that $\sum_{k=1}^{n}\frac{1}{ak+d}$ is never an integer for $a \in \mathbb{N}, d \in \mathbb{N}\cup \{ 0 \}, n > 1$.

A simila problem was discussed at
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98535[/url]
and the problem was solved by rem and zetax; the solution to this problem is similar.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Is it possible to find a group $ G$ and a normal subgroup $ N$ such that $ G/N \cong G$ but $ N$ not trivial?

the group $ G \equal{} \mathbb{Z}/p^{\infty}$ has even the property that $ G/N \cong G$ for every proper subgroup $ N \leq G$.

$ \mathbb{Q}/\mathbb{Z}$ is very different from $ \mathbb{Q}$- every element has finite order. You want quotients of it instead.

sorry I edited it many times .. you're right, we may apply the isomorphism theorem to $ n : \mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$ to optain $ (\mathbb{Q}/\mathbb{Z})/(1/n \mathbb{Q}/\mathbb{Z}) \cong \mathbb{Q}/\mathbb{Z}$.


geometry
rectangle
number theory
prime numbers
national olympiad

Time allowed 3 hours.

[size=150]problem :1[/size]
$ABCD$ is a rectangle, $E$ and $F$ are points choisen on segments  $[AC]$ such that $[AE]=[AD]$, and $[AF]=[AD]$, we denote by $G$ and $H$the projections of the points $E$ and $F$ on side $[AB]$.
prove that $[AG]+[FH]=[AC]$

[size=150]problem :2 [/size]
let $a,b,c$be positive numbers, such that $abc=1$prove that :$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+a} \geq 3$

[size=150]problem:3[/size]
find all prime numbers $p,q$  such that $(p+1)^q$ is a perfect square.

[size=150]problem:4[/size]
$a,b,c$are reals numbers with: $1\leq a\leq,1\leq b \leq ,1\leq c\leq1$
prove that $ab+bc+ca+1 \geq 0$

Post the problems in the forum please . :D

problem 3 : if $q=2$ then $\forall p$ prime the condition is satisfied. If $q \neq 2$ then $p+1$ has to be a perfect square, because if it isn't, $(p+1)^q$, since $q$ is odd, contain at least one prime factor repeated an odd number of times, so it can't be a square. So $p+1=k^2 \rightarrow p=(k+1)(k-1)$ that has solution only $p=3$

For the problems 1 ,4 you should change the question because there are typos.

Problem 3

Clearly $q = 2$ gives $(p, 2)$.
If $q$ is odd then $p + 1 = x^2$ $\Rightarrow$ $p = (x - 1)(x + 1)$. That gives $(3, q)$.

for problem 4
 we have only to see that :

 $ab+bc+ac+1=\frac{1}{2}[(1+a)(1+b)(1+c)+(1-a)(1-b)(1-c)]$ :D

In problem 1, if AE=AD and AF=AD, then E=F.


does anyone know how to find the sum of proper divisors of an integer, without actually writing out the divisors?

[quote="Myth"]There is a formula via prime decomposition: \[\sigma(n)=\frac{p_1^{a_1+1}-1}{p_1-1}\cdot \frac{p_2^{a_2+1}-1}{p_2-1}\cdot...\cdot\frac{p_k^{a_k+1}-1}{p_k-1},\] where $n=p_1^{a_1}...p_k^{a_k}$ is the prime decomposition of $n$.[/quote]

Actually since it is proper divisors it would be
$\sigma(n)=\frac{p_1^{a_1+1}-1}{p_1-1}\cdot \frac{p_2^{a_2+1}-1}{p_2-1}\cdot...\cdot\frac{p_k^{a_k+1}-1}{p_k-1}-n$


What are the last two digits of $ (...(7)^7)^7)...)^7$ if there are $ 1000$  $ 7$s as exponents and only one $ 7$ in the middle?

This is the solution but I didn't understand why $ \mod 4$?:
We could solve this problem using mods. This can also be stated as $ 7^{7^{1000}}$.  After that, we see that $ 7$ is congruent to -1 in [color=red]mod 4[/color], so we can use this fact to replace the $ 7$s with $ \minus{}1$s, because $ 7$ has a pattern of repetitive period $ 4$ for the units digit.  $ (\minus{}1)^{1000}$ is simply $ 1$, so therefore $ 7^1\equal{}7$, which really is the last digit.

$ 7^4 \equiv 1 \pmod{10}$

Thus, $ 7^{7^{1000}} \equiv 7^{N} \pmod{10}$ whenever $ 7^{1000} \equiv N \pmod{4}$

We find that $ N \equiv (\minus{}1)^{1000} \pmod{4} \equiv 1 \pmod{4}$, so $ N\equal{}1$ is one possible $ N$, and we get:

$ 7^{7^{1000}} \equiv 7^1 \pmod{10}$

But this means the units digit of $ 7^{7^{1000}}$ is $ 7$

The reason it's $ \pmod{4}$ is because the units digit of the powers of $ 7$ cycle every $ 4$ powers. You might want to check out [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler's_Totient_Theorem]this[/url] and [url=http://www.artofproblemsolving.com/Wiki/index.php/Carmichael_function]this[/url].

Well, actually, if I read the question correctly, it's to find the last [b]two[/b] digits of $ 7^{7^{1000}}$. In other words, we need to find $ 7^{7^{1000}} \pmod{100}$.
[hide="Slow Solution"]
We know that $ \lambda(100)\equal{}20$, and $ \lambda(20)\equal{}4$, thus
$ 7^{1000}  \equiv (7^{4})^{250}  \equiv 1^{250}  \equiv 1 \pmod{20}$
Therefore,
$ 7^{7^{1000}} \pmod{100} \equiv 7^{1} \pmod{100} \equiv 7 \pmod{100}$.
The last two digits of $ (...(7)^{7})^{7})...)^{7}$ is $ \fbox{07}$[/hide]
[hide="Quicker Solution"]We can notice that $ 7^{4} \equiv 1 \pmod{100}$, so
$ 7^{7^{1000}} \equiv 7^{(\minus{}1)^{1000}} \equiv 7^{1} \equiv 7 \pmod{100}$
The last two digits of $ (...(7)^{7})^{7})...)^{7}$ is $ \fbox{07}$[/hide]


[quote="236factorial"]1. Find the area of a regular hexagon inscribed in a circle having a diameter of 20cm.

2. From a point 2 units from the circle, a tangent segment is drawn. If the radius of the circle is 8, find the length of the tangent segment.

3. A circle having a radius of 6cm is inscribed in a regular hexagon. Another circle whose radius is 6cm is circumscribed about another regular hexagon. Find the ratio of the area of the smaller hexagon to the area of the larger hexagon.[/quote]

[hide="Question 1"]If you divide the hexagon into 6 triangles, each touching the center of the circle, you find you have 6 equilateral triangles, all with sides equal to 10. So the area would be $150\sqrt{3}$[/hide]

[hide="Question 2"]If you draw a radius from the center of the circle to the point of tangency and connect the point outside of the circle to the center of the circle, you have a right triangle, with the right angle at the point of tangency. Using the Pythagorean theorum, the length of the tangent line is $6$
[/hide]

[hide="Question 3"]Using the same strategy as in Question 1, we find the area of the smaller hexagon is $54\sqrt{3}$.

For the larger hexagon, if you draw lines from each of the hexagon's corners to the center of the circle, you have 6 congruent triangles, each with a side that's also a side of the hexagon. Those are all equilateral triangles, with sides equal to the radius of the circle. So each side of the hexagon is equal to 6.

If you draw even more lines, this time from the point of tangencies to the center of the circle, you have 12 congruent triangles. Each of those triangles areas are 9. So the area of the entire larger hexagon is 108.

Therefore, the ratio should be $\frac{54\sqrt{3}}{108}=\frac{\sqrt{3}}{2}$[/hide]

Good, except I don't think the third one's right.

[hide="Question 3"]
The area of the smaller hexagon is $54\sqrt{3}$, as Drunken_Math explained.

You can use the same strategy (breaking hexagon into 6 equilateral triangles) for the larger hexagon as well.
In this case, the height of the equilateral triangle is $6$, which means the base is $6\cdot\frac{2}{\sqrt{3}}=4\sqrt{3}$.

So the area of the larger hexagon is $6\times\frac{(4\sqrt{3})^2\sqrt{3}}{4}=72\sqrt{3}$.

Therefore, the ratio is $\frac{54\sqrt{3}}{72\sqrt{3}}=\boxed{\frac{3}{4}}$[/hide]


calculus
derivative
function
analytic geometry
graphing lines
slope
calculus computations

I only received 1 of 3 points when I solved this equation.  The differential of f(x)=tan^-1(2x) is....
I found the answer to be 2/(4x^2+1).  The grader only marked a (dx) next to my answer.  Is the answer correct and I just needed the (dx)?  If so, why?

Yes.  A [url=http://en.wikipedia.org/wiki/Differential_%28infinitesimal%29]differential[/url] is not a [url=http://en.wikipedia.org/wiki/Derivative]derivative[/url].

Some of my students believe that partial credit should be determined by the [url=http://en.wikipedia.org/wiki/Edit_distance]edit distance[/url] between their solution and a correct solution. I do not subscribe to that theory.

Ok, so my answer is completely wrong?  Well what is the answer then?

Depends on what you mean by "completely wrong."  The answer is $ \frac{2}{4x^2 \plus{} 1} \, dx$, which is (as I have just said) not the same thing as $ \frac{2}{4x^2 \plus{} 1}$.  The latter is simply a function; it gives the value of the slope of the tangent line to $ f(x) \equal{} \arctan 2x$ at each point.

The correct answer, however, is an expression that describes an infinitesimal change in $ f(x) \equal{} \arctan 2x$ at a point.  The differential $ dy$ and the derivative $ \frac{dy}{dx}$ are related by the expression

$ dy \equal{} \frac{dy}{dx} \, dx$

as expected by the fraction notation.  mlok meant that while the two expressions differ only by the $ dx$, the confusion of one for the other implies a greater degree of conceptual misunderstanding than just one term.


prove that the only 3 primes that their difference is 2 are 3,5,7.

Let p.p+2.p+4 the three primes.As.we know (p+4)^2=(p+2)^2mod6 so 6 divide 2(2p+6) so 3 divide p+3 so p=3.

[hide="another soln"]
Assume the primes are $p,p+2,p+4$ since $p>3$, $p\equiv 1 \pmod 3$ or $p\equiv 2 \pmod 3$.  If $p\equiv 1$ then $p+2 \equiv 0 \pmod 3$ and then $p+2$ can't be prime.  If $p\equiv 2 \pmod 3$ then $p+4 \equiv 0 \pmod 3$.  Therefore, one prime must be divisible by 3.

[/hide]

my solution was very similar to calculuslvoer800's. that problem was a lot simpler than i thought it was going to be.
nice problem though  :lol:


Putnam
linear algebra
matrix
trigonometry
limit
induction
function

Let $ d_n$ be the determinant of the $ n\times n$ matrix whose entries, from left to right and then from top to bottom, are $ \cos 1,\cos 2,\dots,\cos n^2.$ (For example, $ d_3 \equal{} \begin{vmatrix}\cos 1 & \cos2 & \cos3 \\
\cos4 & \cos5 & \cos 6 \\
\cos7 & \cos8 & \cos 9\end{vmatrix}.$ The argument of  $ \cos$  is always in radians, not degrees.)

Evaluate $ \lim_{n\to\infty}d_n.$

Rank! The limit is zero, because the determinant is zero for $ n\ge 3.$

Let $ D_n$ be the matrix that we're taking the determinant of. Let $ E_n$ be the matrix
\[ E_n\equal{}\begin{bmatrix}e^{i}&e^{2i}&\cdots&e^{in}\\e^{i(n\plus{}1)}&e^{i(n\plus{}2)}&\cdots&e^{2in}\\\vdots&\vdots&\ddots&\vdots\\ e^{i(n^2\minus{}n\plus{}1)}&e^{i(n^2\minus{}n\plus{}2)}&\cdots&e^{in^2}\end{bmatrix}\equal{}\begin{bmatrix}1\\e^{in}\\\vdots\\e^{i(n^2\minus{}n)}\end{bmatrix}\begin{bmatrix}e^{i}&e^{2i}&\cdots&e^{in}\end{bmatrix}\]
It is evident from the factorization shown that $ E_n$ has rank $ 1.$ So also does its complex conjugate $ \overline{E_n}$ have rank $ 1.$ But then $ D_n\equal{}\frac12E_n\plus{}\frac12\overline{E_n}.$ The rank of a sum of matrices is less than or equal to the sum of the ranks, so $ \text{rank}\,(D_n)\le 2.$ Hence for $ n\ge 3,\ \det(D_n)\equal{}0.$

easiest problem this year.

