probability
USAMTS
articles
function
expected value

Summary (from memory, so I may have messed up):

There are 10 socks, 2 of each color, and a person randomly selects two to make pairs until all of the socks are in pairs. What is the average  number of socks that are in a pair with a sock of matching color?

I didn't understand why the given solution works:

[hide="given solution"] Each pair has a 1/9 chance of matching. There are 5 pairs, so avg. = 5/9.

What I don't understand:
I thought only the first sock would have a 1/9 chance of matching, then it would be 1/7, etc.

Also, if each pair did have a 1/9 chance of matching, I thought expected value = amount of gain * probability of happening, so I thought it would be $ 1*\frac{1}{9}\plus{}2*\frac{1}{9}$ etc. [/hide]

[quote="erdogankerem123"]Summary (from memory, so I may have messed up):

There are 10 socks, 2 of each color, and a person randomly selects two to make pairs until all of the socks are in pairs. What is the average  number of socks that are in a pair with a sock of matching color?

I didn't understand why the given solution works:

[hide="given solution"] Each pair has a 1/9 chance of matching. There are 5 pairs, so avg. = 5/9.

What I don't understand:
I thought only the first sock would have a 1/9 chance of matching, then it would be 1/7, etc.

Also, if each pair did have a 1/9 chance of matching, I thought expected value = amount of gain * probability of happening, so I thought it would be $ 1*\frac {1}{9} \plus{} 2*\frac {1}{9}$ etc. [/hide][/quote]

for some reason i get the feeling the answer is NOT 5/9, at least not by the given solution.  it seems to violate all rules of probability.

The correct answer is in fact 5/9 but it is not nearly that simple...it needs a ton of casework (my friend beasted this and his scratch paper was covered with numbers). I also remember seeing something not so simple on the official solution sheet.
I think you had to find the probability of 1 pair matching, 2, 3, 4 (5), and add, with "weight" accounted for.

The answer is, in fact, that simple: expected value is linear, so the expected value of a sum is the sum of the expected values.  For each first-sock in a pair, the expected value that it's matched with the sock of the same color is $ \frac{1}{9}$.  These $ \frac{1}{9}$ths are very not independent, but for expected value, it doesn't matter.

Let $ E_n = \text{ expected number of matchings given n pairs of sockets}$.

$ E_5 = E(x_1) + E(x_2) + E(x_3) + E(x_4) + E(x_5)$ by linearity of expectation.

(so basically, if $ E$ is the number of socks that match. $ E$ spans the same probability space as the sum of the events - sock pair 1 matching, sock pair 2 matching, etc. )

Where $ x_i = \{\begin{array}{cc} 1 & \text{if sock pair i is matched up correctly} \\
0 & \text{if not}\end{array}$

$ E(x_i) = P(x_i) = 1/9$

Oh oops joel already answered. I read up to the last post. doh.

Thanks for answering, but I'm not exactly what this sentence means:

[quote="JBL"]These $ \frac {1}{9}$ths are very not independent, but for expected value, it doesn't matter.[/quote]

Thanks!

I think this explains linearity of explanation pretty well. Not sure if you can view the whole book though.
[url=http://books.google.com/books?id=FIeZyHyDxfcC&pg=PA245&lpg=PA245&dq=linearity+of+expectation&source=web&ots=wv4yn-ra-5&sig=n1Zoax4tolA5odXgrty4pBoIiAI&hl=en]Book that explains linearity of expectation[/url]

[quote="Begoner"][quote="azjps"]The USAMTS year 17 article on expected value is pretty easy to understand: [url=http://usamts.org/Forms/ExpectedValue.pdf]here[/url].[/quote]

It's a well-written article, but, unfortunately, it isn't consistent throughout.  For example, their solution to the sixth problem is completely flawed.

[quote]"Suppose you and I play a game. You start with 500 dollars and I start with 1000 dollars. We flip a fair coin repeatedly. Each time it comes up heads, I give you a dollar. Each time it comes up tails, you give me a dollar. We continue playing until one of us has all the money. What is the probability you will win this game?"[/quote]

According to the article, the correct answer is 1/3.  In reality, the probability is far lower than this.[/quote]

Really? How so?

[quote="blahblahblah"][quote="Begoner"]According to the article, the correct answer is 1/3.  In reality, the probability is far lower than this.[/quote]

are you sure (http://en.wikipedia.org/wiki/Gambler's_ruin)?[/quote]

Never mind, you're completely correct.  I misread the problem.

[quote="JBL"][quote="erdogankerem123"]Thanks for answering, but I'm not exactly what this sentence means:

[quote="JBL"]These $ \frac {1}{9}$ths are very not independent, but for expected value, it doesn't matter.[/quote][/quote]

This was just an informal restatement of the linearity of expectation.  (Agolsme provided the best explanation of what that means in this case in the fifth post in this thread.)  I was just pointing out that one can take apply linearity regardless of the way that the events $ x_i$ (in agolsme's notation) related to each other -- in particular, the probability of the outcome of $ x_i$ depends on the outcomes of the $ x_j, i \neq j$, but for calculating expected value it doesn't matter.  Of course, you don't have to use linearity of expectation, but you'll probably end up doing something really ugly like serialk11r's friend.

[quote="erdogankerem123"][quote="azjps"]The USAMTS year 17 article on expected value is pretty easy to understand: [url=http://usamts.org/Forms/ExpectedValue.pdf]here[/url].[/quote]

Thanks for replying, but on problem 5 on the link, I thought in order to get expected probability, it must multiply [the probability of happening] * [the gain if it does happen] for every different scenario, but in the solution given, I don't see where it adds in the cases where there are only 18 pairs or less; In the end, I could only see it multiplying [probability of having 19 pairs] * [19] ... Could someone explain this to me?[/quote]

This is just another application of linearity of expectation.  Let me name 19 distinct functions $ x_1, x_2, \ldots, x_{19}$.  Function $ x_i$ maps from an arrangement of 7 Bs and 13 Gs into the set $ \{0, 1\}$.  $ x_i$ takes the value $ 1$ if the $ i$th and $ i \plus{} 1$st people are different letters (either BG or GB), and takes the value $ 0$ otherwise.  Then the number of pairs that are BG or GB for a given arrangment $ A$ is $ \sum_{i \equal{} 1}^{19} x_i(A)$.  The number we want is $ E\left(\sum_{i \equal{} 1}^{19} x_i(A)\right)$.  By linearity of expectation, this is equal to $ \sum_{i \equal{} 1}^{19} E(x_i(A))$, and the expectation of $ x_i(A)$ is just the probability that the $ i$th and $ i \plus{} 1$st letters in the arrangement are different.  The rest is just computation.[/quote]

Thanks for giving such a detailed explanation, but most of the concepts that were talked about were new to me; I haven't learned much about operations on functions and etc. in any of my classes yet, and I also don't know what linearity of expectation is. Is there a good place to start on learning all of this stuff, such as a good website or something?

Also, after reading the various posts on my original question, I think the only thing that I don't understand about the solution is why it says $ /frac{1}{9}$ is the probability of matching for each sock, since I thought this probability changes as the number of socks still unsorted decreases.

[quote="erdogankerem123"]Summary (from memory, so I may have messed up):

There are 10 socks, 2 of each color, and a person randomly selects two to make pairs until all of the socks are in pairs. What is the average  number of socks that are in a pair with a sock of matching color?

I didn't understand why the given solution works:

[hide="given solution"] Each pair has a 1/9 chance of matching. There are 5 pairs, so avg. = 5/9.

What I don't understand:
I thought only the first sock would have a 1/9 chance of matching, then it would be 1/7, etc.

Also, if each pair did have a 1/9 chance of matching, I thought expected value = amount of gain * probability of happening, so I thought it would be $ 1*\frac {1}{9} \plus{} 2*\frac {1}{9}$ etc. [/hide][/quote]
you can't find the probability that the first sock matches because you have nothing to match it with, that's why you find it  for the second sock and then you multiply by the number of pairs there are to find the probability of all 5 pairs matching where you just times probability of one pair times 5.

[quote="erdogankerem123"]I think the only thing that I don't understand about the solution is why it says $ \frac {1}{9}$ is the probability of matching for each sock, since I thought this probability changes as the number of socks still unsorted decreases.[/quote]

Given that the first pair matched, the probability that the second pair matches is better than $ \frac {1}{9}$.  (In fact, it's $ \frac {1}{7}$.)  Given that the first pair didn't match, the probability that the second pair matches is worse than $ \frac {1}{9}$.  (In fact, it's $ \frac {6}{8}\cdot \frac {1}{7} \equal{} \frac {3}{28}$.)  Thus, the probability that the second pair matches is $ \frac {1}{9} \cdot \frac {1}{7} \plus{} \frac {8}{9} \cdot \frac {3}{28} \equal{} \frac {1}{9}$.  In other words, putting two socks together doesn't change the probability that the next two socks match, but knowing the outcome of this first pairing (whether it resulted in a match or not) does.

If this seems counter-intuitive to you, consider the following simplified system: there's a box with two marbles, one blue, one red.  I take a marble at random and hide it behind my back.  Then you reach into the box and choose a marble.  What's the probability that you get a blue marble?  What if I take a marble from the box and show it to you?

Anyhow, the upshot is that the second (or third, or fourth, or fifth) pairs will match $ \frac {1}{9}$th of the time.  The thing you're noticing (that this probability varies if we know the outcomes of previous trials) is what I mean when I say that these probabilities aren't independent.  But averaged over all possible outcomes of previous trials, we still get a pair $ \frac{1}{9}$th of the time, and this is all we need for the expected value.

Any introduction to elementary probability should cover all these issues -- I don't have any particular suggestions, though.

where can u find the official solution sets to the mandelbrot questions


Let $ X$ be a metric space without isolated points, is it always possible to
partition $ X$ into infinitely many dense subsets?

The only difference with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245512]this[/url] is that you need to partition the set of positive integer indices into infinitely many infinite sets instead of just the sets of odd and even numbers to get countably many disjoint dense sets instead of two.


[quote="Silverfalcon"]Solve simultaneously:

\[x+2y+z = 14\\
2x+y+z = 12\\
x+y+2z\]

Basically, we're looking for the values of x,y, and z.[/quote]

On your third equation, should it not have an answer?

Yeah. I should've posted it. :?

Now, it's fixed.

[quote="Silverfalcon"]Solve simultaneously:

\[x+2y+z = 14\\
2x+y+z = 12\\
x+y+2z = 18\]

Basically, we're looking for the values of x,y, and z.[/quote]
[hide]i'm gonna name the first equation a, the second b, and the third c

add the 3 of them up and you get 4x+4y+4z = 44
so x+y+z = 11, and gonna name that d.

subtract d from a and you get y is equal to 3

subtract d form b and you get x is equal to 1

subtract d form c and you get z is equal to 7

I got back and checked the answers... how is it not right? they work in every equation

a = 1+2(3)+7 = 14
b = 2(1)+3+7 = 12
c = 1+3+2(7) = 18[/hide]

??? :? ???

y isn't my solution right?

OBJECTION!!!!!! :D

if you put replace the variables with the actual numbers, it works.. tried all of them and perfectly fitted :?