The terms in the second row have the form $ \cos(n \plus{} k)$.

Then if the second row is multiplied by $ 2\cos{}n$, we will get $ 2\cos{}n{}\cos(n \plus{} k) \equal{} \cos(2n \plus{} k) \plus{} \cos{}k$, which is the sum of the first row and the third row and we are done.

[quote="transseries"][quote="EulerFish"]easiest problem this year.[/quote]

Generally, they put what they think is the easy problem as number 1.  So A1 should be the easiest A, and B1 the easiest B.[/quote]

People have different opinions than the test writers.

This reminds me of problem B5 from the [url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1999.pdf]1999 Putnam[/url].

As I said on the general discussion thread, I think that a significantly larger fraction of the ~3500 students taking the Putnam will get A1 than will get A3. Problem A1 will reward just sitting down and trying something. For A3 you either got it or you didn't - and with the misdirection of asking for a limit, you might have wandered off into estimates that will lead nowhere.

Irregular solution:

Cos(n) =<  a < |1|  where n is an integer. (since Pi is irrational, Pi - n =/= 0)

Regardless of which integers, Lim(Dn) =< lim( 2n*a^n ) = 0  as n -> infinity.

(2n*a^n is used, assuming the determinant is as large as possible you have a^n as the product of any one diagonal, and you have 2n diagonals involved.)

A couple problems with that, unfortunately: there is no constant $ a$ you can select for that purpose across all $ n$ ($ \cos n$ can become arbitrarily close to $ 1$ for $ n$ sufficiently close to an integer multiple of $ 2\pi$); and the determinant formula is on the order of $ n!$, not $ 2n$, 'diagonals' (the formula involves permutations). You may have mistakenly tried to generalize the $ 3 \times 3$ determinant formula.

I tried to show that a subsequence of $ d_n$ went to $ 0$, by taking $ n$ close to a multiple of $ 2\pi$ and then subtracting the first row from the other rows. Of course, that doesn't even establish a limit  :( .

Yes, I did mistakenly generalize the 3x3 formula, and, that disproves me entirely!

Good show!

Alternatively, one could show that \[ det\left(\begin{matrix}\cos a&\cos a\plus{}b&\cos a\plus{}c\\ \cos a\plus{}d&\cos a\plus{}b\plus{}d&\cos a\plus{}c\plus{}d\\ \cos a\plus{}e&\cos a\plus{}b\plus{}e&\cos a\plus{}c\plus{}e\end{matrix}\right)\equal{}0\] for all reals $ a,b,c,d,e$ by expanding and such, but that would take a long time. I know of someone who did that. (The result would follow since every n by n matrix (n>2) can be decomposed into three-by-three matrices)

(Don't quote me on this, I haven't actually verified it  :blush: )

If I just write down like the first 10 terms of the determinant and found it's approaching 0, and say in the end it's 0, would I get 1 mark?

what if I　say because cos can only take on the value of 0 to 1, and the x in cos(x) will repeat the cycle of 2pi, thus every number between 0 to 1 will have a chance to be a term in the determinant. As n becomes really large, every number between 0 to 1 essentially gets the same chance of being a term in the determinant, and because there's equal number of the + sign and - sign in the determinant, in the end the terms cancel each other out. Therefore it will be 0 in the end.
would I get mark from that?

[quote="Eminaz"]If I just write down like the first 10 terms of the determinant and found it's approaching 0, and say in the end it's 0, would I get 1 mark?[/quote]

unlikely

[quote]what if I　say because cos can only take on the value of 0 to 1, and the x in cos(x) will repeat the cycle of 2pi, thus every number between 0 to 1 will have a chance to be a term in the determinant. As n becomes really large, every number between 0 to 1 essentially gets the same chance of being a term in the determinant, and because there's equal number of the + sign and - sign in the determinant, in the end the terms cancel each other out. Therefore it will be 0 in the end.
would I get mark from that?[/quote]

unlikely

[quote="Eminaz"]what if I　say because cos can only take on the value of 0 to 1, and the x in cos(x) will repeat the cycle of 2pi, thus every number between 0 to 1 will have a chance to be a term in the determinant. As n becomes really large, every number between 0 to 1 essentially gets the same chance of being a term in the determinant, and because there's equal number of the + sign and - sign in the determinant, in the end the terms cancel each other out. Therefore it will be 0 in the end.
would I get mark from that?[/quote]

Actually, if you take a matrix where every entry is chosen to be the cosine of a random number in $ [0, 2\pi]$ the determinant will grow roughly like $ \sqrt {n!}*c^{ \minus{} n}$ for some $ c$.   Intuitively this is because the determinant is the sum of $ n!$ "nearly" independent terms each having a magnitude whose expectation is only exponentially small in $ n$  (proving it takes quite a bit more work).  So the fact that this limit is $ 0$ is based on the algebraic structure of the cosine function and the matrix, and not on the fact that cosine sometimes takes on small values.

We could alternatively use the identity $cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$. Setting $z = e^i$, we can divide this determinant into the sum of $2^n$ determinants where the $i$th row is either $z^{(i-1)n+1}, z^{(i-1)n+2}, \dots  z^{ni}$ or $z^{-(i-1)n-1}, z^{-(i-1)n-2}, \dots  z^{-ni}$. Each of these individual determinants will vanish to $0$ as there exist two rows that are multiples of each other.


[quote="hophinhan"]Given $ \triangle ABC$ inscribed in a circle $ (O)$. $ A_1\ ;B_1\ ;C_1$  be the tangency points of the incircle with  $ BC\ ;\ CA\ ;\ AB$ , respectively. $ H$ is orthocenter of $ \triangle ABC$ .  $ A_2\ ;\ B_2\ ;C_2$ be the intersection of $ HA_1\ ; HB_1\ ; HC_1$ with $ (O)$ , respectively. Prove that : $ AA_2\ ;\ BB_2\ ;\ CC_2$ are concurrent.[/quote]
It's not true.[hide]Try it for $ \triangle ABC$ with $ A \equal{} 15^\circ, B \equal{} 45^\circ, C \equal{} 120^\circ.$ [/hide]

What can I say now ...  :o ..I can't believe it  :( ...  $ 178^o < \hat{APA_2} \le 180^o$

I'm so sorry about my mistake in problem...Sorry wasting your time :|

Thanks you very much ! Yetti


Given that the diameter of a convex polygon M is 1 unit.
Show that the area of M is less than 1 square unit.

Remark. The diameter of a convex polygon is defined as the [i]maximum distance [/i]between any two points inside the polygon.

the maximum area is enclosed by the polygon is when when it is circle. and the area is $\pi r^2$.hence maximum area is $\frac \pi_{4}$. which is less than one.

[quote="ashwinrk_jain"][b]the maximum area is enclosed by the polygon is when when it is circle[/b]. and the area is $\pi r^2$.hence maximum area is $\frac \pi_{4}$. which is less than one.[/quote]

The statement looks obvious and this seems to be the key step in your proof. Can you explain why the polygon is enclosed by a circle whose radius is $\frac 1 2$ ?  :)

[hide="Solution"]The circumscribed circle has an area $\ge$ that of the polygon, and that area is itself $\le \frac{\pi}{4}$. [/hide]

The thing luimichael is hinting at is that a polygon of diameter $d$ cannot necessarily be fit inside a circle of diameter $d$.  For instance, a pentagon of diameter (also diagonal) $d$ has a circumscribed circle of diameter $\sqrt{2-\frac2{\sqrt5}}d\approx 1.05 d$.  So that kills both of your arguments so far -- it's not obvious a figure of diamater $d$ will have area less than a circle of diameter $d$.

I was afraid of that. :blush:

[hide="Lets try this then!"]
[b]Claim:[/b]  The polygon fits into a square of side length $1$.

Proof:  Suppose a vertex $P$ on the polygon does not fit into the square.  Find the vertex $Q$ on the polygon farthest from $P$.  If $Q$ does not touch the square, shift the square toward $P$ until it does.  If $Q$ does touch the square, then $PQ > 1$ - contradiction.

It remains to show the polygon cannot attain an area of $1$.  If it fills the square it will clearly have diameter $\sqrt{2} > 1$.  If it is outside the square it will still have diameter $>1$.
[/hide]


Two standard six-sided dice are rolled. What is the probability that the product of the two numbers is 20 or larger?

A.2/9 	B.5/18	 C.5/36	   D.1/9	      E.1/6

[hide=" A: click to reveal hidden content"]

There are $ 36$ possible outcomes; $ 8$ of these, $ (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)$ result in a score of $ 20$ or more. Therefore the required probability is $ \frac {8}{36} \equal{} \frac {2}{9}$, that is option A.

[/hide]


So I'm taking MAT 215 (Analysis in a Single Variable) at Princeton this September.

Has anybody taken it? How is it? I'm incredibly rusty at math, so I'm sort of apprehensive.

Woah, I could be in your class.  That would be awesome.

(Sorry, I can't answer your question.)

Dae, you can do that via WWP? nice ..

Another useless reply, but I may end up in your/Matt's class. As for difficulty level, from asking around I've heard everything from the course being really easy to hard. @mathemonster: MAT 215 is more theoretical and proof-oriented than, say, a BC calc class [which I suppose roughly corresponds to MAT 104].


I heard my name was on one of the questions.

just out of curiousity, is there anyone who has a copy of the test?
is jan still writing the problems?

Dear Geehoon,

I do not have the questions, but the Soph Frosh contest was written by David Linker.  In this set of questions, I believe "Sho the great" was also featured.

Cheers,
Jack

i was just wondering, does sophmores take the soph-fresh contest or the junior contest?

The League does not have any restrictions against younger students taking the more advance competitions  The school themselves can make the decision on whether that's the appropriate thing to do.


abstract algebra
superior algebra
superior algebra unsolved

Let $ I$ be an ideal in a commutative ring $ R$ such that $ I^k \equal{} \{0\}$ for some $ k \geq 1$.
Let $ M$ and $ N$ be $ R$-modules and let $ \phi: M \to N$ be an R-module homomorphism. Show that if the induced map $ \psi: M/IM \to N/IN$ is surjective, then $ \phi$ is surjective.


This is an AMC problem but it's about the difficulty of MC and the MC forums are a lot more active, so I'll just post it.
Points A=(3,9), B= (1,1), C = (5,3) and D=(a,b) lie in the first quadrant and are vertices of quadrilateral ABCD. The quadrilateral formed by joining the midpoints of AB, BC, CD, and DA is a square. What is the sum of the coordinates of point D?

Ok, I thought that if a square is formed by the midpoints of a quadrilateral, that quadrilateral has to be a rhombus. Clearly the vertices listed aren't the vertices of a rhombus. I checked the solution section, and it says that the coordinates for D are (7,3). This doesn't make any sense. Am I missing something?

The quadrilateral doesn't have to be a rhombus. Why would that be necessary?

BTW, (7,3) is absolutely right. Try checking the midpoints of those four points. They'll form a square.

Hm...I see where I went wrong. I assumed the quadrilateral was convex, but it's concave. However, for a square to be formed by the midpoints of a convex quadrilateral, then the quadrilateral must be a square, right?


integration
limit
inequalities
real analysis
real analysis unsolved

How to show that $\mu (x): =Ce^{-1/(1-|x|^{2})}$ as $|x|<1$ and $\mu(0)=0$ elsewhere is $C^\infty$ as $C$ is chosen s.t. $\int_{-\infty}^\infty \mu(x)\,dx=1$

For $\varepsilon>0$ define $\mu_\varepsilon: =\varepsilon^{-1}\mu(\varepsilon^{-1}x)$. Show that
\[\lim_{\varepsilon\to 0}\mu_\varepsilon*f(x): =\lim_{\varepsilon\to 0}\int_{-\infty}^\infty \mu_\epsilon (y)f(x-y)dy=f(x)\]
for all $x\in\mathbb{R}$.

You left out an important hypothesis. What properties is $f$ supposed to have?

If $f$ is continuous on $\mathbb{R},$ then we can conclude that $\mu_{\varepsilon}*f(x)\to f(x)$ pointwise everywhere and uniformly on compact sets.

If $f\in L^{1},$ then we can conclude that $\mu_{\varepsilon}*f\to f$ in $L^{1}$ and pointwise almost everywhere.

Sorry. I forgot the conditions. $f$ is supposed to be continuous and bounded.

Boundedness is not necessary, since you have a compactly supported kernel. Given $x\in\mathbb R$ and $\delta>0$, choose $\epsilon>0$ so that $|f(x')-f(x)|<\delta$ whenever $|x'-x|<\epsilon$. Multiply this inequality by $\mu_{\epsilon}(x-x')$ and integrate over $x'$: the result is $|(\mu_{\epsilon}*f)(x)-f(x)|\le \delta$.

That convergence is pointwise everywhere, but not necessarily uniform. Boundedness doesn't get you uniform convergence either- you would need uniform continuity of $f$ to get uniform convergence of the convolutions.


limit
linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let A=a(i,j) be a (n,n) matrix which satisfies (1) :the spectral radius rho(A) is reached by only one eigenvalue s. Then there exist P, an invertible matrix, and D=diag(s,U) s.t. rho(U)<rho(A), A=PDP^-1.
P=p(i,j),P^-1=q(i,j),A^n=an(i,j) implie an(i,j)~p(i,1).s^n.q(1,j) when n->infinity  (we suppose p(i,1).q(1,j)<>0).
 Thus if we have also p(k,1)q(1,l)<>0 then an(i,j)/an(k,l)->(p(i,1)q(1,j))/(p(k,1)q(1,l)).
Here A satisfies (1) with a real s.
    TIRING METHOD: an(2,1)/an(3,1)->p(2,1)/p(3,1)=y/z=t where x+2y+3z=sy,x+y+2z=sz ([x,y,z] is an s-eigenvector); thus (2) s=(t+1)/(t-1); s is a root of the characteristic polynomial of A: z^3-6z^2+3z-2 and t is a root of the image of this polynomial by the transformation (2): z^3-3. The only real root of this last polynomial is 3^(1/3).....
   MAPLE MAGIC METHOD: If the a(i,j) are in Q  then s is algebraic of degree <=n  over Q. We may choose  the preceding P and U in Q[s]; thus the q(i,j) are in Q[s] and the requested limit of an(i,j)/an(k,l) is in Q[s] and is algebraic of degree<= n over Q.
    You open Maple 8,9 or 10. Take 40 digits "Digits:=40:". Calculate A^100 and the numeric value of a100(2,1)/a100(3,1):
1.44224.....Call PSLQ: "with(IntegerRelations):x:=1.44224....: PSLQ([1,x,x^2,x^3]);"
PSLQ gives you a linear relation with coefficients in Q (if there exists!) between 1,x,….,x^3. If you don't know the degree of the extension then you must try several degrees and if the degree is big you must take sometimes 100 or even 200 digits.
 Here Maple says "[-3,0,0,1]" id est -3+x^3=0…….
   Example with n=4: a=[33,10,7,-89,65,12,-25,-96,50,-60,-42,7,-89,-70,34,-68]  line by line. Digits:=200:  Calculate a^10000: then limit(an(4,2)/an(3,3))=t which is a root of a polynomial of degree 4; Maple gives this polynomial (time<1”).
 (The constant coefficient is -85237248275794968200).
With this method you can even calculate t as a polynomial in s of degree 3 with coefficient in Q.
PS: PSLQ is a variant of LLL due to Lenstra and all.

[quote="loup blanc"][b]...The only real root of this last polynomial is 3^(1/3).....[/b][/quote][b]
Indeed this is the limit, Very nice solution and thanks for informations ! [/b]


prove or dispove  that for $a_1,a_2,...,a_n >0$

$\frac{a_1}{a_2}+\frac{a_2}{a_3}+\frac{a_3}{a_4}+...+\frac{a_n}{a_1}
\geq \frac{a_1+a_2+...+a_n}{\sqrt[n]{a_1a_2...a_n}}$

I hope that at least one of my conjectures become true , just for a change :blush:

I think that it is not true for $n \ge 5$. :)

Yes, I have the same conclusion.

I tried to take a = b = c = x^5, d = 1/x^5, e = 1 and analys this case

Then Sum = 2 + x^10 + 2/x^5 - (3x^5 + 1/x^5 + 1)/x^2
               = 2 + x^20 + 2/x^5 - 3x^3 - 1/x^7 - 1/x^2
Take x = 0.5 ---> Sum < 0.


geometry
inequalities
ratio
geometric series

If the process in the accompanying figure is repeated infinitely many times and the largest equilateral triangle has side length 1, what is the sum of the areas of all of the circles?

47.  [hide]n goes from 1-500 in this problem. Using the inequality given, we get  :sqrt:501 -  :sqrt:500 +  :sqrt:500 ... + :sqrt:2 - :sqrt:1 < S <  :sqrt:500 -  :sqrt:499 +  :sqrt:499 ... -  :sqrt:1 +  :sqrt:1 - :sqrt:0 ;   :sqrt:501 - 1 < S <  :sqrt:500 - 0. It follows that 21 < S < 23, which fits inside the answer choice d.[/hide]

Hints for 1. [hide]Median Concurrence Theorem, [tex]\displaystyle \frac{a}{1-r}[/tex][/hide]

[hide]The first one:  First I found the area of the first circle.  Using A=rs to find the radius I got the area to be [tex]\frac{\pi}{12}[/tex]  Since the radius is 1/3 of the next biggest circle in the next smallest circles the areas will be 1/9 that of the circle preceding it.  Don't forget that there are also 3 times as many circles so the ratio of areas from one step to another is 1/3.  So we have to find [tex]\frac{\pi}{12}+\frac{\pi}{(12)*3}+\frac{\pi}{((12)*3)*3}+\cdots[/tex] Using the sum of a geometric series we get [tex]\frac{\frac{\pi}{12}}{1-1/3}=\frac{\pi}{8}[/tex][/hide]

These are *very nice* problems!  What is the source?

[quote="Robert0201"]If the process in the accompanying figure is repeated infinitely many times[/quote]

I think there is some ambiguity in the statement of this problem.  If "the process" were repeated once more, would the resulting figure contain 40 circles or 22 circles.  I.e., is "the process" applied at the vertices of all of the smallest equilateral triangles or is "the process" applied only at the tip of the three vertices of the largest equilateral triangle.

[Personally, I prefer the former interpretation, but ambiguities should always be avoided.]

I dont think theres any ambiguity rcv.  After each circle draw three lines tangent to it to create 3 more equilateral triangles.  Draw a circle in these triangles and repeat (not just at the vertices of the largest equilateral triangle).


During the year 1985, a convenience store, which was open 7 days a week, sold at least one book each day, and a total of 600 books over the entire year. Must there have been a period of consecutive days when exactly 129 books were sold?
I know how to solve this task by using the pigeonhole principle, but I think there is a solution using euler's phi function. Can anyone help me?

[quote="Andrej_pp"]but I think there is a solution using euler's phi function. Can anyone help me?[/quote]

What would make you think that?   :huh:

If this help to somebody: http://mathworld.wolfram.com/TotientFunction.html


Κι αν υπάρξει χώρος, ίσως μπει και στο βιβλίο.  :)

Λοιπόν, το ζητούμενο γνωστό! Να αποδείξετε ότι $0,999... = 1$, όπου τα  " ... " δηλώνουν άπειρο πλήθος από 9.

[u]Απόδειξη 1:[/u]

Έστω $a=0,999...$. Τότε,
$10a=9,999...$
$10a=9+0,999...$
$10a=9+a$
$9a=9$
$a=1$

[u]Απόδειξη 2:[/u]

$1= \frac{9}{9}= 9 \frac{1}{9}= 9 * 0,111... = 0,999...$
[size=75]Αν και είναι καλό αποδείξεις με κλάσματα να αποφεύγονται, γιατί ανάλογες κινήσεις μπορεί να μας οδηγήσουν σε λάθος συμπεράσματα.[/size]

[u]Απόδειξη 3:[/u]

[color=red]edit:[/color] Μετά από μία υπόδειξη που μου έγινε, θα προσθέσω έναν αυστηρό ορισμό για το τί εννοούμε γράφοντας $0,999...$ .

Έστω η ακολουθία με γενικό όρο $a_{n}= 9 * 10^{-n}, n=1,2,...$ . Τότε ορίζω τη σειρά της $\{ a_{n}\}$, με n-οστό όρο $S_{n}= a_{1}+a_{2}+...+a_{n}$.

Αν δεχτούμε ότι μία σειρά συγκλίνει στο $s$ εάν $\lim_{n \to \infty}S_{n}= s \in \Re$, τότε μπορούμε να συμβολίσουμε $\sum_{n=1}^{\infty}a_{n}= s$. Το $s$ καλείται το [i]άθροισμα[/i] της σειράς $\sum_{n=1}^{\infty}a_{n}$.

Αν συμφωνούμε ότι το άθροισμα αυτής της σειράς είναι ο αριθμός $0,999...$, έχουμε ότι
$0,999... = \lim_{n \to \infty}[9*(10^{-1}+10^{-2}+...+10^{-n})] =$
    $= 9 \lim_{n \to \infty}(\frac{1-10^{-n}}{10-1}) =$
$= \frac{9}{9}\lim_{n \to \infty}(1-10^{-n}) =$
$= 1-\lim_{n \to \infty}10^{-n}=$
  $= 1$

[u]Απόδειξη 4:[/u]

Και μία ακόμα απόδειξη, όταν είχα ρωτήσει πώς μπορούμε να δείξουμε ότι ο $0,999...$ δεν ανήκει στο $(0,1)$.

Συμβολίζουμε με $x$ τον 0,999... και με $x_{n}$ τον 0,999....9 (n εννιάρια).
Στο σημείο αυτό, δεχόμαστε ότι $x>x_{n}$ για κάθε φυσικό $n$. Δηλαδή, ότι $x>1-\frac{1}{10^{n}}$.

Έστω ότι $x<1$.
Τότε έχουμε $(\frac{1}{10})^{n}> 1-x$ ή ισοδύναμα ότι $10^{n}< \frac{1}{1-x}.$
Γνωρίζουμε ότι $10^{n}> n$ για κάθε φυσικό $n$.
Οπότε μπορώ να γράψω $n < 10^{n}< \frac{1}{1-x}$ και τελικά $n < \frac{1}{1-x}$.
Το τελευταίο λέει ότι κάθε φυσικός αριθμός $n$ θα είναι πάντα μικρότερος από το $\frac{1}{1-x}$.
Αν όμως δεχτούμε ως αληθές ότι δεν υπάρχει άνω φράγμα στους φυσικούς, τότε οδηγούμαστε σε άτοπο.

Άρα, θα έχουμε ότι $x \ge 1$. Δε νομίζω να θέλει κανείς να συζητήσουμε αν $x > 1$.  :huh:

Σχετικό άρθρο από Wikipedia: [url]http://en.wikipedia.org/wiki/0.999...[/url]

μπορει να μπει εισαγωγη ή και εξωφυλλο , ποιος ειναι μεσα ? :lol:


geometry
AMC
AIME
USA(J)MO
USAMO
inequalities
geometric transformation
\/closed

We will soon be scheduling our Geometry Olympiad Problem Series, which we expect to offer starting at the end of February.  If you are considering taking the course, please list here what times/days of the week you are available (with time zone, please).

This course will examine olympiad-level geometry problems, and is most appropriate for students preparing for the AIME & USAMO (most problems will be olympiad-level in difficulty).  If you have any questions about the course, please post them here.

thanks you very much. i've been hoping for something like this (and something for inequalities) for a long time. i am free Mon, Tue, Fri evenings. weekends - umm...saturday evenings would be great. sunday evenings too would probably be good. if those don't work, i could probably free up some other times (heck i'd come on during lunch at school but one hour's probably not long enough)

For more information on the course:

The classes will be 2 hours long, once a week.  The topics covered will be similar triangles, power of a point, cyclic quadrilaterals, homothecy, locus, construction, vectors, inversion, collinearity, concurrency, and more.  At the moment, the times under consideration are 1-3PM PST on Sundays, and 4:30-6:30 PM PST Monday, Tues, or Wed.  The course will start in February and will be scheduled in the next two weeks.

Sundays would be best for me because I live in MA, and I don't get out of school until 3:30 pm.

I am more or less comfortable with most of the geometry in AoPS I. I can understand most of the geometry in AoPS II, but I can't use it to do a problem unless it was made extremely obvious.

The class was already scheduled before the last two posts were put up.  The class is on Monday evenings...

[quote="interesting_move"]I can't do it on Sunday because I have another math class at that exact time.  Other times are fine though.