Those should be correct. I don't know what jili is rambling about.


function
induction
real analysis
real analysis solved

Hi,
  I want to find all real functions f which verify  f(3X)+f(2x)+f(x)=0 and f differentiable in 0 .
  I think the only possibility is f=0 but I can't prove it.
  Thanks

One can prove by induction that f(x)= \sum (k=0,1,...,n) C(n,k)*f((1/3)^k*(2/3)^(n-k)x) for even n and minus that expression for odd n. This is easy, since we have f(x)=-f(x/3)-f(2x/3) and to finish the inductive step we take in thos expressions x/3 and 2x/3 and sum up. Thus, f(x)= \sum (k=0,2n) C(2n,k)f((1/3)^k*(2/3)^(2n-k)x). In another recent topic (some problem from our county olympiad) I have shown that this expression tends to f'(0)*x when n tends to infinity. THus, f(x)=xf'(0) and f'(0)=0 from where f is trivial.

We have f(0)=f'(0)=0. Let  \varepsilon > 0 and \delta>0 such that |f(x)|<= \varepsilon |x| if |x|\leq \delta. Then if \delta \leq |y| \leq 3\delta /2, x=y/3 verifies |x|\leq \delta and |2x|\leq \delta, so |f(y)|=|-f(x)-f(2x)|\leq|f(x)|+|f(2x)|\leq \varepsilon |x|+ \varepsilon|2x|=\varepsilon |3x|=\varepsilon |y|
i.e. |f(x)|\leq \varepsilon |x| is true if |x|\leq 3\delta/2. It's true for all x in R by induction. As \varepsilon can be chosen as small as we want, f=0


Find all functions $f: \mathbb{Q}\to \mathbb{R}$ such that for all $x,y\in \mathbb{Q}$: \[f(xy)=f(x)f(y)-f(x+y)+1.\]

[quote="pohoatza"]Setting $ m=1$ and $ n=t$ , where $ t$ is a positive integer, therefore we obtain $ f(t+1)=f(t)+1$, so $ f(t)=t+1$.

Now similar setting $ m=-1$ and $ n=1$, we obtain that $ f(-1)=0$, and by taking $ m=-1$ and $ y=t$ it gives $ f(-t)=-f(t-1)+1=-t+1$.

So $ f(z)=z+1$, for each integer $ z$.

By plugging $ m=t$ and $ n=\frac{1}{t}$, where $ t$ is an integer., we get
$ f(1)=(t+1)f(\frac{1}{t})-f(t+\frac{1}{t})+1$

Furthermore for $ m=1$ and $ y=p+\frac{1}{q}$, $ p,q$ are integers , we have that.

$ f(p+1+\frac{1}{q}) = f(p+\frac{1}{q})+1$, hence by induction $ f(p+\frac{1}{q}) = p+f(\frac{1}{q})$.

Therefore, $ f(\frac{1}{t}) = \frac{1}{t}+1$ , for each integer $ t$.

So, $ f(q)=q+1$, for all rational $ q$.[/quote]

The solution posted above is not complete; it misses the solution $f(x) = 1$.  I believe it is assuming that $f(1) = 2$.

Let $P(x,y)$ denote the given assertion. $P(0,0)\implies f(0)=1.$
Let $f(1)=c+1,$ where $c$ is any constant. $P(x,1)\implies f(x+1)=cf(x)+1.$
Making some basic substitutions one gets that $c=-1,0,1.$
a. $c=-1$: yields a contradiction which is not hard to see.
b. $c=0$: Then $f$ is identically 1.
c. $c=1$: Then $P(1,-1)\implies f(-1)=0.$ If $a,b$ are integers then comparing $P(a,1)$ and $P(a,-1)$ we get $f(a)=a+1.$ And $P(a/b,b)\implies 1+a=f(a/b)+bf(a/b)-f(a/b+b)+1.$ But we have that $f(a+b)=f(a)+b,$ so $f(x)=1+x,$ for any rational $x.$
All of the solutions work.

[quote=Arefe]It's exactly K17 just with a little difference in the situation that f(1)=0 , in that case we see that f(2x)=0 but now because we work with rational numbers we get f(x)=0 for all rational x but we know f(0)=1 , a contradiction so this situation doesn't have any answers . the else it's the same K17 .[/quote]

K17 is the same except the function is defined on Z.


In order to draw a graph of $f(x) = ax^2 + bx + c$, a table of values was constructed.  These values of the function for a set of equally spaced increasing values of $x$ were 3844, 3969, 4096, 4227, 4356, 4489, 4624, and 4761.  The one which is incorrect is:
\[ \text{(A)}\ 4096 \qquad \text{(B)}\ 4356 \qquad \text{(C)}\ 4489 \qquad \text{(D)}\ 4761 \qquad \text{(E) none of these} \]

[hide]Since it's a quadratic function, the answer must be a perfect square, and all four of the listed values are perfect squares. Therefore, the answer is $E$.[/hide]

[hide]I wrote out the numbers and found the differences between the terms, and got:
125, 127, 131, 129, 133, 135, 137.  Obviously the 131 and 129 are out of place, after finding out what the terms actually should be, I found that $4227$ should instead of $4225$.  So the answer is [b]E[/b].[/hide]

Jesusfreak, I'm not understanding something about your solution.  Just because something is a quadratic function doesn't mean $f(x)$ has to be a perfect square....for example if $f(x)=2x^2+2x+1$ then $f(2)=8+4+1=13$, which isn't a perfect square.

[hide]
I get E as well. But I merely subtracted each of the numbers from the one above/below and found 4227 is the one with problems
[/hide]

I also don't get what JesusFreak means because not all quadratic equations are perfect squares...

yes, you have to look at the differences, and Eryaman is right, being a quadratice doesn't mean you have to have a perfect square...

Hmm, let me revise that. For [b]this[/b] quadratic equation, it must be a perfect square; six of the seven given values are, and only $4227$ is wrong, because it should be $4225$, which is also a perfect square.


geometry
analytic geometry
graphing lines
slope
ratio

In a right triangle $ ABC$ with angle $ A \equal{} 90$ we draw the antitude $ AD$ and the median $ AM$. On $ AM$ we take point $ E$. The parallel through $ D$ to $ AC$ meets segment $ EB$ at point $ Z$. Prove that $ CE \perp AZ$.

[color=green][size=117][b]It is a nice and easy problem for the analytical geometry ![/b][/size][/color]

[quote="stergiu"][color=darkred]In a $ \triangle\ ABC$ with $ AB\perp AC$ denote $ \|\begin{array}{ccc} M\in (AC) & , & MA = MC \\
\ D\in (BC) & , & AD\perp BC\end{array}$.
For $ E\in AM$ denote $ Z\in BE$ for which $ DZ\parallel AC$. Prove that $ CE \perp AZ$.[/color][/quote]
[color=darkblue][b]Proof I (with [u]analytical geometry[/u]).[/b] $ \|\begin{array}{c} A(\ 0\ ,\ 0\ ) \\
\ B(\ 0\ ,\ b\ ) \\
\ C(\ c\ ,\ 0\ )\end{array}\|$ and $ m\in\mathcal R$  $ \implies$

$ \|\begin{array}{c} M(\ \frac c2\ ,\ \frac b2\ ) \\
\ D(\ \frac {b^2c}{b^2 + c^2}\ ,\ \frac {bc^2}{b^2 + c^2}\ ) \\
\ E(\ mc\ ,\ mb\ )\end{array}\|$ $ \implies$ $ Z(\ - \frac {mb^2c}{(b^2 + c^2)(m - 1)}\ ,\ \frac {bc^2}{b^2 + c^2}\ )\ .$ Therefore,

the slopes $ \|\begin{array}{c} s(CE) = \frac {m}{m - 1}\cdot\frac bc \\
 \\
s(AZ) = - \frac {m - 1}{m}\cdot\frac cb\end{array}\|$ $ \implies$ $ s(CE)\cdot s(AZ) = - 1$, i.e. $ CE\perp AZ\ .$[/color]

[quote="stergiu"][color=darkred]In a $ \triangle\ ABC$ with $ AB\perp AC$ denote $ \|\begin{array}{ccc} M\in (AC) & , & MA = MC \\
\ D\in (AC) & , & AD\perp AC\end{array}$.
For $ E\in AM$ denote $ Z\in BE$ for which $ DZ\parallel AC$. Prove that $ CE \perp AZ$.[/color][/quote]
[color=darkblue][b]Proof I (with [u]analytical geometry[/u]).[/b] $ \|\begin{array}{c} A(\ 0\ ,\ 0\ ) \\
\ B(\ 0\ ,\ b\ ) \\
\ C(\ c\ ,\ 0\ )\end{array}\|$ and $ m\in\mathcal R$  $ \implies$

$ \|\begin{array}{c} M(\ \frac c2\ ,\ \frac b2\ ) \\
\ D(\ \frac {b^2c}{b^2 + c^2}\ ,\ \frac {bc^2}{b^2 + c^2}\ ) \\
\ E(\ mc\ ,\ mb\ )\end{array}\|$ $ \implies$ $ Z(\ - \frac {mb^2c}{(b^2 + c^2)(m - 1)}\ ,\ \frac {bc^2}{b^2 + c^2}\ )\ .$ Therefore,

the slopes $ \|\begin{array}{c} s(CE) = \frac {m}{m - 1}\cdot\frac bc \\
 \\
s(AZ) = - \frac {m - 1}{m}\cdot\frac cb\end{array}\|$ $ \implies$ $ s(CE)\cdot s(AZ) = - 1$, i.e. $ CE\perp AZ\ .$

[b]Proof II ([u]synthetically[/u]).[/b] Denote $ \|\begin{array}{c} N\in AZ\cap BC \\
\ P\in AD\cap CE\end{array}$ . Observe that $ DZ\parallel AC$ $ \implies$

$ \frac {ZA}{ZN} = \frac {DC}{DN}$ . Apply the [b]Menelaus' theorem[/b] to the transversals :

$ \|\begin{array}{cccc} \overline {CEP}\mathrm {\ in\ }\triangle\ ADM\ : & \frac {CD}{CM}\cdot\frac {EM}{EA}\cdot\frac {PA}{PD} = 1 & \implies & \frac {PA}{PD} = \frac {CM}{CD}\cdot \frac {EA}{EM} \\
 \\
\overline {BZE}\mathrm {\ in\ }\triangle\ AMN\ : & \|\begin{array}{c} \frac {BN}{BM}\cdot\frac {EM}{EA}\cdot\frac {ZA}{ZN} = 1 \\
 \\
(\ \frac {ZA}{ZN} = \frac {DC}{DN}\ )\end{array} & \implies & \frac {NB}{ND} = \frac {BM}{CD}\cdot \frac {EA}{EM} \end{array}\|$ $ \implies$

$ \frac {PA}{PD} = \frac {NB}{ND}$ $ \implies$ $ NP\parallel AB$ $ \implies$ $ NP\perp AC\ .$ Therefore, $ \|\begin{array}{c} AD\perp NC \\
\ NP\perp AC\end{array}\|$ $ \implies$

the point $ P$ is the orthocenter of the triangle $ ANC$ $ \implies$ $ CP\perp AN$ $ \implies$ $ CE\perp AZ\ .$[/color]

 [quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u] In a $ \triangle\ ABC$ denote the midpoint $ M$ of the side $ [BC]$.