-interesting_move[/quote]

what kind of a math clats do you have on a sunday?


[quote="mathwizarddude"]Find the first digit before and after the decimal point in the number:

$ {(\sqrt {2} \plus{} \sqrt {5})}^{1000}$[/quote]

$ {(\sqrt {2} \plus{} \sqrt {5})}^{1000}\equal{}{(7 \plus{} 2 \sqrt {10})}^{1000}$

Now sequence $ a_n\equal{}{(7 \plus{} 2 \sqrt {10})}^{n}\plus{}{(7 \minus{} 2 \sqrt {10})}^{n}$ is given by

$ a_0\equal{}2, a_1\equal{}14, a_{n\plus{}2}\equal{}14 a_{n\plus{}1}\minus{} 9a_n$

their unit digits forms periodic sequence $ 2,4,8,6, \ \ 2,4,8,6, \ \ 2,4,8,6 \ \ ......$

So unit digit of $ a_{1000}\equal{}2$.

As $ {(7 \minus{} 2 \sqrt {10})}^{1000}<\frac{1}{10}$ we have the first digit before the decimal point $ \equal{}1$ and that after the decimal point $ \equal{}9$


Let $ \sum a_n$ a series with real terms. Let $ B\equal{}\{\varphi: \mathbb N\to\mathbb N,\,\varphi\text{ is a bijection}\},$ and let $ f: B\to\mathbb R$ defined by $ f(\varphi)\equal{}\sum a_{\varphi_(n)}.$ Does exist a series $ \sum a_n$ such that $ f$ is a surjection?

I wanted to know if this is okay:

Consider a conditional convergent series, then by the Riemann Series Theorem, then given $ b\in\mathbb R,$ exists a rearrangement $ \sum b_n$ of $ \sum a_n$ that satisfies $ \sum b_n\equal{}b$ and also a bijection $ B$ (already given), then $ f(\varphi)\equal{}\sum a_{\varphi_(n)}\equal{}\sum b_n\equal{}b,$ which is the definition of a surjection.

That's correct. At least as correct as an answer to this strange question can be. (The function $ f$ is not defined on the entire set $ B$ when $ \sum a_n$ is conditionally convergent.)


A circular disc is partitioned into $n$ equal sectors .Each sector is to be coloured with one of the four different colors provided that no two adjacent sectors are of same color.If $a_{n}$ is the number of ways to colour the disc with $n$ sectors. Find an explicit formula for  $a_{n}$

Recursion?
Seems like it will be a messy function though. Looks like good problem :)


- http://www.payvand.com/news/07/jul/1157.html
- http://www.persianstudents.org/

Sad news in the middle of the ongoing competition.
[url=http://www.ipho2007.ir/News/NewsE/Item.asp?RowNo=1&CurPage=1&NewsID=43&qsFromSearch=]Here[/url] is an other link.


induction
ARML
search
Euler
algebra
difference of squares
special factorizations

Whats a nice way to prove this:

Prove that, if $n$ is composite, then $\displaystyle 2^n - 1 $ is not a prime number.

Induction I suppose.

Or can I use contraction here?

Because it's really easy when you use contraction.

[hide="Contraction"]

Let's assume $2^n - 1$ IS a prime for composite $n$.

Then taking $n=1$, we get: $2^1 - 1 \Rightarrow 2 - 1 = 1$.

But by the definition of prime number, 1 is not prime since it only has one divisor (remember, the prime number is when the divisor is itself and 1 so two numbers but for 1, itself = 1, so only one divisor).

This contradicts our assumption so $2^n - 1$ IS NOT a prime for composite $n$.[/hide]

I don't really think you proved anything by saying that... it says for some composite $n$ so $1$ isn't even one of them.

[hide]
If $n$ is even, it follows from Fermat's Little Theorem.
$2^n-1 \equiv 0 \mod{(n+1)}$.

If $n$ is odd, I'm not sure yet.[/hide]

1 isn't composite ;), but overlooking that...

Your counterexample only proves that $2^n-1$ is not always prime for composite $n$, not that $2^n-1$ is always not prime.

One way to do contradiction is assume $2^n-1$ is prime, and show that for any composite $n$, it has another factor besides 1 and itself.

Another problem from the Chicago ARML homework. Tsk tsk... anyways, here's a (big) hint:

[hide]${x^{ab}-1 = (x^a-1)(x^{a(b-1)}}+x^{a(b-2)}+\cdots + 1)$[/hide]

Figure out why that's true, then you should be able to solve your problem.  :)

if n=ab  (a,b>1)  then you can factor 2^ab - 1 as
   2^a=x    x^b -1 = (x-1)(x^b-1 + x^b-2 + ... + x + 1)

You guys are making this [b]way[/b] too hard!

[hide]Write 2^(xy)-1 in base xy (here I use x=4,y=3 as an example):

aaaaaaaaaaaa (xy times) Break this up into:
aaaa aaaa aaaa (each sequence is x long and there are y such sequences) Therefore,
2^(xy)-1 (base 10)=aaaa*100010001 (base xy).

Note that a appears x times (4), and each sequence of 100... has a 1 and x-1 (3) 0s. This appears y-1 (2) times, followed by a 1.[/hide]

[hide]
Well you know that if n is even, it is a difference of squares, and if a multiple of 3, it can be factored easily as well.  In fact, for any a^n-1, you can factor to
(a-1)({a^[n-1]}+{a^[n-2]}+...+1)
Normally in 2^n-1, if you did that, a-1 would be 1.  But if n is composite, then 2^n-1 can be expressed as
(2^a)^b-1
Which is
(2^a-1)({(2^a)^(b-1)}+...+1).  Therefore, if n is composite, 2^n-1 is also composite.[/hide]

Never heard of "contraction" as a proof method but anyways... :D

[hide="My solution"]
[i]Case 1:[/i] $n$ is even

If this is the case then we have $n=2k$ for some positive integer $k$.  And thus:

\begin{eqnarray*}
2^n-1 &=& 2^{2k}-1\\
&=& 4^k-1
\end{eqnarray*}

Looking at this $\mod 3$ we get $1^k-1$ which is $0$ and thus the number is a multiple of 3.

[i]Case 2:[/i] $n$ is odd

Since $n$ is odd in this case and it is given in the problem that $n$ is composite we conclude that $n$ must be equal to the product of two odd numbers.  Thus:

\[n=(2k_1+1)(2k_2+1)\]
for some nonnegative integers $k_1$ and $k_2$.  Thus:

\[\begin{eqnarray*}
2^n-1 &=& 2^{(2k_1+1)(2k_2+1)}-1\\
&=& 4\cdot 4^{k_1}\cdot 4^{k_2}-1\\
&\equiv & 1 -1\\
& \equiv & 0 \mod 3\]

QED
[/hide]

[quote="seamusoboyle"]For F*** sake guys!!! :)

Just use the fact that

$(a-b)$  divides  $(a^n-b^n)$

Not that hard!!!! :lol:[/quote]

That doesn't actually help since a-b=1.  :P

When n is composite we can raise the 2 to some factor of n and still be left with a power.

Ah, sorry. I thought you meant do that directly.

But yeah, I think that's what K81o7 did.

Yeah...
Now how would one prove that it's prime if n is a prime?  That's another thing to think about.  The proof is probably really simple though.

[quote="JBL"][quote="amirhtlusa"]that is not ture. if $n$ is prime number then $2^n-1$ is not always a prime.
example:
$2^{11}-1=2047=23*89$ but for many prime $n$'s, $2^{n}-1$ is a prime number these numbers are called Mersenne numbers ($M_n$).[/quote]

It is not known if there are infinitely many Mersenne primes.  However, most of the largest known prime numbers are Mersenne primes.  The Great Internet Mersenne Prime Search uses a distributed network to test possible primes -- you can download a program to have your computer take part, if you like!  Just do a search for "Great Internet Mersenne Prime Search" and it'll pop up.[/quote]
did i say there are infinte mersenne primes???
 i said there are many (not infinite, since it's unknown), prime $n$'s that M_n is a prime!!

I wasn't contradicting you, I was adding additional information that I thought was relevant.

whoops that's not what i meant.  There's this one thing though...
Maybe it's
2^([2^n]-1)-1 is prime or something like that.

hmm...that looks awfully familier.
i don't know if this relates or not but euler had proved that the number $2^{2^{n}}+1$ is not a prime for all primes $n$.(because $2^{32}+1$ devides $641$ ). and your number $2^{{2^n}-1}-1$can be rewritten as $\frac {(2^{2^{n}}+1)-3}{2}$

There is something or the other...or maybe i'm just confusing things...

isn't it too long for such an easy problem? :)  i mean legth of the discussion

the original problem was indeed easy, but you might want to read all of the responses first, before submititng a reply, (we are talking about a new problem here now!! ;) )

amirhtlusa
         Mersenne numbers are too well known to discuss them so much :)

well then prove this:
[quote="seamusoboyle"]What if $n$ is a Merenese prime!?! Is it always prime then? convince me! :P[/quote]

all Mersenne numbers are prime, but not all primes are mersenne numbers
  mersenne thought that there are infinitely many such numbers, but it is still not known was he true or no


calculus
integration
trigonometry
algebra
polynomial
inequalities
calculus computations

Integral: [x + 1/x]^1/2  dx

PLEAS....
   TRY AT LEAST

[quote="Kent Merryfield"]Try rewriting it at

$\int\sqrt{x+\frac1x}\,dx=\int\frac{\sqrt{x^2+1}}x\,dx.$[/quote] :?

Oops - bad algebra. Forget that I said that.

Maple says that it has no elementary antiderivative.

well blahblahblah,
   I know that it seems it has no antiderivative thats why i asked the people here.
also its a challenge of an A+ in my class.

and the Dr. said ask who ever you know if any one can..

and as you see am trying

[quote="lost"]thx Kunny u helped

and it was exactlu given to me as the 1st one U wrote ( the left one)[/quote]

$\sqrt{x+\frac{1}{x}} \not\equiv \frac{\sqrt{x^2+1}}{x}$.

The integral doen't have an elementary antiderivative, was it given to you as an indefinite integral or did it have limits?

well, it may not have an elementary primitive.

Mathematica gives
$\int \sqrt{x+\frac{1}{x}} \ dx = \frac{1}{3(1+x^2)}\left(2x\sqrt{\frac{1}{x}+x}\left(1+x^2+2(-1)^{1/4}\sqrt{1+\frac{1}{x^2}}\sqrt{x}F\left(i\sinh^{-1}\left(\frac{(-1)^{1/4}}{\sqrt{x}}\right),-1\right)\right)\right)$

where $F$ denotes the elliptic integral of the first kind.

From mathworld
$F(\phi, k)=\int_0^\phi \frac{d\theta}{\sqrt{1-k^2\sin^2(\theta)}}$

Indeed Jimmy. After some simplification though you can reduce the problem to $x.EllipticF\left(\sqrt{1-x.I},\frac{1}{\sqrt{2}}\right)$

And hey... what happens now if we do a polynomial approximation (taylor) around x=-1? :) yes indeed, we get a simple but perfect(=infinite) approximation:

simplify(taylor(f,x=-1,10),symbolic);

[code]series((-EllipticF((1+I)^(1/2),1/2*2^(1/2)))+1/4*2^(1/2)*(EllipticF((
1+I)^(1/2),1/2*2^(1/2))*(1+I)^(1/2)*2^(1/2)*(2-2*I)^(1/2)-1+I)*(x+1)+(
1/8-1/8*I)*2^(1/2)*(x+1)^2+(1/16-1/16*I)*2^(1/2)*(x+1)^3+(1/32-1/32*I)
*2^(1/2)*(x+1)^4+(9/640-9/640*I)*2^(1/2)*(x+1)^5+(7/1280-7/1280*I)*2^(
1/2)*(x+1)^6+(1/512-1/512*I)*2^(1/2)*(x+1)^7+(1/1024-1/1024*I)*2^(1/2)
*(x+1)^8+(23/24576-23/24576*I)*2^(1/2)*(x+1)^9+O((x+1)^10),x=-1,10);[/code]

Now if we could just find out the pattern in the coefficients there, and we find the infinite-summation for it, then we have solved this unsolvable problem :D

In any case we have reduced it to an elementary problem with some non-elementary constant, which we can call a or something.

[quote="lucipopa"]Here we have an binom integral:
$\int{x^m}{(ax^n+b)}^p{dx}$
If we have that $p$,$\frac{m+1}{n}$,$\frac{m+1}{n}+p\notin{\mathbb{Z}}$
then this integral can't be solved by an elementary method.
Cebisev proved this for the first time[/quote]
Chebyshev

[quote]Chebyshev[/quote]
There are many different versions of the name around. I wouldn't take any as definitive except the one that uses the Cyrillic alphabet.

yes but cebisev isn't one of the options as far as I know ;)


advanced fields
advanced fields unsolved

I was wondering whether anyone knew the answer to these two questions:

1. When using the Wilcoxon Signed-Rank test to determine whether a population mean is 10, what are the steps to be performed and what is their order?

2. Are there any assumptions made for the Wilcoxon Signed-Rank test? For the Wilcoxon Rank-Sum Test?

Thanks for your help.

For #1, I got the following answer:

1) Compute the number of observations that are below 10 ($ W^ \minus{}$)
2) Compute the number of observations that are above 10. ($ W^ \plus{}$)
3) Define X as the minimum between $ W^ \minus{}$ and $ W^ \plus{}$.
4) If the sample size is large enough, compute

$ z \equal{} {\frac {X \plus{} 0.5 \minus{} \frac{n}{2}}{\frac {\sqrt {n}}{2}}}$

and then compute the critical value $ z_c$.

(5) If the sample size is not large enough, we a special for the critical value for X.

For #2, I got this answer:

 The Wilcoxon Signed-Rank test requires the differences to be independent and drawn from a symmetrical distribution.
The Wilcoxon Rank-Sum Test requires that the subjects within samples are independent and are identically distributed

Are these correct or is there anything I need to add to them? Thanks.


calculus integration geometry conics ellipse function trigonometry

I'm pretty new to calculus, but am currently building on a code that someone has given me to model plasma physics. Unfortunately, I'm not completely familiar with all of the calculus issues involved. So my question is, when you calculate an elliptic integral (is that the right way to say it?) whether it be the first or second type of elliptic integral, is the result merely the area of said elliptical?

Thanks

well, elliptic integrals originally started with the problem of finding the circumference of an ellipse, and cannot be expressed in terms of the various elementary functions we are familiar with.

i don't know much about the subject, all i know is evaluating certain values of the jacobi form of the [b]complete [/b] ([u]not[/u] incomplete) elliptic integrals of the [i]first [/i]and [i]second [/i]kind in terms of the gamma function (whittaker and watson), i.e.

$ \int_{0}^{1}\frac{\textbf dx}{\sqrt{(1\minus{}x^{2})\cdot(1\minus{}k^{2}\,x^{2})}}$

and

$ \int_{0}^{1}\frac{\sqrt{1\minus{}k^{2}x^{2}}}{\sqrt{1\minus{}x^{2}}}\;\;\textbf dx$

i know next to nothing about the third kind though...  :|

The area of an ellipse is easy, and does not involve an elliptic integral. There are many forms that come up; some are algebraic and some are trigonometric. The important thing to realize is that if the usual Calc 2 methods don't work, it's probably impossible to evaluate by finding an elementary antiderivative.

The most basic form, and the easiest to see the name in, is the arclength integral for an ellipse: $ \int_{0}^{2\pi}\sqrt{a^{2}\sin^{2}\theta\plus{}b^{2}\cos^{2}\theta}\,d\theta$

A physical example that leads to an elliptic integral: the swing period of a pendulum.
To simplify a bit, consider a rigid pendulum with all of its mass $ m$ concentrated at a distance $ L$ from the point of attachment (If we don't make this assumption, there are two different distances to worry about). When this center is at a height $ y$, the potential energy (compared to the bottom) is $ mg(y\plus{}L)$, and the kinetic energy is then $ E\minus{}mg(y\plus{}L)\equal{}mg(y_{0}\minus{}y)$ for some initial energy $ E$ or initial height $ y_{0}$. The speed is then $ \sqrt{2g(y_{0}\minus{}y)}$, and $ \frac{dy}{dt}\equal{}\pm\sqrt{2g(y_{0}\minus{}y)}\cdot\frac{\sqrt{L^{2}\minus{}y^{2}}}{L}$

To get the period, we integrate the reciprocal of $ \frac{dy}{dt}$: the period is $ 4\int_{\minus{}L}^{y_{0}}\frac{L}{\sqrt{2g(y_{0}\minus{}y)(L^{2}\minus{}y^{2})}}\,dy$ if $ y_{0}<L$ and it doesn't go over the top, or $ 2\int_{\minus{}L}^{L}\frac{L}{\sqrt{2g(y_{0}\minus{}y)(L^{2}\minus{}y^{2})}}\,dy$ if $ y_{0}>L$ and it does.
That square root of a cubic is a sure sign of an elliptic integral.
If you use angle rather than height as a parameter, you get a trigonometric version instead.

so what you are saying is an elliptic integral would get me the change in position of a point following an elliptic path? Or would it get me the circumference of that ellipse? Sorry, I'm just new to calculus.

That arclength integral came from the parametrization $ x\equal{}a\cos\theta,y\equal{}b\sin\theta$. You don't need integrals to get at the position.


Let a1 = 2006, and for n ≥ 2.
a1 + ... + an = n^2an. What is the value of 2005a2005 ?

Thank all


geometry
circumcircle
Math Prize for Girls

Let $ S$ be a set of $ 100$ points in the plane.  The distance between every pair of points in $ S$ is different, with the largest distance being $ 30$.  Let $ A$ be one of the points in $ S$, let $ B$ be the point in $ S$ farthest from $ A$, and let $ C$ be the point in $ S$ farthest from $ B$.  Let $ d$ be the distance between $ B$ and $ C$ rounded to the nearest integer.  What is the smallest possible value of $ d$?

Math Prize for Girls #19

I believe the answer is 17, but I'm not sure that I have a solid proof. Perhaps another reader can take what I write here and tighten it up?

First, there is some extraneous information in this problem.

Given that S is a set of 100 points, only 5 points are really important, and we can do away with some of those.

Here's why: we are seeking the smallest possible value of d, the integer approximation of a distance determined by choice of point A which ranges over the 100 points of set S. Just set A to be the choice that gives the absolutely minimum distance between the resulting points B and C.

The other two points that are important are X and Y, the two points that are a distance 30 apart, the largest distance between any two points of set S.

So really only points X, Y, A, B, and C are important.

Another piece of extraneous information is that the distance between every pair of points is different. Since the final answer is rounded to the nearest integer, we could consider the general question about any 100 points with greatest distance 30. If two distances are the same, or two points coincide, we could perturb the configuration slightly to satisfy the original conditions without changing the value of d, the integer approximation to the distance between points B and C.

What about the other 95 points? Can you convince yourself that the value of d would not change if we remove those 95 other points? If you like, we could replace them with 95 points extremely close to X without changing the choice of A which yields the smallest possible value of d.

Point C is also extraneous. Given points X, Y, A, B and C as above, what would happen by moving C to a new location extremely close to X? Then either the new distance BC (now extremely close to BX) becomes smaller, or stays the same. In any event, we could find a new set of 100 points whose value of d is smaller or equal to the original value of d.

Let's rephrase the problem:
We are given points X and Y in the plane that are 30 units apart. Where can we place points A and B so that
(1) Each distance between two of A, B, X and Y is no more than 30 units.
(2) Distance AB is greater than (or equal to) distance AX and AY.
(3) The largest of distances BA, BX and BY (which we call d) is as small as possible?

The points B, X and Y determine three regions of the plane: P_B is the region of points furthest from B, P_X is the region of points furthest from X and P_Y is the region of points furthest from Y. These three regions meet at the circumcenter O of triangle BXY, and point A must be in region P_B. If we move point A toward point O, then distance d either stays the same or decreases. This observation lets us assume that A is the circumcenter of triangle BXY.

We can now rephrase the problem again: Points B, X and Y are on the circumference of a circle with center A. Distance XY is 30, and no other distance between two of A, B, X and Y is greater than 30. What is the least possible value for the largest of the distances BA, BX and BY?

Since no other distance is greater than distance XY, point B must be on minor arc XY. Name angle XAY as theta. Since distance XY is no less than the radius of the circle, angle theta must be at least 60 degrees.

If theta is between 60 and 120 degrees: we can decrease the maximum of BX and BY by moving B to the center of minor arc XY. Then the maximum of lengths BX, BY and BA is BA, the radius of the circumcircle. Moving point A towards segment XY, i.e. increasing angle XAY, will decrease the length of the radius, and thus d. Thus, increasing theta will decrease d.

If theta is between 120 and 180 degrees: In this case as well, moving B to the center of arc XY will decrease the maximum of BX and BY, so assume that B is at the center of arc XY. The equal lengths BX and BY are both greater than the radius BA, so d will be the length of BX. Decreasing theta will also decrease length BX, thus d.

From the above two arguments, length d will be minimized when segment AB is perpendicular to segment XY, with angle XAY equal to 120 degrees. In this situation, triangles XAB and YAB are both equilateral, with height 15. That is, length d is the side of an equilateral triangle with height 15, or 10sqrt{3} = 17.32, thus the closest integer is 17.

It should not be difficult now to find a set of 100 points that match the original specifications, yielding a value of d equal to 17.

Regards,

Al Gebra

17 was my answer too, also coming from $ 10 \sqrt {3}$ rounded to an integer.  Sorry, I didn't get a chance to study your argument; it may be perfectly fine.  My argument was different.


integration
real analysis
real analysis theorems

$n$ is a positive integer. Then:
$\int u^{n}\sqrt{a+bu}du=\frac{2u^{n}(a+bu)^{\frac{3}{2}}}{b(2n+3)}-\frac{2an}{b(2n+3)}\int u^{n-1}\sqrt{a+bu}du$

[quote="sinajackson"]Duuuuuuuuuuuuuuude, where did you get that from?!!!!!!! :ninja:[/quote]

Tks. They are my own formulae. I deduced they integrating by parts in a similar way that the known one $\int u^{n}\sqrt{u^{2}\pm a^{2}}\cdot du$

[quote="Virgil Nicula"]Please, allow me a joke : from the viewpoint of the software, for that the your recurrence be "roundly", i.e. completely, must appears and the "start", i.e. $I_{r,0}\ .$[/quote]

Every joke is wellcame   :)


I got a TI-89 Titanium today (yaaaaaaaaaaaaaaaaaaaaaaaay!) and I was wondering if it was possible to write a program on my computer instead of on the calculator (because I can't type that fast on it) and then somehow compile the program and send it over.

I'm pretty sure you can; you can certainly do that for the TI-84s since they come with USB connection cables as well as software (although I think you can write ASM programs - I'm not sure about BASIC).

I think that for TI-89, you can write in TI-Basic, Assembly, or even C with a special compiler.
Of course, the best way to find out would be with a google search...

You can try Ti-Graph Link. It allows you to write programs on your computer and then send them to your calc.

You can write calculator programs in C?  I love C.  It is my native language(actually obj-c is )- I am a mac user.

Please Do Not Revive Old Topics. Thank You For Your Cooperation.

well really on my ti 85

I just press the stml button and program.


Let $ \phi(x)\equal{}e^{\minus{}x^2}$. We all know that this function does not have an elementary antiderivative. But if we rescale it: $ \phi_t(x)\equal{}t^{\minus{}1}\phi(x/t)$ and differentiate with respect to $ t$, the result has a very simple antiderivative with respect to $ x$.

(a) find this antiderivative [1 pt]
(b) tell me why this happens [99 pts]

It's the heat equation, after the substitution $ 2t\mapsto t^2$ (and an arbitrary scale factor). For your $ \phi(x,t)\equal{}\frac1te^{\minus{}\frac{x^2}{t^2}}$, we have the variant $ \phi_{xx}\equal{}\frac2t\phi_t$.
With the original version $ f(x,t)\equal{}\frac{1}{\sqrt{2\pi}}\frac1{\sqrt{t}}e^{\minus{}\frac{x^2}{2t}}$ (satisfying $ f_{xx}\equal{}f_t$), we're looking at what happens to a unit amount of heat initially concentrated at the origin. Both the Gaussian shape and the variance increasing linearly with time are predicted by the Central Limit Theorem- the model for heat spread is essentially a random walk.