For $ \|\begin{array}{c} D\in BC \\
 \\
E\in AM\end{array}$ denote $ \|\begin{array}{ccc} Z\in BE\ ,\ DZ\parallel AC \\
 \\
N\in AZ\cap BC \\
 \\
P\in AD\cap CE\end{array}$. Prove that $ NP\parallel AB\ .$[/b][/color][/quote]

[quote="archimedes1"][hide="Synthetic--I think this works"]Extend $ DZ$ to meet $ AB$ at $ P$, extend $ BE$ to meet $ AC$ at $ R$, and extend $ CE$ to meet $ AB$ at $ Q$.
We have $ \triangle DPA$ is similar and has sides perpendicular to $ \triangle BAC$.  So if we can show that $ PZ/ZD \equal{} AQ/QB$, then we are done!
Now $ PZ/ZD \equal{} AR/RC$, so Ceva gives us $ \frac {AR}{RC} \cdot \frac {CM}{MB} \cdot \frac {BQ}{QA} \equal{} 1$.  This reduces to $ AR/RC \equal{} AQ/QB$, so we are done.[/hide][/quote]

 Yes, it works  :) because then $ \triangle APZ \sim  \triangle AQC$ , so $ \angle ACQ \equal{} \angle BAZ$ ect.

Let $N=DZ\cap AB, P=BE\cap AC, Q=CE\cap AB$. Since $E$ belongs to median $AM\implies\frac{AQ}{QB}=\frac{AP}{PC}\ (\ 1\ )$.
Since $ND\parallel AC,\implies\frac{NZ}{ZD}=\frac{AP}{PC}\ (\ 2\ )$.
But $\triangle NDA\sim\triangle ABC$ and have the homologous sides perpendicular, for our case $ND\bot AB$. Since, from $(1)\ \wedge \ (2)$ the points $Q,Z$ divide the homologous sides $AB, ND$ into the same ratio, i.e. $\frac{AQ}{BQ}=\frac{NZ}{ZD}\implies AZ\bot CQ\iff AZ\bot CE$, done.

Best regards,
sunken rock

[color=darkred][b][u]Extension[/u].[/b] For the midpoint $M$ of $[BC]$ in $\triangle ABC$ denote  $ \left\|\begin{array}{c}
\begin{array}{c}
D\in BC\ \searrow\\\\
E\in AM\ \nearrow\end{array}\ P\in AD\cap CE\\\\
X\in BE\ \ \rightarrow\ \ N\in AX\cap BC\end{array}\right\|$  . Prove that $ DX\parallel AC\iff PN\parallel AB\ .$[/color]

[color=darkblue][b][u]Proof[/u].[/b] [b]Menelaus's theorem[/b] $\ :\ \left\{\begin{array}{ccc}
\overline{CEP}/\triangle ADM & \implies & \frac {CM}{CD}\cdot\frac {PD}{PA}\cdot\frac {EA}{EM}=1\\\\
\overline{XBE}/\triangle ANM & \implies & \frac {XA}{XN}\cdot\frac {BN}{BM}\cdot\frac {EM}{EA}=1\end{array}\right\|\ \bigodot$ $\implies$ $\boxed{\frac {PA}{PD}=\frac {XA}{XN}\cdot\frac {BN}{CD}}\ (*)$ .

Hence  $DX\parallel AC\iff$ $\frac {XA}{XN}=\frac {DC}{DN}\stackrel{(*)}{\iff}$ $\frac {PA}{PD}=\frac {NB}{ND}\iff$ $PN\parallel AB$ . In conclusion, $ DX\parallel AC\iff PN\parallel AB\ .$[/color]

Dear Mathlinkers,
another way...

1. F the point of intersection of DZ and BM
2. the circle with diameter BC goes through F (angle chasing)
3. X the point of intersection of BZ and CE
4. by a converse of Pascal’s theorem (hexagone AAXCDFA), X is on the circle…
and we are done.

Sincerely
Jean-Louis


trigonometry
algebra
system of equations
algebra unsolved

Find all the real solutions $(x, y, z)$ to the following system of equations
$5 ( x + \frac{1}{x}) = 12 ( y + \frac{1}{y}) = 13 ( z + \frac{1}{z})$,

$xy + yz + zx = 1$.

Observe that $x,y,z$ have the same sign and that if $(x,y,z)$ is a solution, then $(-x,-y,-z)$ is a solution as well. Therefore, we can assume that $x,y,z$ are positive reals.
It is known that in a triangle $ABC$ we have
\[\tan {\frac{A}{2}}\cdot \tan {\frac{B}{2}}+ \tan {\frac{A}{2}}\cdot \tan {\frac{C}{2}}+\tan {\frac{B}{2}}\cdot \tan {\frac{C}{2}} =1,
\]
so take $x=\tan {\frac{A}{2}}$, etc. Then
\[x+\frac1x =\frac{2}{\sin{A}},
\]etc. and using the sine law we find that $ABC$ is a right triangle. Finally, we obtain $x=\frac15, y=\frac23 ,z=1$, so the solutions of the system are $(\frac15 ,\frac23 ,1)$ and $(-\frac15 ,-\frac23 ,-1)$.


complex analysis
function
complex analysis unsolved

1. If $ f(z)$ is a holomorphic and periodic function prove or disprove that the equation $ f(z)\minus{}z\equal{}0$ has at least one root.
2. If $ f(z)$ is a holomorphic function in the region {$ z|z<1$} and $ f(z)$ has real value in the region {$ z|z\equal{}1$}, what can you conclude about $ f(z)$ ?

I have an interesting problem which will be simply created  by adding a short phrase to 2)  :wink:

$ f(z)\equal{}f(\frac 1z)$ for $ z \not \equal{} 0$. Prove that $ f(z)$ is real for all real $ z \not \equal{} 0$.


i have been looking for graduate schools that provide a PhD in mathematical finance or financial engineering, however, most of the schools i have seen only have masters programs for these degrees. I was wondering if someone could tell me why this is so or if i am just not looking at the right schools.

Finance isn't exactly a strictly academic field, so I'm not sure if PhD programs for it even exist.

would my best bet be to go to graduate school for applied mathematics then?

You don't need to have a PhD to work in finance. A lot of the top companies either are indifferent or prefer to hire people right from undergrad (because they are younger and easier to train, I suppose).

one of the reasons i am thinking about graduate school is because i will not be graduating from a big name school like MIT or Stanford, and i think this might hurt my chances of getting a job at a top financial firm.

You can try to get a summer internship, so you can do some networking and have some experience. A good recommendation can be extremely valuable when you're looking for that kind of position -- the recruiters have to go through tons of applications, so if someone they respect and trust says "hire this guy," they may feel that they can save a lot of time with little risk (hiring someone that turns out to suck can cost them a lot) by hiring the guy with the good rec.

Princeton certainly has a Ph.D. program in finance (http://orfe.princeton.edu/). But I believe that other places also have similar programs. If you are interested in going to the industry after obtaining your degree, you might also consider getting a Ph.D. in physics. (AFAIK, many hedge funds are eager to hire PhDs in physics. And they don't really expect the applicants to have any background in finance.)


category theory
advanced fields
advanced fields unsolved

Hello,

I am extremely unexperienced in category theory, but I asked my professor and she did not yet know, and I have been searching for a long time

suppose you have a category C and you have two representants of a functor $F: C\rightarrow$ Set

$(A_1,a_1)$ and $(A_2,a_2)$

here $A_i$ is an object in C, and $a_i\in FA_i$

Now representing object means that there is a natural isomorphism from the $Hom(A_1,-)$ functor to $F$

where we map $g\in Hom(A_i,X)$ onto $F(g)(a_i)\in FX$

now what can I say about $A_1$ and $A_2$?
I guess they will be isomorphic objects

A really really nice result would be, there is a unique isomorphism $q: A_1\rightarrow A_2$ such that
$F(q)(a_1)=a_2$

Really, all thoughts, corrections, hints are welcome

Hi,

Indeed, your conjecture is true! :-) The proof is easy:

Since $(A_i,a_i)$ is a representing object of $F$, for all objects $X$ of $C$ and for all $y \in F(X)$ there's a unique morphism $g : A_i \to X$ such that $y = F(g)(a_i)$. For $i=1,X=A_2,y=a_2$ this yields a unique morphism $g : A_1 \to A_2$ such that $a_2 = F(g)(a_1)$, and for $i=2, X = A_1 , y = a_1$ a unique morphism $h : A_2 \to A_1$ such that $a_1 = F(h)(a_2)$. Now, $g \circ h$ is a morphism $A_2 \to A_2$ such that $F(g \circ h)(a_2)=a_2$. Then, by setting $i=2, X=A_2 , y = a_2$, our assumption implies $g \circ h = \text{id}_{A_2}$, and dually we derive $h \circ g = \text{id}_{A_1}$.


A monkey inserts a text on a keyboard with z symbols, the two parenthesis symbols ( and ) are included. The text written by the monkey has n symbols. Each symbol occurs equally often in average. Determine the probability that the text written by the monkey makes correct (proper) use of the parenthesis.

Note P(n) the asked probability, and consider the following events :
A = <symbol [>< correct text having n-2 symbols >< symbol ]>    ===> Pr(A) = P(n-2)*1/z^2
B = <other symbol than [ or ]><correct text having n-1 symbols>  ===> Pr(B) = P(n-1)*(1-1/z)
C = <correct text having n-1 symbols> <other symbol than [ or ]> ===> Pr(C) = P(n-1)*(1-1/z)
Manifestly P(n) = P(AuBuC) . We can observe that A\capB ={empty} , A\capC = {empty}  and
B\capC = <other symbol than [ or ]> <correct text having n-2 symbols> <other symbol than [ or ]>  ===> P(B\capC) = P(n-2)*(1-2/z)2
So P(n) = P(A) + P(B) + P(C) - P(B\capC) = 2*(1-1/z)*P(n-1) + (1/z2 - (1-2/z)2)*P(n-2) . It's a discret linear recursive equation.
The roots of its characteristic equation are (1-1/z),(1-3/z), so P(n) = C1*(1-1/z)n  + C2*(1-3/z)n
The C1,C2 constants are determined from P(0)=1 , P(1) = 1-2/z
Finally P(n) = ((1-1/z)n + (1-3/z)n)/2
 :cool:

ugh? what does proper use mean? I thouhg it just means there are no ) before ( so you'd get on a -1 level of parenthesis depth; but (A u (B \ C)) and stuff like it is allowed.

So all you need for this is the number of them, n?? :?

So I must have misunderstood your question :(

[quote="Diogene"]
A = <symbol [>< correct text having n-2 symbols >< symbol ]>    ===> Pr(A) = P(n-2)*1/z^2
B = <other symbol than [ or ]><correct text having n-1 symbols>  ===> Pr(B) = P(n-1)*(1-1/z)
C = <correct text having n-1 symbols> <other symbol than [ or ]> ===> Pr(C) = P(n-1)*(1-1/z)[/quote]
And what about "(word)(word)" ? This text is correct but it doesn't belong to A, B and C.
BTW, Pr(C) = P(n-1)*(1-2/z) and Pr(B) = P(n-1)*(1-2/z).

My approach is the following. Reserve $2k$ places where we will place paranthesis. There are $\frac{1}{k+1}C_{2k}^{k}$ possibilities to distribute $k$ pairs of paranthesis, and also $(z-2)^{n-2k}$ possibilities to fill other letters.
Therefore the answer is \[\frac{1}{z^{n}}\sum_{k=0}^{[n/2]}\frac{(z-2)^{n-2k}}{k+1}C_{2k}^{k}.\] I think there is a closed form for it.


let a,b,c>0 such that abc=1  prove that
$ \sum\frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{1\plus{}b\plus{}c\plus{}a} \geq 1$

Easy ;) .Expand ;)
<=>$ q^2\minus{}2qp\plus{}p^3\minus{}5p\minus{}3 \ge 0$
$ \Delta'\equal{}p^2\plus{}3\plus{}5p\minus{}p^3\equal{}\minus{}(p\minus{}3)(1\plus{}p)^2 \le 0$
Done ;)


Prove that if a is odd and n:ge:3, then a^(2^(n-2)) = 1 mod 2^n.

A solution using Euler theorem :

[hide]

phi (2^n) = 2^(n-1)

by Euler theorem a^(phi (2^n)) = 1 mod 2^n

b = a^(2^(n-2)), b:^2: = 1 mod 2^n so b = :pm:1 mod 2^n

a^(2^(n-2)) = a^(2^(n-3)):^2:

a^(2^(n-3)) is odd = 2A+1

If (2A+1):^2: = -1 mod 2^n then 2(2A(A+1) + 1) = 0 mod 2^n which is impossible because (2A(A+1) + 1) is odd.

[/hide]

your statement: if b:^2::equiv:1, b:equiv::pm:1 is not necessarily true, I think. take the example of b=3 or 5 and n=3.

You're [color=red]right[/color], my answer is full wrong  :oops:  :(

yeah, i know, i tried that too before running into that little obstacle... I think all the square roots of 1 mod (2^n) are all on either sides of powers of 2, though, which could still be useful...

for example, 3 and 5 in the counter-example i gave are on either sides of 4, which is a power of 2. i'm wondering if the square roots of 1 will always be on either side of a power of 2.

I'm pretty sure you guys are barking up the wrong tree.  Actually, I bet you could prove the Mystic's claim, but it doesn't look easy.  Instead, use the fact that [hide]every problem with the letter "n" in it can be solved using induction.[/hide]

[color=cyan]Well, maybe not [i]every[/i] problem.  But I think it's a good line.[/color]

It's actually not that hard (and surprise! induction DOES work!  :D): the problem is equivalent to proving that 2^n|a^(2^(n-2))-1 (*). We prove it for n=3 and then assume it's true for n and let's try to prove it for n+1. a is odd, so 2|a^(2^(n-2))+1, and by multiplying this with (*) we get 2^(n+1)|a^(2^(n-1))-1. Here I used the formula (A-B)(A+B)=A^2-B^2 for B=1 and A=a^(2^(n-2)).

[color=cyan]You squaring the expression right away works, too, because you get (powerof2)^2 - 2(powerof2) + 1, and the extra 2 comes in that way.[/color]

however, i do have a salvage to the one part....

i believe the solution relied on proving

a^(2^(n-2)) == +/- 1 (mod 2^n) right?

this is actually fairly easy to prove.
consider the order of a in the group of units (mod 2^n)
since phi(2^n) = 2^(n-1), the order of 2 divides 2^(n-1) i.e. the order k = 2^c for some c <= n-1. If c <= n-2, then the LHS == 1. (This is fairly clear)

otherwise then c is 2^(n-1). Then a is a primitive root of unity (mod 2^n) Therefore, -1 = a^k for some unique k taken modulo 2^(n-1).
But then 1 = a^2k. But then 2k = 2^(n-1) => k = 2^(n-2).

Therefore a^(2^(n-2)) == -1.
Proceed to desired contradiction from before.

Thats the end of that. ( I hope )


I know things like .989898989898...can be expressed as 98/99.

How do you express something like .99999999999 (goes on infinity) as a fraction?

[size=59][color=red]Please preview your posts and try to use correct grammar/spelling, etc.
b-flat[/color][/size]

[quote="rebirth"]i know things like .989898989898...can be expressed as 98/99

How do you express something like .99999999999 (goes on infinity) as a fraction?[/quote]

Refer to [url]http://en.wikipedia.org/wiki/0.999...[/url]

Predictably, because $0.\overline{98}= \frac{98}{99}$ we have the corresponding equality

$0.\overline{99}= \frac{99}{99}= 1$

There are many ways to show that $0.\overline{9}=1$.

Sample (and my favorite):

Let $x=0.\overline9$.

Then, we have:

$0.9+\frac{x}{10}=x$
$\frac{9x}{10}=0.9$
$x=1$.

$\mathbb{Q.E.D}$

Careful...

When you say that $\sum_{k=0}\frac{1}{2^{k}}=2$, equality means something different that when you say $2=2$. The former definition uses the idea of a limit (to learn more about rigirous definitions of limits, investigate: delta-epsilon proofs).

[quote="rebirth"]How do you express something like .99999999999 (goes on infinity) as a fraction?[/quote]

[hide]

Well,

$x=.\overline{9}$ can be expressed as

$0.9+0.09+0.009$...

$\to$

$\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}$

This is an geometric sequence, where r is $\frac{1}{10}$. It can also be expressed as $\sum_{n=0}^{\infty}\frac{9}{10}\left(\frac{1}{10}\right)^{n-1}$.

To solve an infinite series, use $\frac{a_{1}}{1-r}$. In this case, since $a_{1}= \frac{9}{10}$, we get

$\frac{\frac{9}{10}}{1-\frac{1}{10}}$

$\frac{\frac{9}{10}}{\frac{9}{10}}$

$\frac{9}{9}$.

This means that $.\overline{9}=1$ so therefore, any fraction where the numerator equals the denominator (except when they are 0) will equal $.\overline{9}$.

[/hide]


LaTeX
inequalities
inequalities unsolved

Let $x,\;y,\;z$ be positive real numbers such that $xyz = 1$. Prove that

$\frac{1}{x^3 - x^2 + x} + \frac{1}{y^3 - y^2 + y} + \frac{1}{z^3 - z^2 + z} \geq 3$

Can some one change it lo LATEX please...  :) I don't know how to write it in LATEX.  :blush: hehehe sorry...  :P I'm still a beginner.  :lol:

Let $x,y,z \in \mathbb{R}_{+}$ such that $xyz=1$ Prove that
\[\displaystyle \frac{1}{x^3-x^2+x}+\frac{1}{y^3-y^2+y}+\frac{1}{z^3-z^2+z} \geq 3\]

Thanx!!! ;)
but you forget to write in latex that the x,y,z are positive reals and the product is one.

i'm very upset.... :mad:
No one try to solve my problem!!!

hehehe just kiddin'   :D

c'mon everybody, try my problem!!!
it's not that hard...

thanx for writing the latex version!!!

$\frac{1}{x^3-x^2+x} = \frac{1}{x(x^2-x+1)} \ge \frac{4}{(x^2+1)^2}$
Replace $x^2$ by a we have to prove
$abc=1$ then $\sum \frac{1}{(1+a)^2} \ge \frac{3}{4}$
But this is well known. If we have $abcd=1$, then
$\sum \frac{1}{(1+a)^2} \ge \frac{1}{1+ab}+\frac{1}{1+cd} = 1$
Now, put d=1 and we can get the desired inequality.

Excellent Soarer!!!
The first two steps are similar to my solution.
Except i don't introduce the variable d.
Instead, i use the fact that $xyz = 1$ to change the ineq into a one variabel inequality which is very easy to prove.

Namely,

$\frac{4}{(x^2+1)^2} + \frac{4}{(y^2+1)^2} + \frac{4}{(z^2+1)^2} \geq \frac{4}{(x^2y^2 + 1)} + \frac{4}{(z^2 + 1)^2} = \frac{4z^2}{(z^2 + 1)} + \frac{4}{(z^2 + 1)^2} = \frac{(4z^4 + 4z^2 + 4)}{(z^2 + 1)^2} \geq 3$

Yaaaaayyyyyy my first latex typing!!!!!!!!!
Really beautiful!!!!!!!!!!!
Thank you very much arne!!!!!!
I really appreciate it!!!!!
hahahaha
yaaaaayyyyyyy!!!!!!! from now on i can write in latex!!!!!  :rotfl:

Maybe I should answer your question with "no", Detective, since then you really HAVE to learn it. It doesn't take more than 5 minutes.

That's right arne.

I have to learn it because i need to learn it too.

Just tell me where i can learn it okay.
or maybe you're willing to teach me??? :lol:
then i'll be very glad.
hehehe
anyway thank you very much arne.


vector
inequalities unsolved
inequalities

For $a_k,b_k,c_k$ where $    k=1,..,n$ prove that

$( \sum a_kb_kc_k)^2 \leq \sum(a_k)^2\sum(b_k)^2\sum(c_k)^2 $

This is Cauchy again:
First if $\sum a_k^2. \sum b_k^2. \sum c_k^2 =0$ then it is trivial.
If not,we have:
[tex]\sum \frac{|a_k|}{\sqrt{\sum a_i^2}}\frac{|b_k|}{\sqrt{\sum b_i^2}}\frac{|c_k|}{\sqrt{\sum c_i^2}} \leq \sum \frac{|a_k|}{\sqrt{\sum a_i^2}}\frac{|b_k|}{\sqrt{\sum b_i^2}}[/tex]
[tex]\leq \sum (\frac{a_k^2}{\sum 2a_i^2}+\frac{b_k^2}{\sum 2b_i^2})=1 [/tex]
[tex]\Leftrightarrow (\sum a_kb_kc_k)^2\leq\sum a_k^2 \sum b_k^2 \sum c_k^2[/tex]. :)


The sum of the lengths of the twelve edges of a rectangular box is $140$, and the distance from one corner of the box to the farthest corner is $21$. The total surface area of the box is

$\text{(A)}\ 776 \qquad \text{(B)}\ 784 \qquad \text{(C)}\ 798 \qquad \text{(D)}\ 800 \qquad \text{(E)}\ 812$

[hide="Solution"]Let $x$, $y$, and $z$ be the dimensions of the box. We are given $4(x+y+z)=140$, and $21=\sqrt{x^2+y^2+z^2}$, and we're trying to find the value of $2(xy+xz+yz)$.

Notice that $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$, so that we can easily use substitution to find the surface area.

$35^2=21^2+2(xy+xz+yz)$, and $2(xy+xz+yz)=1225-441=784$.[/hide]


Believe it or not, I just played it for the first time yesterday. Post what you think of it, strategies, etc. here. In short, discuss.

Such an awesome game. If you go to the Wikipedia page you can find a bunch of links to strategy stuff if I remember correctly. If not try google.

I generally go ore-grain or single-resource and port. Anyone play online (Sea3D)?

Hmm, no one is discussing... Maybe people would be more interested if they knew what it was? I don't know, anyway, [url=http://en.wikipedia.org/wiki/Settlers_of_catan]here's[/url] the wiki link.

[quote="worthawholebean"]I generally go ore-grain or single-resource and port. Anyone play online (Sea3D)?[/quote]Yeah, I discovered this game back at the IMC in 2003. Ever since then I've played it at almost every math competition I was at, and even online.

Worthawholebean, a better online site (with more players, ladders, etc) is http://ladder.jsettlers.com/index.html (you can also play training games, including one with computer only).

Wow, thanks, I never knew about that. But I have yet to learn how to use it... What does ntrr7 or something mean? Yeah, I'm noob I guess. Also, how do you play against computers?

I've never gotten beaten by a computer :)

How good is the AI for settlers of catan right now, anyway?