1)$sin\theta=\theta$

The above equatipon will have:-

a)1 real root
b)2 real roots
c)3 real roos
d)none of the above

2)Which of teh following signs are appropriate for  $sin 1 \boxed{}sin1degress$

a)>
b)<
c)=
d):(

 :D ofcourse,methods will be appreciated

[hide="2"]A
Is it $\sin{1}$ compared to $\sin(1 degree)$? First one is in radians.
If so, it would be $>$ since $\sin(1 degree)$ is very small. 1 radian is still in the first quadrant, so it can't be negative or anything. :D [/hide]

[quote="lotrgreengrapes7926"][hide="1"] I just drew a graph. Umm... start with: the intersection must be between $-1<\theta<1$ Otherwise $\theta$ would be greater than $1$, but $\sin{\theta}$ will never be greater than 1. I sketched the graph and guessed there would be 2 intersections between 0 and 1. This implies that there are another 2 intersections between -1 and 0. So the answer is 4 or D. I don't know if this is right.  :? [/hide] EDIT: Oh, yeah. Mine is wrong. Ignore it. And the solution would have been[hide]3 if they did intersect more than once. :blush: [/hide][/quote]

lotofgreengrapes7926, I think you are correct. I haven't graphed it yet, but just thinking about the graphs of $y=x$ and $y=\sin{x}$, they should intersect three times (once at the origin). Unfortunately, I didn't think about graphing until just after i posted my answer ;)

[hide="click text"]
Well first off, 0 is a solution.
The rate of change in sinx is usually less than 1 and at most equal to 1. Therefore it doens't intersect x at any other points.
So the answer is A)1.

However if the equation was
sinx-ax=0 where a is any number less than 1, then the answer would be 3 or more.[/hide]

For number 1, when I tried values for $x$ and $\sin{x}$, I always found that $x>\sin{x}$ except for when x=0. So the graphs have 1 intersection?

yes thats what I was trying to say, although I can't think of anyway to prove it without using calculus.

The strange thing is that if it was any number less than 1, for eample sinx=.99999x, then there would be more than 1 intersection.

[hide="how to do 1 without calc"]
$\theta =0$ is a solution. If there is a solution to $\sin \theta =\theta$ for some $\theta \neq 0$, then since $\theta$ and $\sin \theta$ are both odd, we only need to look at $\theta >0$ because the solutions for $\theta <0$ can be found by symmetry.

$\sin \theta \leq 1\Rightarrow \theta \leq 1<\frac{\pi}{2}$, so $\theta$ will have to be in the first quadrant. However, drawing a sector with radius 1 and angle $\theta$ instantly reveals $\sin \theta<\theta$ because the area of the segment of the sector is $\frac 12 (\theta -\sin \theta)$ which must be positive. So there can be no other solutions $\Rightarrow \boxed{A}$.
[/hide]


counting
derangement
floor function
probability

What is the simplest way of fnding derangement of say 21 objects??? Is there a better way of calculating than using the formulae???? the formulae is going to be tedious.......
n![1 - 1/1! + 1/2!........ +(-1)^n(1/n!)]

In terms of practicality, you might be better off using the recursion.  I think the most important relevant question is, "Why would you ever need to explicitly calculate $D_{21}$?"

[quote="JBL"]I think the most important relevant question is, "Why would you ever need to explicitly calculate $D_{21}$?"[/quote]
For my cryptography problem set, this is one of the questions. It's calculating the number of derangements, right?
http://www.math.brown.edu/~jhoff/math158/MathCrypto.pdf

Yes, part (a) asks for the derrangements of 26 letters.  Oy vey.  (Do you think they'd appreciate your use of the formula without further justification?  :) )  It's a number with a couple dozen digits (actually, 27 digits, I think).  Have fun!

The simplest numerical formula is probably

$!n = \lfloor \frac{n!}{e}\rfloor$

This formula is only correct if you make a small correction (as it is, try $n=2, 4$).

[hide="correct one"]$!n = \left\lfloor \frac{n!}{e}+\frac{1}{2}\right\rfloor$[/hide]

[quote="me@home"]What's the definition of derangements?[/quote]

A derangement, in simple terms, is a permutation that "messes everything up".

Take {1, 2, 3, 4}.

The permutation {1, 3, 2, 4} is not a derangement because it keeps 1 and 4 in their original positions.

However, the permutation {4, 3, 2, 1} is one because no object ends up where it was in the first place.

Unless you want the rigorous abstract-algebra definition like "A derangement is an automorphism that does not map any member to itself" or something like that.

Oh wait a second now... there was a problem in AoPS vol 2 at the end of the counting chapter which went something like: 6 people have name cards to sit at but they completely ignore it, what is the probability that no one is in the right spot
is that derangements/

Yup...thats derangement.....21 was just a figure out of thin air... i was actually doing derangement of 8 people on a similair problem and thought what would i do if it was 21???? :oops:

Is there a derangement formula, like the n choose k formula?

Yes.  It was given in the first post.  ;)

ooops hehe :blush:
I guess I'll look it up on mathworld for more info...


function
vector
algebra
polynomial
linear algebra
matrix
linear algebra unsolved

let E a linear space on a field K of finite dimension and U a linear function from E to E. We define C(v)=vect(v, U(v), U^2(v), U^3(v), ... ) for any vector v. Prove that if there exist an irreducible polynomial P over K such that P(U)=0 then there exist some vectors v1,v2,..,vk so that:
   E=C(v1)+C(v2)+..+C(vn) and the sum is direct ..

bye.

I'm not quite sure if this is your question, but I'll post this answer anyway.

Rational Canonical Form: If $E$ is a linear operator from a finite-dimensional vector space $V$ to itself, we can decompose $V$ into a direct sum of spaces $V_r$ with basis $v_{r,k}$, $0 \le k \le c_r$, with $E(v_{r,k})=v_{r,k+1}$ if $k < c_r$ and $\displaystyle E(v_{r,c_r})=\sum_{k=0}^{c_r}a_kv_{r,k}$ where the polynomial $p(x)=a_0+ a_1x+\dots + a_{c_r}x^{c_r} -x^{c_r+1}$ is a divisor of the minimal polynomial of $E$. (In the actual canonical form, we choose the first polynomial equal to the minimal polynomial of $E$, and each succeeding polynomial a divisor of the preceding)
The matrix of E over this basis is easily described; a series of diagonal blocks, with zeros everywhere else. In each block, there are 1's just below the main diagonal, the $a_k$ in the rightmost column, and zeros everywhere else.

I won't prove the general case (too deep in technical details), but in the case that the minimal polynomial $P$ is irreducible, it becomes simpler.
Let $v_0$ be an arbitrary nonzero vector and consider the span of $v_0$, $E(v_0)$, $E^2(v_0)$, ...  . This is a space of dimension $j\le \dim E$, and the first $j$ of the vectors must form a basis for the space, since if $E^k(v_0)$ is a linear combination of the previous vectors, we have $Q(E)(v_0)=0$ for some polynomial $Q$ of degree $k$, and $Q(E)(E^i(v_0))=E^iQ(E)(v_0)=0$, making $E^{k+i}(v_0)$ a linear combination of the previous vectors.
We therefore take $v_1=E(v_0), \dots, v_{j-1}=E^{j-1}(v_0)$.
Let $v_j$ be an arbitrary vector outside the span of what we already have, and repeat the process until the entire space is covered.

Since the minimal polynomial of a restriction of a linear operator must obviously divide the minimal polynomial of the full operator and the only nontrivial divisor of $P$ is itself, the polynomial associated with each block must be $P$, each block must have dimension equal to the degree $j$ of $P$. Finally, we consider the intersection of any two blocks: this intersection is a vector space of dimension less than $j$ and $E$ sends the intersection to itself, so the minimal polynomial of $E$ on this subspace must have degree less than $j$. The only factor of $P$ with degree less tan $j$ is a constant, so the intersection has dimension zero and the full space is the direct sum of the blocks.

if P(u)=0 then P is a multiple of M the minimal polynomial. But P is irreducible so
P=a*M where a is in the field K isn't it. So M is irreducible . so you've solved the pb , isn't it ?

in fact you've just prove that the intersection of the C(vi) is empty but not that it's a direct sum .. i think it can be proved as the following:
if P(u)=0 and P is irreducible, if we assume that u<>0, then u is invertible (0 is not en eigenvalue). Hence if we have
(u^p)o( v_(i+1) )  =  (P1(u))o(v1) + (P2(u))o(v1) +..+ (Pi(u))o(vi)
then, because u is invertible, we have:
v_(i+1)   = (u^(-p))o[P1(u))o(v1) + (P2(u))o(v1) +..+ (Pi(u))o(vi)]
 but u^(-p)=Q(u) where Q is a polynomial.So
v_(i+1)   = Qo[P1(u))o(v1) + (P2(u))o(v1) +..+ (Pi(u))o(vi)]
which is impossible. So (u^p)o( v_(i+1) ) does not belong to the space spans by
C(v1)+C(v2)+..+C(vi).

it was the first question of the exercise. The second question was:
does the conclusion still hold if P is irreducible and P^2(u)=0 ?


search
analytic geometry
graphing lines
slope
geometry
angle bisector
geometry proposed

Let ABC be a right triangle with a right angle at vertex C. Let
CXPY be a square with P on the hypotenuse, and X, Y on the sides.
Show that the length t of a side of this square is given by
$\frac{1}{t}=\frac{1}{a}+\frac{1}{b}$

[hide]
Suppose that $X\in AC$ and $Y\in BC$.

The distances from $P$ to $AC,BC$ are $PX=PY$, so $CP$ is the angle bisector of $\angle C$

$\frac{BP}{PA}= \frac{a}{b}\Rightarrow \frac{c}{PA}= \frac{a+b}{b}$

From Thales $\frac{a}{CY}= \frac{c}{PA}$

So $\frac{a}{CY}= \frac{a+b}{b}\Rightarrow \frac{1}{CY}= \frac{a+b}{ab}\Rightarrow$

$\frac{1}{CY}= \frac{1}{a}+\frac{1}{b}$[/hide]

[hide="probably the easiest solution"]we have $(ABC)=(APC)+(BPC)\Longrightarrow \frac12ab=\frac12tb+\frac12ta\Longrightarrow \frac1t=\frac1a+\frac1b$.[/hide]

[quote="junggi"][hide]Let $X\in BC,Y\in CA$
Since $XP\parallel CA,$ $\Delta BXP\sim\Delta BCA.$
Thus,
$1-\frac{t}{a}\frac{a-t}{a}=\frac{BX}{BC}=\frac{XP}{CA}=\frac{t}{b}$
$\iff 1=\frac{t}{a}+\frac{t}{b}$
$\iff \frac{1}{t}=\frac{1}{a}+\frac{1}{b}$
[/hide][/quote]

Let me dear [b][size=100]junggi[/size][/b] to rewrite your nice solution.

From $PX\parallel BC$ and $PY\parallel AC$ $\Longrightarrow$ $\frac{t}{a}= \frac{AP}{AB}$ $,(1)$ and $\frac{t}{b}= \frac{BP}{AB}$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $\frac{t}{a}+\frac{t}{b}= \frac{AP+BP}{AB}= 1$ $\Longrightarrow$ $\frac{1}{t}= \frac{1}{a}+\frac{1}{b}$

Kostas Vittas.

Three points $B,\ P,\ A$ are colinear in this order, thus the slope of the line $BP$ equals to that of the line $BA$, yielding $\frac{t}{a-t}=\frac{b}{a}.$
Solving this for $t$ gives the given equation.


Prove that if $ \{v_1,v_2,...,v_n\}$ is a basis for $ V$ and $ w_1,w_2,...,w_n$ are vectors in $ W$, not necessarily distinct, then there exists a linear transformation $ T: V\mapsto W$ such that

$ T(v_1)\equal{}w_1,T(v_2)\equal{}w_2,...,T(v_n)\equal{}w_n$

The easiest way to prove that something exists is to write it down.  Why is this true if $ v_1, ... v_n$ is the standard basis?  How do you change coordinates to make it true in general?

Given $ v\in V$, we may write $ v\equal{}\sum_i \alpha_i v_i$ in a unique way, and we define $ T(v)\triangleq \sum_i \alpha_i T(v_i) \equal{} \sum_i \alpha_i w_i$ (i.e., we extend $ T$ by linearity).  Trivially this agrees with the previous definition of $ T$ on $ \{v_1,\ldots,v_n\}$.  Why is this well-defined on all of $ V$?  Why is it a linear transformation?


induction
blogs
combinatorics unsolved
combinatorics

Ana and Bora are each given a sufficiently long paper strip, one with letter $A$ written , and the other with letter $B$. Every minute, one of them (not necessarily one after another) writes either on the left or on the right to the word on his/her strip the word written on the other strip. Prove that the day after, one will be able to cut word on Ana's strip into two words and exchange their places, obtaining a palindromic word.