Unfortunately my parents won't let me get a java account or give someone my email address to play a game.  I didn't ask yet, but I know they won't let me.  My mom is already angry at me because I'm addicted to FTW, which I'm not.


In $ABC$ draw medians $AD$, $BE$, and $CF$. Then pick $G$ on $AC$ so that $BG$ is perpendicular to $AC$. Prove that $\angle FGD = \angle FED$.

We know that $FE\parallel BD$ and $ED\parallel AB \longrightarrow \angle FED=\angle ABC$
now $\angle BGC=90$ and $BD=DC \longrightarrow GD=BD=DC \longrightarrow \angle BGD=\angle GBD$ and similiary we get that:$\angle BGF=\angle GBF$
$\longrightarrow \angle FGD=\angle ABC=\angle FED$ :)

[hide]We know that $BF=ED$ and $BD=EF$.
Since $\triangle BAG$ is right, and $FG$ is the median to the hypotenuse, $FG$ is congruent to half the hypotenuse, $FG=BF$. Thus $FG=ED$.
Similarly, since $\triangle BCG$ is right, $DG=BD=EF$.
By SSS, $\triangle FGD \cong \triangle DEF \implies \angle FGD=\angle FED$.[/hide]


Let $ABCD$ be a parallelogram with $\angle ABC = 120^\circ, AB = 16$ and $BC = 10$. Extend $\overline{CD}$ through $D$ to $E$ so that $DE = 4$. If $\overline{BE}$ intersects $\overline{AD}$ at $F$, then $FD$ is closest to

$\text{(A)} \ 1 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 4 \qquad \text{(E)} \ 5$

[hide]When you extend and draw, FD || BC.

Using ratios, ED/EC = 4/20 = FD/BC = x/10.  X = 2 so the answer would be (B).[/hide]

[hide]We know angle BFA is congruent to angle EFD by the Vert. Angles Thm. Angle ABF is congruent to angle DEF by PAI. Triangle ABF ~triangle DEF by AA ~.We can see from sides AB and DE that the sides are in a ratio of 1:4.This means DF:AF=1:4=2:8. DF is closest to 2, answer B.[/hide]


probability
vector
inequalities
geometry
3D geometry
sphere
real analysis

I have seen a solution that uses Turan's theorem from graph theory.

Here's the basic idea.  Given two vectors in the Euclidean space, draw an edge between them if the angle between them is greater than $120^\circ$.  The key geometric fact is that there are no triangles in the resulting graph.

We first prove the inequality when the probability distribution is a uniform distribution on a finite set of points.  But that case is quickly settled by Turan's theorem, which says that a triangle-free graph on $n$ vertices has at most $\frac{1}{4}n^2$ edges.

For more general probability distributions, we can approximate them as closely as we want by uniform distributions, so by a continuity argument we are done.

[quote="sam-n"]i dont know am i right about the forum of this problem.
Let $X$ and $Y$ be independent identically distributed random variables taking values in euclidean space .prove that $P(|X+Y|\geq x)\geq \frac12 P(|X|\geq x)^2$ for every $x\geq 0$.[/quote]
Let $\alpha = P(X \geq x/2), \, \beta = P(X \leq -x/2)$.
Then  $P(|X+Y|\geq x)\geq P(X \geq \frac{x}{2}, \, Y \geq \frac{x}{2}) + P(X \leq \frac{-x}{2}, \, Y \leq \frac{-x}{2}) = \alpha^2 + \beta^2$.

Since $a^2 + b^2 \geq \frac{1}{2}(a+b)^2$, it follows that $P(|X+Y|\geq x)\geq \frac{1}{2}(\alpha + \beta)^2 = \frac{1}{2}P \left(|X| \geq \frac{x}{2}\right)^2$, which is a stronger version of the desired inequality.

[quote="Ravi B"]Your solution assumes that $X$ and $Y$ have real values.  But actually $X$ and $Y$ are vectors in a Euclidean space.[/quote]

Ah yes ... Euclidean, with inner product  $<u,v> = u^Tv$. Then

$P(|X+Y|\geq x) = P(|X+Y|^2\geq x^2) = P(|X|^2 + 2X^TY + |Y|^2 \geq x^2)$
$... \geq P(|X|^2 + |Y|^2 \geq x^2,\, X^TY \geq 0)$
$... \geq P(|X| \geq x, \, |Y| \geq x, \, X^TY \geq 0)$
$... = P(X^TY \geq 0 \big| |X| \geq x, \, |Y| \geq x) P(|X| \geq x, \, P(|Y|\geq x)$
$... = P(X^TY \geq 0 \big| |X| \geq x, \, |Y| \geq x) P(|X| \geq x)^2$.

It therefore suffices to show that $P(U^TV \geq 0) \geq \frac{1}{2}$, if  $U$ and $V$ are i.i.d. random variables.

Now let $u$ be an arbitrary unit vector and let $A = \{z|z^Tu \geq 0\}, \, B = A^c, \, P(U \in A) = \alpha, \, P(U \in B) = 1 - \alpha$. Then

$1 = P(U, \, V \in A) + P(U \in A, \, V \in B) + P(U \in B, \, V \in A) + P(U, \, V \in B)$
$... = \alpha^2 + 2 \alpha(1-\alpha) + (1-\alpha)^2$ .

If $U, \, V \in A$ or $U, \, V \in B$, then $U^TV \geq 0$. Therefore $P(U^TV \geq 0) \geq \alpha^2 + (1-\alpha)^2 \geq \frac12$.   :)

[quote="Ravi B"][quote="hpe"]It therefore suffices to show that $P(U^TV \geq 0) \geq \frac{1}{2}$, if  $U$ and $V$ are i.i.d. random variables.[/quote]Unfortunately, this lemma is false.  For example, in two dimensions, consider the uniform distribution on the set of three cube-roots of unity.  Then the probability is $1/3$.

[quote="hpe"]If $U, \, V \in A$ or $U, \, V \in B$, then $U^TV \geq 0$.[/quote]Unfortunately, this claim is false, which invalidates your proof of the lemma.[/quote]

You got me again  :oops:

The lemma to prove is: If $U, \, V$ are iid rv's, then $P(U^TV \geq -\frac12 |U||V|) \geq \frac12$. That is, the median angle between two independently chosen random vectors on the n-sphere (regardless of the distribution) is no more than $120^o$.


geometry
circumcircle
perimeter
geometry proposed

$O$ is a point in triangle $ABC$. We draw perpendicular from $O$ to $BC,AC,AB$ which intersect $BC,AC,AB$ at $A_{1},B_{1},C_{1}$. Prove that $O$ is circumcenter of triangle $ABC$ iff perimeter of $ABC$ is not less than perimeter of triangles $AB_{1}C_{1},BC_{1}A_{1},CB_{1}A_{1}$.


2.  Using the Laws of Boolean Algebra, simplify the following expressions.  List the specific law used for every simplification.

a)  Z =  P.Q.R + P.R

b)  Z = (P+Q).(R+S) + P.(R + ~R)

These are just practice homework questions but I am having a hell of a time simplifying these. I learn well by example so if anyone would care to take a stab at these and show step by step that would be great.

Where can I get some help with this then?


trigonometry
arithmetic sequence
algebra unsolved
algebra

[quote="April"]Let $ u_1$, $ u_2$, $ \ldots$, $ u_{1987}$ be an arithmetic progression with $ u_1 \equal{} \frac {\pi}{1987}$ and the common difference $ \frac {\pi}{3974}$. Evaluate
\[ S \equal{} \sum_{\epsilon_i\in\left\{ \minus{} 1, 1\right\}}\cos\left(\epsilon_1 u_1 \plus{} \epsilon_2 u_2 \plus{} \cdots \plus{} \epsilon_{1987} u_{1987}\right)
\]
[/quote]
Doesn't it follow from $ cos(x) \plus{} cos(\minus{}x) \equal{} 0$ that $ S \equal{} 0$. Since if $ \cos\left(\epsilon_1 u_1 \plus{} \epsilon_2 u_2 \plus{} \cdots \plus{} \epsilon_{1987} u_{1987}\right)$ is part of the sum then $ \cos\left(\minus{}\epsilon_1 u_1 \plus{} \minus{}\epsilon_2 u_2 \plus{} \cdots \plus{} \minus{}\epsilon_{1987} u_{1987}\right)$ is also part of it.

$ \cos(x)$ is even, not odd  :wink:
I like this problem - its first part is a generalization of the trig identity for sum of $ \cos(x)$. We'll use the representation $ \cos(x) \equal{} \frac {e^{i\pi x} \plus{} e^{ \minus{} i\pi x}}{2}$:
$ S \equal{} \sum_{\epsilon \in \{ \minus{} 1,1 \} ^{n}} \frac {e^{i\pi \sum \epsilon_i u_i} \plus{} e^{ \minus{} i\pi \sum \epsilon_i u_i}}{2} \equal{}$
because $ f( \epsilon) \equal{} \minus{} \epsilon$ is bijective:
$ \sum_{\epsilon \in \{ \minus{} 1,1 \} ^{n}} e^{i\pi \sum \epsilon_i u_i}\equal{}$
from expanding the following product:
$ \prod_{i \equal{} 1}^{1987} (e^{i\pi u_i} \plus{} e^{ \minus{} i\pi u_i}) \equal{} \prod_{i \equal{} 1}^{1987} 2\cos(u_i) \equal{} 2^{1987}\prod_{i \equal{} 1}^{1987} \cos(u_i)$
But $ u_{1986} \equal{} \frac {\pi}{2}$, so the product is 0.

[quote="bambaman"]$ \cos(x)$ is even, not odd  :wink:[/quote]
You are right.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose that $ |G|$ and $ |aut{G}|$ are relatively prime (both finite). Prove that $ G$ is Abelian.

Look at the group of inner automorphisms, $ T_g$ that sends $ x \in G$ to $ gxg^{\minus{}1}$

[quote="t0rajir0u"]Correct me if I'm wrong:

We must have $ |T_g| | |G|$ and $ |T_g| | |\text{Aut}(G)| \implies |T_g| \equal{} 1$, so $ T_g$ contains only the identity element $ x \mapsto x$ and we conclude that $ gxg^{ \minus{} 1} \equal{} x \Leftrightarrow gx \equal{} xg \forall x, g \in G$.[/quote]

Why would we have $ |T_g| \mid |G|$?  $ T_g \leq \text{Aut} G$ but $ T_g$ is not a subgroup of $ G$, so you can't apply Lagrange to get $ |T_g| \mid |G|$.  I think...

Yes, but $ T_g$ is isomorphic to a quotient of $ G$ (with its center), so its order still divides $ |G|$.


Hmm.. not sure were to ask this, but since it comes up in the AMC series alot(supposedly) and since this forum is popular and I will get my response quicker :lol: , What is cauchy? I think it was mentioned as an inequality or something.

Rep123max wrote:If  are real numbers with , then what is the maximum value of ? (Source: Mock AMC D)

#1-[hide]

Make two sequences, with sequence  being , and sequence  being . Using Cauchy's inequality, we may write this as:

The maximum value of the given equation occurs when there is equality. Thus, it's maximum value is .[/hide]

Quote:If real numbers  satisfy  and , compute the maximum value of . (Source: 1978 USAMO)

#2-[hide]Since you want , you want to set up an equation in termso f the other four values, and then replace them. Thus, we start by rearranging the two equations into  and . Using Chauchy's with the two sequences  and , we can set up an inequality:

This equality only holds true in the set . Thus, the maximum value of  is .[/hide]

#2 seemed far too easy to be a USAMO problem.