I thought I'd be interested in it at some point, but I had some trouble understanding the text:

What do you mean by "the day after"? How many days does the game last? And how does it work, exactly? In every minute one of them performs a move as described, or can they both perform a move at the same time?

Thanks.

I don't understand what you mean, but I can answer your last question:

[quote]In every minute one of them performs a move as described, or can they both perform a move at the same time?
[/quote]

as I wrote: every minute, [i]one of them[/i] (not necessarily one after another) and non of them could do anything together.

But that doesn't seem to be true!

Suppose Ana changes her card to $AB$ after the first minute, but for every move thereafter it is Bora that writes to the left on her strip.  The next day Ana will have $AB$, which can't possibly be cut up to make a palindrome (Bora's would still be cutable though).

It is a problem from All-Russian Olympiad 2003 (problem 11.4)! but not from "Serbia and Montanagro 2004"!!!!

And also there is a mistake in the statement of the problem. Here is a correct statement:

"...
Prove that the day after, one will be able to cut word on Ana's strip into two words and exchange their places, [b]obtaining the same word but turning from left to the right.[/b]"

This means that if Ana's word is $w=w_1 \ldots w_n$ then we have to prove that we can cut it into $w=uv$ such that $vu=w_n \ldots w_1$.

Clearly, it is sufficient to prove that we can cut $w$ into two palindromes. And this statement could be proved by induction. I have a complete solution, but I don't want to post it right now. Let me know if somebody needs a solution (and I would post it).

P.S. I am sorry about my poor English.

how can I solve it by induction

I applied induction and obtained some thing

but at last I failed it mmm

Is there someone to solve it by induction??????mm

yugrey told me to do this, so here is my solution I came up with. Note the correct problem statement is along the lines of this:
[quote="math154"]
Ana and Bora start with the letters $A$ and $B$, respectively. Every minute, one of them either prepends or appends to his/her own word the other person's word (not necessarily operating one after another). Prove that Ana's word can always be partitioned into two palindromes.
[/quote]
My first instinct was to apply induction (i.e. if two strings $T,U$ can be split up into two palindromes, then $TU$ can) but this statement seems hopeless and far too difficult to prove as you immediately enter a barrage of casework. So we go the "reverse" path, i.e. lets say we have a string of $A$'s and $B$'s such that things work, then changing $A \to AB$ and $B \to B$ or $A \to BA, B \to B$, etc. all produce strings which can still be split up into two palindromes and we induct on the number of transformations we apply. Note that this is just playing the game in reverse effectively, where at the $n^{th}$ step of our induction we are treating our string as a string of letters where our alphabet is the strings from the $n^{th}$ minute before.

Now, the base of $n=1$ is trivial. Now suppose we have made $n$ transformations and our string can be split up into $XY$ with $X,Y$ palindromes. Now we will deal with the case of $A \to AB$, the other cases are analogous. Define $f$ as the mapping that acts on a string of letters $A,B$ and sends $A$ to $AB$ and $B$ to $B$. Note that if $\mathcal{P}$ is a palindrome, then $Bf(\mathcal{P})$ is a palindrome by easy induction, and same with $f(\mathcal{P}) - B$ (i.e. strip a $B$ off the end). Now note that $f(XY) = f(X)f(Y) = [f(X)-B][Bf(Y)]$ which is clearly a split into two palindromes. Thus the inductive step is complete and therefore we are done.

I did this problem on my blog here.

[url]http://www.artofproblemsolving.com/blog/80508[/url]

I also did this problem on my blog here:
[url]http://www.artofproblemsolving.com/blog/80547[/url] (the second one)

I did it a MUCH different way that is less clever and more messy, but (I think) it worked (although you start having to deal with "negative palindromes" if you insist in doing certain annoying cases :P )

Note that the original statement seems to be indeed the right one (see [url=http://www.iis.it-hiroshima.ac.jp/~ohkawa/AoPS.pdf/The%20IMO%20Compendium/RusMO03.pdf]here[/url]), and apparently equivalent whenever it is correct, but not always correct :O
[hide="Demonstrating equivalence / incorrectness"]
Start with the statement dinoboy, yugrey, and I used. If one of our palindromes is even-length, split half of it off and move it to the other side, obtaining a palindrome.

If you start with a palindrome and split something off, say split $X'$ off from $XY(X')$, then you get $X'XY$. $X'X$ is a palindrome, as is $Y$. But $X'X$ has even length...

So if you do this:
$(A,B) \implies (AB,B)\implies (ABB,B)\implies (ABBB,B)$
Clearly $ABBB$ cannot be permuted into a palindrome (even length, odd # of A's and B's) :lol:
[/hide]

[quote="dinoboy"]Note that if $\mathcal{P}$ is a palindrome, then $Bf(\mathcal{P})$ is a palindrome by easy induction, and same with $f(\mathcal{P}) - B$ (i.e. strip a $B$ off the end).[/quote]
OK, IMO here's a more conceptual/algebraic perspective in terms of the free group on $\{A,B\}$: let $\overline{X}$ denote the reverse of $X$; then we have $Bf(A)B^{-1} = B(AB)B^{-1} = BA = \overline{AB} = \overline{f(A)}$, and of course $Bf(B)B^{-1} = B(B)B^{-1} = B = \overline{f(B)}$. It follows that if $X = w_1w_2\cdots w_n$ is a palindrome (i.e. $X = \overline{X}$), then \[ Bf(X)B^{-1} = \prod Bf(w_i)B^{-1} = \prod \overline{f(w_i)} = \overline{\prod f(w_{n+1-i})} = \overline{f(\overline{X})} = \overline{f(X)}, \] so $Bf(X)$ and $f(X)B^{-1}$ are palindromes.

I'll probably look at antimonyarsenide's solution more carefully the next time I teach at MOP. :P

[quote=dinoboy] So we go the "reverse" path, i.e. lets say we have a string of $A$'s and $B$'s such that things work, then changing $A \to AB$ and $B \to B$ or $A \to BA, B \to B$, etc. all produce strings which can still be split up into two palindromes .... [/quote]

What's the motivation behind noticing that changing $A \rightarrow AB$ and $B \rightarrow B$ in a bipalindromic string would keep it bipalindromic ? I thought on this problem for four hours but couldn't notice this :(

[quote=Yugrey's solution in his blog] Start with $A_n$ and then convert it to $A_{n-1}$'s and $B_{n-1}$'s, then write it in terms of $A_{n-2}$ and $B_{n-2}$.

Call each of these things where you shift the index down by $1$ a step.

It is clear that if you have some sequence a step is taking one of $A$ or $B$ and replacing every it by one of $AB$ or $BA$, so there are $4$ kinds of steps. [/quote]

Consider the game: $(A, B), (AB, B), (AB, BAB), (ABBAB, BAB), (ABBABBAB, BAB)$. Now, I don't understand what he means by "Start with $A_n$ and then convert it to $A_{n-1}$'s and $B_{n-1}$'s, then write it in terms of $A_{n-2}$ and $B_{n-2}$", but assuming it's some sequence of the short $\cdots \rightarrow ABBABBAB \rightarrow ABBAB|BAB \rightarrow AB|BAB|BAB \rightarrow \cdots$, how "a step is taking one of $A$ or $B$ and replacing every it by one of $AB$ or $BA$" holds ?
--------------------------------------
[b]EDIT:[/b]  Ooops I'm very dumb, after 10 minutes of posting this I realized the answer to both questions What Yugrey meant by replacing is in my example $A_5 = A_4B_4 = A_3B_3B_3 = A_2B_2A_2B_2A_2 = A_1B_1B_1A_1B_1B_1A_1B_1$. And the motivation for considering the change $A \rightarrow AB, B \rightarrow B$ (or similar) is gained instantly by looking at the above steps: at each step you change something like that to get $A_n$ in terms of $A_{n-1}B_{n-1}$, so inparticular if you can prove that doing that in a bipalindromic words keeps it bipalindromic, then you're done by induction, and it happens to be true, so you're done.

Solved with [b]magicarrow[/b] and [b]tapir1729[/b].

Induct on the length of the number of minutes, with base cases trivial. WLOG, the words go to $A,AB$. Denote $AB=B'$, and follow the process. By inductive hypothesis, the word obtained written in terms of $B'$s can be split into two palindrome. Thus, it suffices to show that if a word formed from $A$ and $B'$ can be broken into two palindromes, then replacing all $B'$s with $AB$s maintains this property.

Indeed, applying this operation to a palindrome, we see it remains a palindrome if we either remove an $A$ from the left, or add an $A$ to the right. At this point, if our word split into two palindromes, we just apply the operation, and shift our divider by one to the right. This completes the proof.


If posssible, define the function at 1 so that it becomes continuous at 1.

f(x) = 1 / (x-1)

You have to find the limit of the function as it approaches 1. If it was just a hole or something, then define the function at that x such as the function is equal to the limit. However, such a limit does not exist because the left limit is negative infinty while the right limit is positive infinity. So there is no solution to the problem.


A,B and C in E.
prove that:

AΔ(BΔC) = (AΔB)ΔC
 :P

I'm a little bit confused about your notation: what's $ E$ and what's $ \Delta$?

Δ is the  symetric difference.
$ E$ is a  ensemble

The most obvious way to show this is to consider an arbitrary element of $ E$ and analyze 8 cases: when it is and isn't in $ A$, when it is and isn't in $ B$, and when it is and isn't in $ C$. The LHS and RHS will correspond for all 8.

Or alternatively, we can prove $ a\oplus(b\oplus c) \equal{} (a\oplus b)\oplus c$, where $ \oplus$ is "exclusive or". Some simplification can come from the facts that $ x\oplus 0 \equal{} 0\oplus x \equal{} x$ and $ x\oplus 1 \equal{} 1\oplus x \equal{} \bar x$. Hence:

1. If $ a \equal{} 0$, then the LHS is $ b\oplus c$ and the RHS is $ b\oplus c$

2. If $ a \equal{} 1$, then the LHS is $ \overline{b\oplus c}$ and the RHS is $ \bar b\oplus c$. Now if $ c \equal{} 0$, we get $ \bar b \equal{} \bar b$, and if $ c \equal{} 1$, we get $ b \equal{} b$

Is "ensemble" the French word for "set"?


One dimension of a cube is increased by $ 1$, another is decreased by $ 1$, and the third is left unchanged. The volume of the new rectangular solid is $ 5$ less than that of the cube. What was the volume of the cube?

$ \textbf{(A)}\ 8 \qquad
\textbf{(B)}\ 27 \qquad
\textbf{(C)}\ 64 \qquad
\textbf{(D)}\ 125 \qquad
\textbf{(E)}\ 216$

if we let each dimension equal x, than the increased side of the cube is modeled as x+1, the decreased size is x-1, and the regular one is just x.

to find the voluime u would multiply the 3 dimensions, (x) by (x-1) by (x+1) and you would get x^3-x.

the volume of the original cube would be x^3, but the problem say 5 less than so it would be x^3-5.

youd then make these two equivilant, giving you X^3-x=x^3-5

from here subtract x^3 from both sides, then divide by -1, and youd find x to be 5.

then plug 5 as the x value for the volume of the original cube, x^3, so 5x5x5 which equals 125.

:-)

hope i helped


LaTeX
number theory unsolved
number theory

Find all integers $ x$ such that

$ x^3\plus{}x^2\plus{}8x\plus{}16$ is a perfect square.

It is again the standart tecnique! take $ x^3\plus{}x^2\plus{}8x\plus{}16\equal{}m^2$ and add both sides 4. we get (x+2)($ x^2\minus{}x\minus{}10$)=$ m^2$+4, and we know sth about primes in the factorisation of $ m^2$+1, they will be of the form 4k+1. So just proceed with case analysis for x odd and even and obtain a contradiction. I am not so good at latex and I will not write a full solution but I hope to have explained the way clearly! :)


function
integration
real analysis
real analysis unsolved

For a bounded linear functional $ F$ on $ L^{p}[0,1]$, $ 1 \leq p<\infty$, there is a function $ g \in L^{q}$ such that $ F(f)\equal{}\int fg$ and $ \parallel{}F\parallel{}\equal{}\parallel{}g\parallel{}$.

What would be an elementary proof that $ g$ is unique?