[quote]#2 seemed far too easy to be a USAMO problem.[/quote]

When I looked at some older olympiads the other day, I was quite surprised at how much more difficult USAMO/IMO problems have gotten over the years.  Take a look at the very first (1959) IMO, problem #1, for example (available on the Kalva website).  Then compare it to #1 of the 2003 IMO.

Yakko


salam bacheha.ye moddate inja kheili fa`aliat kam shode va be har hal nazdike marhale 2 hastim.pishnahadam ine ke too in post har kasi,age soale khoobi dare,hala dar sathe marhale 2 ya balatar bezare (soala tori bashan ke halleshoon vase marhale2 mofid bashe) . nemidoonam chaghad Steghbal mikonid!
man ba ye soale nazarie va ye soale hendese shooroo mikonam.
bad har kas soalio hal mikone kheili kootah faghat raveshe karesho bege,ke vaghtesho nagire! :wink:

1.dar mosallase $ ABC$ , $ T$ mahalle barkhorde nimsaze rase $ A$ ba dayere mohiti,va $ X$ mahalle barkhorde ertefae nazire rase $ A$ ba dayere mohitist. $ TX$ , $ BC$ ra dar $ A'$ ghat mikonad.sabet konid $ A'$,$ B'$ va $ C'$ hamkhattand va in khat bar $ HI$ amood ast.
2.tamame adade avale p,q,r ra biabid ke $ 25pq\plus{}r\equal{}2004$ shavad va $ pqr\plus{}1$ morabbae kamel bashad. :)

[b]1:
Ba mohasebeye zavie va ba inke gharineye H nesbat be BC rouye dayereye mohiti hast va AH amoud bar BC, be rahati A' mahalle barkorde amoude az H bar nimsaze A mishavad. hala kolli 4zelie mohati va kolli zavie darim. kafist esbat konim IH bar A'B' amoud ast. va be rahati zavaya masale ra hal mikonand! :maybe: !! :lol:

2:
ruye in rah fekr nakardam vali shayad A'B'C' ghotbie H nesbat be dayereye mohati bashad!!!

Kholase tar nemishod :) [/b]

Bebakhshid soalo dir didam!

manam ba rahe 1 hal kardam :wink:
khob khodetoonam soal bezarin dg.ye zare inja ro active konin

[quote="goodgirl_2278"]1.dar mosallase $ ABC$ , $ T$ mahalle barkhorde nimsaze rase $ A$ ba dayere mohiti,va $ X$ mahalle barkhorde ertefae nazire rase $ A$ ba dayere mohitist. $ TX$ , $ BC$ ra dar $ A'$ ghat mikonad.sabet konid $ A'$,$ B'$ va $ C'$ hamkhattand va in khat bar $ HI$ amood ast.
2.tamame adade avale p,q,r ra biabid ke $ 25pq \plus{} r \equal{} 2004$ shavad va $ pqr \plus{} 1$ morabbae kamel bashad. :)[/quote]

avvalan be vozooh $ r$ nemitone $ 2$ bashe pas $ p$ va $ q$ har do fardand .Darim :

$ pqr \equal{} (a\minus{}1)(a\plus{}1)$  amma  $ (a\minus{}1,a\plus{}1) \equal{} 1$ pas yeki az halaate zir be vojood miad :

1- $ pq \equal{} r \plus{} 2$ dar in soorat $ 26r \equal{} 1954$ ke choon $ 1954$ bar $ 13$ bakhshpazir nist pas javab nadare .

2-$ r \equal{} pq \plus{} 2$ dar in halat $ 26pq \equal{} 2002$ pas $ r\equal{}79$ va $ pq\equal{}77$

3-$ pr \equal{} q \plus{} 2$ pas
$ 25p(pr\minus{}2) \plus{} r \equal{} 2004 \Rightarrow  r(25p^2 \plus{}1) \equal{} 2004 \plus{} 50p$
$ \Rightarrow  25p^2 \plus{} 1 | 2004 \plus{} 50p \Rightarrow  25p^2 \plus{} 1 | 2004p \plus{} 50p^2 \Rightarrow  25p^2 \plus{} 1 | 2004p \minus{} 2 \Rightarrow   25p^2 \plus{} 1 | 2004p^2 \minus{} 2p \Rightarrow  25p^2 \plus{} 1 | 4p^2 \minus{} 2p \minus{} 80 \Rightarrow  2p^2 \minus{} p \minus{} 40 \le 0  \Rightarrow  p<5 \Rightarrow  p \equal{}3$ ke be sadegi mishe did gheire momkene ke $ p\equal{}3$
4-$ pr \plus{} 2 \equal{} q$ ba tekrare hamin marahele bala bazam mibinam ke javabi nadarim .
5, 6 = $ qr \plus{} 2 \equal{} p$ va $ p \plus{} 2 \equal{} qr$ ham daghighan mesle 2taye balayiie pas javabha ebaratand az :
$ (p,q,r) \equal{} (11,7,79) , (7,11,79)$
dorost bud ?

inam  soal :
1-yek jadvake $ n*n$ darim ke har yek az adade $ 1$ ta $ n$ daghighan $ n$ bar toosh oomadan sabet konid satr ya stooni ba haddeaghal $ \sqrt n$ adade motamayez vojjod darad .

[quote="Toxic_Frog"]
1-yek jadvake $ n*n$ darim ke har yek az adade $ 1$ ta $ n$ daghighan $ n$ bar toosh oomadan sabet konid satr ya stooni ba haddeaghal $ \sqrt n$ adade motamayez vojjod darad .[/quote]

[url]http://www.mathlinks.ro/viewtopic.php?t=222368[/url]


Anyone interested in making an AoPS record by scheduling a time where everyone gets on at one time so we can set the "Most users on at one time" at the bottom of the "Home" Page

How about St. Pat's day?

P.S. When is that?

Hmm beating 593 would be hard but probably manageable. Just name the time and date

Okay, everyone from now on let's say times in CENTRAL TIME (U.S.)   Okay, i just thought that you guys might name the dates. Anyhow, how about Febuary 17 (Sunday) 11 A.M (Central Time)?

ok, let's do it

I'll get other AoPSers from my school to participate if possible

That's a good idea.

Let me check my time zone... GMT+8

Let me check the central time zone... GMT-6

So, the time in my time zone is... 1am on Monday :o ... i will be sleeping then

I believe there was over 800 at one point (due to a server problem)?

Um... If I knew what time it was in Greenwich right now (when I post this) I could tell you yes or no.

Can somebody post time zone conversions?  I'm too lazy to.

Sounds... interesting. But I'm not sure about the time zones.

If you just left the window open all day, would it count as you being on?
If so,
count me in.
Else,
Too lazy!  :lol:
BTW, check out the sailboat!
_/)

I think some time like 7 or 8 is best because most people don't get home until around then.  Besides, you can tell your parents what's going on.

[quote="miyomiyo"]Friday at 7 or 8 doesn't work?[/quote]

Will be at a Math Circle meeting which is three hours long.

How about SATURDAY?!?!?!?!?!?!??!?!!?!?!?!?!?!??!?!?!?!?!?!?!??!?!?!?!?!?!

Guys I don't even think we can pull it off. We'll maybe get 100 users if we're lucky but that's pretty much it.

I agree with 7h3.D3m0n.117
I doubt others will care

True...  :D

Alright should we post the link to this topic in our signatures...?
How else can we get more ppl
and what exactly is the date anyways  :huh:

That's why we get 100 users when the place already has lots of users on.

I have an idea. How about on pi day we just get on AoPS as much as we possibly can. That way, time won't be that much of an issue.

... can't do pi day... school celebration. Cant go on... Want to be part of record.

Okay, everyone put a link to this post in a signature...i'll do it later g2g

No, somebody should make a cleaner, neater topic with just one post.  I'll do it in about half an hour.

there are people who go to church...there are people who need to eat dinner at a table with the family...it's normal to have problems :)  we'll try and see..

Nightmare, stop posting on old threads and look at the annoucements. Not to mention the links on like everyone's sigs.


function
inequalities proposed
inequalities

Here is a nice problem, generalizing both [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17207]GM_15[/url] and Nesbitt:

Let a, b, c be three positive reals. Prove that the function

$f\left( t\right) =\frac{a^t}{b^t+c^t}+\frac{b^t}{c^t+a^t}+\frac{c^t}{a^t+b^t}$

is monotonically increasing for $t\geq 0$.

I wonder why nobody has noted this one, especially since the solution is so easy with Harazi's idea...

  Darij

Thanks, darij, but I have obtained this result long time ago... Anyway, who dares an even larger generalization?


$A',B',C'$ are on the sides $BC, CA$ and $AB$ of a $\triangle ABC$ with area $F$, such that $\frac{AC'}{AB}=\frac{BA'}{BC}=\frac{CB'}{CA}=p$ where $0<p<1$. Find
$1)$ the area $S$ of the $\triangle A'B'C'$. $f(p)=S$;
$2)$ $p$ when $S$ is minimal;
$3)$ the locus of all points $T=A'X/\ C'Y$ - parallels to $AB, AC$ when $p\in (0;1)$

Will [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=divide&t=40828]this[/url] give you some ideas?


Find the quadratic equation $ax^{2}+bx+c=0$ that has a real

double root $r$ that satisfies $2r^{2}+7r+3=0$,

where $a, b,$ and $c$ are integers and

 $\frac{b}{a}=1$

What's a real double root?

Thank you :)

I don't know how to define it, but I can give an example:
4 is a double root of $x^{2}-8x+16=0=(x-4)^{2}$ because the exponent of $(x-4)$ is 2.

[hide="Solution"]$2r^{2}+7r+3=(2r+1)(r+3)$ so the roots are $-\frac{1}{2},-3$  If $ax^{2}+bx+c=0$ has a double root, it must be in the form $(x-r)^{2}=x^{2}-2rx+r^{2}$ with a double root r.  Since $\frac{b}{a}=1$, we have that $a=b\rightarrow 1=-2r\rightarrow r=-\frac{1}{2}$.  This works since it is one of the roots of the quadratic, so the answer is the quadratic $x^{2}+x+\frac{1}{4}$[/hide]

[quote="mstoenescu"][quote="myc"]What's a real double root? [/quote]

The equation $x^{2}-2x+1=0$ has the double root $x=1$ that means the roots are equals ![/quote]
$x^{2}-2x+1=0;\ x=1$

$x^{2}+2x+1=0;\ x=-1$

[hide="My Solution"]

First solve $2r^{2}+7r+3=0$:

$(x+3)(2x+1)=0\Longrightarrow x=-3,-\frac{1}{2}$

Now try the two results:

$(2x+1)^{2}=4x^{2}+4x+1$ <=answer.

[/hide]


AMC
USA(J)MO
USAMO
geometry
summer program
PROMYS

How did everyone do on the usamo over the years? scores are higher or lower each year?

10th: 6. Only knew of USAMO's existence before qualifying (I was first at my school to do so). My preparation consisted of learning the definitions of the triangle centers, and that was about the only geometry I knew.

11th: 19. I attribute a lot of this to PROMYS for teaching me rigor, but I had also learned a lot of things over the past months to actually have a decent shot at doing well.

8th:3...didnt know a lot of the symbols on the test  :blush:
9th:11...i studied

10th: 10

Making me like... tied for 7th within my school >_<


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a group of prime order $ p^a$, with $ a > 0$. By considering the conjugation action of $ G$, show that the centre $ Z$ of $ G$ is non-trivial.
Show that any group of order $ p^2$ is Abelian, and that there are (up to isomorphism) just two groups of that order for each prime $ p$.

The hint is pretty clear.  Write down the class formula.

Let $ c_1, ... c_n$ denote the conjugacy classes of $ G$.  Since these are a partition of $ G$, it follows that $ |G| \equal{} |c_1| \plus{} ... \plus{} |c_n|$, which is the class formula.  What does it mean if $ |c_i| \equal{} 1$?  What do you know about $ c_i$ if $ |c_i| > 1$?

It means the element in $ c_i$ is conjugate only to itself - so it's a member of the center of $ G$?

Yep.  What happens if you assume that the center is trivial?

Then $ c_i$ must equal $ e$, and there is only one such $ c_i$. I'm not sure how to proceed from here - is it true that the center is a union of conjugacy classes of size 1?

Yes; it is the union of all such conjugacy classes.  Given that $ G$ is a $ p$-group, what can you infer about the other conjugacy classes?

I know subgroups of $ G$ have orders of the form $ p^k, k < a$, but I don't know anything special about conjugacy classes of a group with order $ p^a$  :maybe:  If we somehow assume (I don't know if this is true) the number of elements of a conjugacy class must divide the order of the group, then all conjugacy classes have orders $ p^k$ for some $ k$, and the sum of the orders must be $ p^a$.

Then we can have a contradiction because $ p$ must divide both sides of the equation, so the center cannot be trivial?

[quote="aidan"]If we somehow assume (I don't know if this is true) the number of elements of a conjugacy class must divide the order of the group, then all conjugacy classes have orders $ p^k$ for some $ k$, and the sum of the orders must be $ p^a$.

Then we can have a contradiction because $ p$ must divide both sides of the equation, so the center cannot be trivial?[/quote]
Yes.  What do you know about orbits of group actions?

Orbits of group actions form a partition of the set the group is acting on.. can't thinkg of anything else significant

Do you know the orbit-stabilizer theorem?

I think that's $ |G| \equal{} |\mathcal{O}_P|\cdot|G_P|$ where $ G_P$ is the stabilizer group of the element $ P$.

On the LHS we have $ p^a$, on the RHS we have $ |\mathcal{O}_e| \equal{} 1$, where $ e$ is the identity element, and $ G_e$ must be equal to $ p^a$?

I'm not sure where you're getting $ e$ from.  What does the orbit-stabilizer theorem tell you about the possible sizes of conjugacy classes of a $ p$-group?

That was probably a wrong direction on my part..

All conjugacy classes have size $ 1$ or are a power of $ p$?

Yes.  And now your previous argument works.


integration
trigonometry
calculus
function
calculus computations

If $ \int_{0}^{\pi}f(x)\sin x dx\equal{}0$ and $ \int_{0}^{\pi}f(x)\cos x dx\equal{}0$,then prove that $ f(x)\equal{}0$ must have atleast two real roots on the interval $ (0,\pi)$.

With the additional hypothesis that $ f$ is continuous on $ [0,\pi],$ right? Otherwise it makes no sense.

Assume $ f$ is not identically zero, else the result is trivial.

Since $ \sin x>0$ on $ (0,\pi),$ the hypothesis that $ \int_0^{\pi}f(x)\sin x\,dx\equal{}0$ implies that $ f$ takes on both some positive values and some negative values in the interval. By the IVT, $ f$ has at least one zero in the open interval $ (0,\pi).$

Suppose now that $ f(x)\equal{}0$ has only one root in this interval, $ f(x_0)\equal{}0.$ By what I said about positive and negative values, $ f$ must change signs at $ x_0.$ WLOG, we can say that $ f(x)<0$ for $ x<x_0$ and $ f(x)>0$ for $ x>x_0.$ (Else we replace $ f$ by $ \minus{}f.$)

Consider the integral $ \int_0^{\pi}f(x)\sin(x\minus{}x_0)\,dx.$

On the one hand, since $ \sin(x\minus{}x_0)$ is a linear combination of $ \sin x$ and $ \cos x,$ this integral is zero by hypothesis.

On the other hand, $ \sin(x\minus{}x_0)<0$ on $ (0,x_0)$ and $ \sin(x\minus{}x_0)>0$ on $ (x_0,\pi).$ Hence $ f(x)\sin(x\minus{}x_0)\ge0$ on $ (0,\pi)$ and is continuous and not the zero function, meaning that its integral is strictly positive.

That's a contradiction, so our supposition that $ f(x)\equal{}0$ has only one root must be false.

Thank you,Kent.Do you know the source of this problem.

Actually i did not understand prof . Kent's solution so i tried this myself .
please tell me whether the following argument is correct .

$ \int_{0}^{\pi}f(x)\sin x dx\equal{}0$  therefore applying the property of definite integral we have $ \int_{0}^{\pi}f( \pi \minus{} x )\sin x dx\equal{}0$

Consider $ f$ to be not identically zero .

suppose it has no roots therefore $ f(x) > 0$ or  $ f(x) < 0$  .

suppose $ f(x) > 0$ therefore $ \int_{0}^{\pi}f(x)\sin x dx$ cannot be zero .

similarly in the case of $ f(x) < 0$ $ \int_{0}^{\pi}f(x)\sin x dx$ cannot be zero .

Suppose $ f$ has exactly 1 root . therefore   $ f'(x) > 0$ or $ f'(x) < 0$ . If  $ x_0$ is the

root  then either $ f( x_0 ) > f( \pi \minus{} x_0)$  or  $ f(x_0) < f( \pi \minus{} x_0 )$

but in either cases $ \int_{0}^{\pi}f(x)\sin x dx$ and $ \int_{0}^{\pi}f(\pi \minus{} x)\sin x dx$

cannot be both zero.

even if  $ x_0 \equal{} \frac{ \pi}{2}$ its not possible

Therefore we can conclude that it has atleast 2 roots .

is this argument correct ?? if not please tell me where i go wrong ..

Also can we find $ f$ ?

You cannot assume that $ f(x)$ is differentiable. (I stopped reading when I saw $ f'(x)$ in your proof. Sorry).

To understand the Professor's solution I suggest you:

1) draw a graph and follow what he is saying.
2) use the identity $ sin(a\plus{}b) \equal{} sin(a)cos(b) \plus{} cos(b)sin(a)$.

That argument cannot possibly be correct, since you never used the hypothesis that $ \int_0^{\pi}f(x)\cos x\,dx\equal{}0.$ That hypothesis is vital - without it, the result would not be true. (Just try setting $ f(x)\equal{}\cos x$ for an example.)

but $ \cos x$ has only one root in $ [ 0, \pi ]$ so how can we use that  :(  ?
Also i make use of the other hypothesis when i say $ x_0 \equal{} \frac{ \pi}{2}$ is not possible .

To aryabhata .

Ok i don't assume it is differentiable , but i can instead anyway say that $ f$ is either increasing or decreasing , which i think  is sufficient condition  for rest of my  solution .

Anyways can we find $ f$  ?

[quote="vijaymenon"]...but in either cases $ \int_{0}^{\pi}f(x)\sin x dx$ and $ \int_{0}^{\pi}f(\pi \minus{} x)\sin x dx$

cannot be both zero.

even if  $ x_0 \equal{} \frac {\pi}{2}$ its not possible [/quote]
Why is it so?

Moreover why should $ f$ be either increasing or decreasing? It can very well oscillate and cross x-axis only once!

Oh thats definitely a problem . sorry i didn ' t think of it but then also isn't this true ? that is $ f( x_{0}) > f(\pi\minus{}x_{0})$ or $ f(x_{0}) < f(\pi\minus{}x_{0})$ .?

Yes the above is true. But then why $ \int_{0}^{\pi}f(x)\sin x dx$ and $ \int_{0}^{\pi}f(\pi \minus{} x)\sin x dx$  cannot be both zero?

[quote="vijaymenon"]but $ \cos x$ has only one root in $ [ 0, \pi ]$ so how can we use that  :(  ?
[/quote]

Your proof only uses the fact that $ \int_{0}^{\pi} f(x)\sin x dx \equal{} 0$. If that is so, then $ f(x) \equal{} \cos x$ satisfies the hypothesis of the problem, but has only one root. Since your conclusion is that $ f(x)$ must have two roots, it follows that $ \cos x$ must have two roots too (if your proof was right), but that is false.

Also, you cannot assume $ f$ is a monotonic function, even if you are given that it is differentiable.

ya i am sorry  , did not realise that .Really sorry ..


function
algebra
functional equation
real analysis
real analysis unsolved

Let $n>1$ and $a_1,...,a_n\in (0,1)$ which add up to $1$. Find all real functions $f$ for which $\frac{f(x)}{x}$ goes to 1 in 0 and $f(x)=\sum_{i=1}^{n}{f(a_i x)}$.

Let $a=\max(a_1,\ldots,a_n)$.

Given some arbitrarily small $\varepsilon>0$, we can find $\delta>0$ s.t. $(1-\varepsilon)x<f(x)<(1+\varepsilon)x,\ \forall x\in(-\delta,\delta)$. For every $x\in\left(-\frac\delta a,\frac\delta a\right)$ we have $a_1x,a_2x,\ldots,a_nx\in(-\delta,\delta)$, and so the functional equation will give $(1-\varepsilon)x<f(x)<(1+\varepsilon)x,\ \forall x\in\left(-\frac\delta a,\frac\delta a\right)$. We repeat this to get $(1-\varepsilon)x<f(x)<(1+\varepsilon)x,\ \forall x\in\left(-\frac\delta {a^n},\frac\delta {a^n}\right)$ for all $n\in\mathbb N$, or, in other words, $(1-\varepsilon)x<f(x)<(1+\varepsilon)x,\ \forall x\in\mathbb R$. Since this holds for arbitrary $\varepsilon$, we get $f(x)=x,\ \forall x$.

[quote="grobber"]Let $a=\max(a_1,\ldots,a_n)$.
....
 We repeat this to get $(1-\varepsilon)x<f(x)<(1+\varepsilon)x,\ \forall x\in\left(-\frac\delta {a^n},\frac\delta {a^n}\right)$ for all $n\in\mathbb N$...[/quote]

$a=\max(a_1,\ldots,a_n)$ so $a=b(n)$ depend with $n$

How do you know the magnitude of $a^n=b(n)^n$ when $n$ tends +oo, to deduce that

$(1-\epsilon)x<f(x)<(1+\epsilon)x$ for all $x \in R$

I think there's a misunderstanding. I used $n$ with two different meanings. First of all, $a_1,\ldots,a_n$ are finitely many reals in $(0,1)$. This $n$ is fixed. Then, I used $n$ in $a^n\to 0$. This $n$ is not fixed. I agree that it can be a little confusing, but not [b]that[/b] confusing! :)

$a$ is simply one of $a_1,\ldots,a_n$, and since they're all in $(0,1)$, you get $\frac\delta{a^k}\to\infty$ as $k\to\infty$.