I did something like that:

Suppose there exist $ g,h \in L^{q}$ such that $ L(f)\equal{}\int fg\equal{}\int fh$ for all $ f \in L^{p}$. Choose $ f\equal{}|g\minus{}h|^{q\minus{}2}(g\minus{}h)$. Then $ |f|^{p}\equal{}|g\minus{}h|^{p(q\minus{}1)}\equal{}|g\minus{}h|^{q}$ so $ f \in L^{p}$. Thus $ \int fg\equal{}\int fh$ means $ \int g|g\minus{}h|^{q\minus{}2}(g\minus{}h)\equal{}\int h|g\minus{}h|^{q\minus{}2}(g\minus{}h)$, which means $ \int |g\minus{}h|^{q}\equal{}0$ and so $ g\equal{}h$ a.e.

I wonder if this is fine.

In general, the proof must be divide into 2 cases $ p \equal{} 1$ and $ 1 < p < \plus{} \infty$ since we are using the reflexive of $ L^p$ spaces which does not hold true for the case $ p\equal{}1$: ,

To obtain the uniquenessm try to use: if $ f \in L_{loc}^1 (\Omega)$ and $ \int_\Omega {f(x) u(x) dx} \equal{} 0$ for all $ u \in C_c(\Omega)$ then $ f \equiv 0$ a.e. Moreover, $ C_c(\Omega)$ is dense in $ L^p(\Omega)$ for all $ 1 \leqq p <\plus{}\infty$. Lastly, for $ \Omega \equal{} [0,1]$ we have $ L^p \subset L^1 \subset L_{loc}^1$.

Where is reflexivity being used? I think we only need $ (L^1)^{*} \equal{} L^{\infty}$, which is true.


function
calculus
integration
real analysis
real analysis unsolved

Let $a<b$ amd $f: [a,b]\rightarrow R$ be a bounded function. Suppose that for all $\epsilon > 0$ there is a $\delta >0$such that if $P$ and $Q$ are partitions of $[a,b]$ with $\| P \|<\delta$ and $\| Q \| <\delta$ then $| R(f,P)- R(f,Q) | <\epsilon$ for any Riemann sums $R(f,P)$ and $R(f,Q)$ of $f$ with respect to the partitions $P$ and $Q$. ( the two Riemann sums do not ncecssarily involve the same sample points)

Show that $f$ is integrable

Pick any sequence $R(P_{n})$ where $||P_{n}|| \rightarrow 0$. Since this sequence is Cauchy, it converges to $L \in \mathbb{R}$. Now there exists $N$ such that $|R(P_{N})-L|<\frac{\epsilon}{2}$ (*). Also for every subdivision $P$ whose mesh is small enough, we have $|R(P)-L|<\epsilon$, which solves the question. In order to prove this last point, you need to use (*) [plus somehow some Darboux sums argument] and the mesh $||P||$ must be small enough so that you minimize the riemann sum over those intervals partitioning $P$ which contain the points delimiting the partition $||P_{N}||$. This is rather straightforward but very tedious to write down. Trying to write everything correctly you will notice that maybe it is less awkward to prove the function is Darboux-integrable first, which also means Riemann-integrable.

Thx, I got an idea what you talked about,  but I have not learned "Darboux-integrable" which you mentioned... Could you be kindly explain in a simple detail way? Thx alot.

The Darboux integral is equivalent to the Riemann integral: $f$ bounded is Darboux-integrable iff the lower and upper Darboux sums have same finite limit as the mesh of $P$ goes to $0$.

got confused about " there exists $N$ such that $|R(P_{N})-L|<\frac{\epsilon}{2}$ (*). Also for every subdivision $P$ whose mesh is small enough, we have $|R(P)-L|<\epsilon$" "

Could u be more specific to show me how? Thx

Need more explaination, please. Thx in advance.


Consider the function $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ and let $ I\equal{}\{(x,y)\in \mathbb{R}^2 |y\leq ax^2\plus{}bx\plus{}c\}$ and $ S\equal{}\{(x,y)\in \mathbb{R}^2 |y\geq ax^2\plus{}bx\plus{}c\}.$

Among the affirmations:

I) $ (0,0)\in I$ if and only if $ c>0.$

II) If $ b^2\minus{}4ac<0$ and $ c>0$, then $ (0,0)\in S.$

III) If $ a>0, (0,0)\in S$ if and only if  $ b^2\minus{}4ac>0.$

IV) If $ a>0$ and $ c<0,$ then $ (0,0)\in S.$

V) If $ b^2\minus{}4ac<0$ and $ p\equal{}(x,0),$ then, $ p$ never belongs to $ I$ and $ S$ at the same time.

Hence:

a) I and V are false.
b) II and IV are false.
c) Only IV and V are correct.
d) Only I and III are false.
e) Only III and IV are correct.

Since $ f(0) = c$ if $ c < 0\longrightarrow (0,0) \in S$, If $ c > 0 \longrightarrow (0,0) \in I$, If $ c = 0 \longrightarrow (0,0) \in I, %Error. "union" is a bad command.
S$. All other information is superfluous.

I) False; because if c=0 then $ (0,0) \in I$

II) False; $ c > 0 \longrightarrow (0,0) \in I$

III) False; If $ b^2 - 4ac = 0$ then $ (0,0) \in S$

IV) True; $ c < 0 \longrightarrow (0,0) \in S$

V) True;

 :( But this isn't an option...

EDIT: HAHA, C! i would hate to do that in a test... :roll:

[quote="ocha"]
 :( But this isn't an option...[/quote]

C?  :)

I think this belongs in HSB, but I suppose it's borderline.


When I was reading a trigonometry section, they gave me like 30 identites to memorize. Which ones are probably the most important? I sure don't want to memorize 30 identities  ;)

By identies, I mean something like this:

$1+\tan^2{x} = \sec^2{x}$

The double angle identities and the pythagorean identities, I think, are most important.  You can derive a lot of the identities with these.

Offhand, I would say:

$\cos 2\theta=2\cos ^2\theta -1=\cos ^2\theta - \sin ^2\theta$
$\sin 2\theta=2\sin \theta \cos \theta$
$\sec ^2\theta -1=\tan ^2\theta \Rightarrow \tan ^2\theta +1=\sec ^2\theta$
$\sin ^2\theta + \cos ^2\theta=1$
$\sin \theta = \cos (90-\theta)$
$\sin \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{2}}$
$\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}$

Some of these are rather obvious and/or well-known, but that means you'll definitely need to know them. There are also probably a few other useful formulas, but this is all I can think of, unless $i$ is included in trig, as some do... in which case, you should also try to know:

$e^{i\theta}=\cos \theta +i\sin \theta$

Perhaps memorize a certain few and then derive them as needed.
You could also just memorize certain patterns like for:
$\sin (x \pm y)$ or something.

Right, that's an important one. It's where $\sin 2\theta$ is derived from. To give you the formula for that:

$\sin (\alpha +\beta)=\sin \alpha \cos \beta + \sin \beta \cos \alpha$
$\sin (\alpha -\beta)=\sin \alpha \cos (-\beta) +\sin (-\beta) \cos \alpha = \sin \alpha \cos \beta - \sin \beta \cos \alpha$
$\cos (\alpha +\beta)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$
$\cos (\alpha -\beta)=\cos \alpha \cos (-\beta) -\sin \alpha \sin (-\beta)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$

From there, you can easily see:

$\alpha =\beta \Rightarrow \sin (\alpha +\alpha)=\sin 2\alpha =\sin \alpha \cos \alpha +\sin \alpha \cos \alpha =2\sin \alpha \cos \alpha$
$\cos (\alpha +\alpha)= \cos ^2\alpha - \sin ^2\alpha =\cos ^2\alpha - (1-\cos ^2\alpha)=2\cos ^2\alpha -1$

Unfortunately, I don't know the $\tan (\alpha +\beta)$ formula.

Well $\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}$ = $\tan(\alpha+\beta)$

:)So just combine the formulae and if you want it to get nicer just mess with a few things.

So yeah, $\sin(\alpha+\beta)$ and $\cos(\alpha+\beta)$  are good ones to know, b/c you can derive a few others from those, along with the pythagorean identities (which is just knowing that $\sin^2\alpha + \cos^2\alpha = 1$ and the others can be derived from that.

So there isn't too much to actually memorize, you just gotta know how to derive it..

What about $\sin x+\sin y=2\sin()\cos()$ and such formulas? And once I saw a problem based on the identity about writing arctan of something as sum/difference of two other arctans... and the series became telescopic suddenly.

Well of course there are many formulae for sin and cos that will be very useful in particular situations, I was just commenting for the common trig identities.

Also, for stuff that you seem to be talking about, the hardest part is finding out how to apply trigonometry to the situation rather than memorizing the formula. You can memorize all of the formulas but when you get to the problem that doesn't state that you need to use trig outright and don't know which one to use, they don't come in handy.

$\sin (a+b)=\sin a\cos b+\cos a\sin b$
$\sin (a-b)=\sin a\cos b-\cos a\sin b$
$\sin (a+b)+\sin (a-b)=2\sin a\cos b$
Let $x=a+b$ and $y=a-b$.
Then $a=\frac{x+y}{2}$ and $b=\frac{x-y}{2}$
So $\sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$

You can most of what you will ever need from the sum identities and from the Pythagorean identity. Just try to remember the general idea of deriving what you need from these basic ones.

tan(A+B) = (tan A + tan B)/ (1 - tanAtanB)

tan(A-B) = (tan A - tan B)/ (1 + tanAtanB)

cheers

Those two bother me because they are quite long.

simple way to remember them :

always remember that when u have A + B, then the top is tanA + tan B. the bottom is the opposite.

then, when u have A - B, the top is tanA - tan B. the bottom is STILL the opposite. u see any patterns?

that means, the numerator follows the exact sign of the angle, while the denominator does the opposite. is that a good way? i dunno.. haha.. i use this method to remember tho.

That's actually how I remember them too.

Anyone interested in the derivation?
\begin{eqnarray}\tan(a\pm b)&=&\frac{\sin(a\pm b)}{\cos(a\pm b)}\\
&=&\frac{\sin a\cos b\pm\cos a\sin b}{\cos a\cos b\mp\sin a\sin b}\\
&=&\frac{\frac{\sin a}{\cos a}\pm\frac{\sin b}{\cos b}}{1\mp\frac{\sin a\sin b}{\cos a\cos b}}\\
\tan(a\pm b)&=&\frac{\tan a\pm\tan b}{1\mp\tan a\tan b}
\end{eqnarray}

From $(2)$ to $(3)$, factor out and cancel $\cos a\cos b$.

Isn't it nice to know where things come from?


[quote="radio"]Prove that for any integer $ n\geq 2$ there exist a set $ A_{n}$ of $ n$ distinct positive integers such that for any two distinct elements $ i,j\in A_{n}$, $ |i-j|$ divides $ i^{2}+j^{2}$.[/quote]
for $ n\ge 2$, we take $ A_{n}=\{i(n!)|\ 1\le i\le n\}$
and we have $ \forall (i,j)\in A_{n}^{2},\exists (a,b)\in [|1,n|]^{2},\ such\ that\ i=a(n!),j=b(n!)$
so, $ i^{2}+j^{2}=(a^{2}+b^{2})(n!)^{2}=(a^{2}+b^{2})(n!)(a-b)\frac{n!}{a-b}=(i-j)(a^{2}+b^{2})\frac{n!}{a-b}$
then $ i-j|i^{2}+j^{2}$


[quote="bos1234"]A force of $ 160N$ stretches a spring of 0.050m beyond its unstretched length.

(a)What magnitude of force is required to stretch the spring 0.015m beyond its unstreched length?[/quote]
[hide="Hint"]
The relation you need is Hooke's law:
\[ F \equal{} k\Delta x\]
[/hide]


Welcome to the "knot the rite word game!" This is where you spell out a sentence using mostly homonyms that aren't the right one to use in that sentence! You may still use regular words, but try to use as many homonyms as you can! Example:

I dew knot no what yue are saying.

Please note that individual letters are not homonyms. "I C U" would be invalid in this case. Numbers are okay if you write them as a word. Example: I eight some, two!

As always, refrain from using inappropriate words (the only acceptable variation of "do" is "dew") and follow the forum guidelines! Ok! Go ahead!  :D

Eye am sew sew sew tired.
Are yue tired, two?

knot sew tired today.
But eye no what it's like.

sow dew eye. watever eye em verie bord rite now...

ewe arrrr? I went two the fare two day :rotfl:

Lucky. Eye tink eye am going on vacation...


$43/197<a/b<17/77$;

[b]a[/b] and [b]b[/b]  are positive integers. Then find the

minimum value of b.

That's a very tedious method. Isin't it.
 :)