Let $AB$ be a diameter of circle $\omega$. $\ell$ is the tangent line to $\omega$ at $B$. Take two points $C$, $D$ on $\ell$ such that $B$ is between $C$ and $D$. $E$, $F$ are the intersections of $\omega$ and $AC$, $AD$, respectively, and $G$, $H$ are the intersections of $\omega$ and $CF$, $DE$, respectively. Prove that $AH=AG$.

We Have : $AB^{2}= AF.AD = AE.AC$ $\longrightarrow$ $CEFD$ is cyclic . now we have : $\angle CFD = \angle CED$
$\longrightarrow 90+\frac{\angle BAH}{2}=90+\frac{\angle BAG}{2}\longrightarrow \angle BAH=\angle BAG \longrightarrow AH=AG$

[quote="khashi70"]We Have : $ AB^{2} \equal{} AF.AD \equal{} AE.AC$[/quote]Why is it so? Then, why does it follow that $ CEFD$ is cyclic.

It's straightforward, since $ \angle ABD\equal{}\angle AFB\equal{}\angle 90^\circ$, so from a well-known result we have $ AB^2\equal{}AD \cdot AF$ in the right-angled triangle $ ABD$. Similarly, $ AB^2\equal{}AC \cdot AE$, so that $ AD \cdot AF \equal{} AC \cdot AE$, and using power of a point it follows $ CEFD$ must be cyclic.

On the other hand, to prove $ CEFD$ is cyclic, it might be simpler to notice that $ \angle ADB\equal{}\angle ABF\equal{}\angle AEF$. :D

Let $AG$ meet $CD$ at $M$ and $AH$ meet $CD$ at $N$.
$AE\cdot AC$ = $AB^2$ = $AF\cdot AD$.
So, EFDC is cyclic.
$\angle AGF$ = $\angle AEF$ = $\angle ADC$
Thus, GMDF is cyclic. Applying power of a poiny, we find that EMDA is also cyclic.
So, $\angle AMN$ = $\angle AED$.
Similarly, $\angle ANM$ = $\angle AFC$.
But, $\angle AFC$ = $\angle AED$.
Hence, $\angle AMN$ = $\angle ANM$.
Since, $AB$ is perpendicular to $CD$,
so, $\Delta ABM$ is congruent to $\Delta ABN$.
From here, we can easily deduce the result.

$\angle ACD=\angle ABE=\angle AFE=180^{\circ}-\angle DFE$ $\implies$ Points $C,D,E,F$ are concyclic $\implies$ $\angle EDC=\angle EFC$ $=$ $\angle EHG$ $\implies$ $HG||CD$ $\implies$ $AB \perp HG$ $\implies $ $AH=HG$


AMC
USA(J)MO
USAMO
algebra
difference of squares
special factorizations

Set $ x\equal{}\frac{5^{125}\minus{}1}{5^{25}\minus{}1}.$

(1) Prove that $ x\in\mathbb{N}.$
(2) Prove that x is composite.

Let $ a\equal{}5^{25}$, we have $ x\equal{}\frac{a^5\minus{}1}{a\minus{}1}.$

[quote="kunny"]Let $ a \equal{} 5^{25}$, we have $ x \equal{} \frac {a^5 \minus{} 1}{a \minus{} 1}.$[/quote]

I do hope that your solution continues, because I don't think this sentence would give good marks in a competition. ;)
It trivializes (1) though.

This solution has no motivation whatsoever.
$ a^{4}\plus{}a^{3}\plus{}a^{2}\plus{}a\plus{}1\equal{}(a^{2}\plus{}3a\plus{}1)^{2}\minus{}5a(a\plus{}1)^{2}$ where $ a\equal{}5^{25}$. Since the value is a difference of squares (5a is a square), we're done.

[quote="t0rajir0u"]That technique is similar to USAMO 2007 #5 (which also has no motivation :P).  I've seen this problem, though, so I believe it predates that one.[/quote]
This problem is in Engel, so anyone who saw it before USAMO had a large advantage.

The motivation behind this is http://www.mathlinks.ro/viewtopic.php?t=146020 and http://www.mathlinks.ro/Forum/viewtopic.php?t=169135 .


Hi everyone,

   I am a new high school student in Canada.  I just came here this year.  I know that the SIN exam will be in May.  Most probably, I won't be ready to join this year.  However, I am really interested in trying to tackle some of the problems.  But I don't know where I can find the past exams, like from 1980 to 1990s.  I tried to search for different websites, but it was in vain.

   Would you please introduce me some websites and/or reference on past exams?  Or, does anyone have them?  Please help me!

   Thanks a lot.

There is a Sir Isaac Newton Exam? I've never heard of it. Is it very prestigious?

-rts2007

You can order the book "A Decade of SIN (Plus Sixteen)" which covers the first 26 years. All contests after that until 2008 can be found in "A Collection of SINs".
I ordered them via Canada Post. Go here for address and all that http://sin.uwaterloo.ca./FAQ.php .

Do you have now? I also want to know. Could someone help me?


abstract algebra
linear algebra
matrix
linear algebra theorems

Let T be a linear map T:V->W:
In the most common proof of the rank nullity theorem we take a basis of ImT, expand to a basis for W, take pre-images of those elements, and extend with the kernel elements to a basis for V. Then, the matrix for T would look like:
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 0
0 0 0 0 0
But then, wouldn't this mean that if we apply this theorem to an endomorphism, that any square matrix is diagonalizable?
Where do I make a false conclusion?

Every matrix becomes diagonalizable when you are allowed to choose [b]two different bases[/b], i.e., to multiply from the left and the right by two unrelated invertible matrices, not just by $ C$ and $ C^{\minus{}1}$.


MATHCOUNTS
number theory
prime factorization

just go to the mathcounts website and click on state solutions which should be up till April

Yeah, but in general, the mc solutions suck. I think this one's is fine, though, and I'm nt in the mod to post a solution. btw, i made nats last year but not this year.

factor 336 to get 2^4*3*7
then find possible cmobinations of the factors that multiply to 336, noticing that seven cannot be multiplied by anything but 1 or it will result in a 2 digit number. thus you know 7 must be a # in the phone #
you have:
7,2,3,8
7,4,4,3
7,2,4,6
7,8,6,1
then find the number of distinct arrangements for each of the above and you get 24,12,24,24 respectively
add them to get 84.

just go to mathcounts.org and click on 2006 state solutions.

[quote="rd5493"]Is there anyone that can help me find the answer to this problem?

Derek’s phone number, 336-7624, has the property that the
three-digit prefix, 336, equals the product of the last four
digits, 7 × 6 × 2 × 4. How many seven-digit phone numbers
beginning with 336 have this property?

Thanks.[/quote]
first the prime factorization of 336 is 2^4*3*7. so since 3 and 7 is only to the power of 1 it is up to 2 to make the most of it. So,
the possible numb combinations: 8, 3, 2 7
4, 4, 3, 7
2, 4, 6, 7
8, 6, 7, 1

(24)3 + 12= 84.
84  phone numbers[/hide]

I didn't make nat's last year, but I made it this year.

ya well there is only one problem, its down until may


شرمنده من انگلیسیم اونقدر خوب نیست که بتونم انگلیسی سوالم رو مطرح کنم!
سوالم اینه:

میخواستم بدونم هیچ راهی هست که بشه باهاش   توان های کسری مثلا یک سوم.  رو  راحت حساب کرد؟
میدونم یکی از راه حلاش اینه که زیر رادیکال ببریم ولی باز اینجوری خیلی سخته
کسی راه بهتری سراغ نداره؟ ممنون

[quote="P.H"]man fekr mikonam ke aslan maenie tavane kasri hamine ke radical begirio baed be tavane soorat kasr beresooni.masalan hamoon tavane 2/3 yaeni aval ebarat ro be tavane 2 beresoon baed azesh radicale ba forjeye se begir .banabarin rahe dige nabayad sadetar az in bashe. :P[/quote]
che natije giriye jalebi!!!  :P

___________________________________

[b][color=blue]Lemma[/color][/b]:[color=darkblue]farz konid $\lim_{x\rightarrow 0}f(x)=0$.
sabet konid:
$[1+f(x)]^{p}-1\sim p.f(x)$
va dar halate khas:$\sqrt[n]{1+f(x)}-1\sim \frac{f(x)}{n}$[/color]

masalan mikhahim meghdar $\sqrt[3]{1042}$ ro hesab konim:

$\sqrt[3]{1042}=10\times \sqrt[3]{1+0.042}\stackrel{\text{ from Lemma }}{\longrightarrow}\sqrt[3]{1042}\approx 10\times (\frac{1}{3}\times 0.042+1)=10.14$

$10.14^{3}=1042.5907$

Dar morede tavan haye gong ham fek konid jalebe!!!


geometry
3D geometry
calculus
integration
number theory
greatest common divisor
Ring Theory

26 is between a perfect cube, and a perfect square.
What other numbers are there like this?

There are no other numbers like this.  :P
Thats why 26 is a very special number...

An easy solution (though not appropriate for the Intermediate forum) is provided by working in $\mathbb{Z}[\sqrt{-2}]$. We want to find all $x$ such that $a^{2}<x<b^{3}$ satisfying $x=a^{2}+1=b^{3}-1\implies b^{3}=a^{2}+2$.  Now find all integral solutions $(a,b)$ to the equation $b^{3}=a^{2}+2$.  For $x=26$, $(a,b)=(5,3)$.

yea the two cases are, for integers $ a,b$

$ a^{2}= b^{3}-2$ and $ a^{2}= b^{3}+2$

now how to show that $ (5,3)$ is the only solution to the first one, and that there are no solutions to the second ... i dont think boxedexe really showed how to do it

[quote="boxedexe"]An easy solution (though not appropriate for the Intermediate forum) is provided by working in $ \mathbb{Z}[\sqrt{-2}]$. We want to find all $ x$ such that $ a^{2}<x<b^{3}$ satisfying $ x=a^{2}+1=b^{3}-1\implies b^{3}=a^{2}+2$.  Now find all integral solutions $ (a,b)$ to the equation $ b^{3}=a^{2}+2$.  For $ x=26$, $ (a,b)=(5,3)$.[/quote]

I don't think that $ \mathbb{Z}[i\sqrt{2}]$ has the euclidean algorithm...[and consequently no gcd, and no UFD].

I don't think it is that obvious which quadratic fields have UFD. If we look at: http://www.cut-the-knot.org/arithmetic/int_domain4.shtml, we have that $ \mathbb{Z}[\sqrt{2}]$, $ \mathbb{Z}[i\sqrt{2}]$ both have UFD.

That means that we can look at $ (a+\sqrt{2})(a-\sqrt{2})$, and note that $ (a+\sqrt{2},a-\sqrt{2})=(a+\sqrt{2},2\sqrt{2})=1,\sqrt{2},2,2\sqrt{2}$. Then we have like one factor is a perfect cube and continue from there.


function
recurrence relation
algebra
recursion
IMO Shortlist

Let $ a, b \in \mathbb{N}$  with $ 1 \leq a \leq b,$ and $ M \equal{} \left[\frac {a \plus{} b}{2} \right].$ Define a function $ f: \mathbb{Z} \mapsto \mathbb{Z}$ by
\[ f(n) \equal{} \begin{cases} n \plus{} a, & \text{if } n \leq M, \\
n \minus{} b, & \text{if } n >M. \end{cases}
\]
Let $ f^1(n) \equal{} f(n),$ $ f_{i \plus{} 1}(n) \equal{} f(f^i(n)),$ $ i \equal{} 1, 2, \ldots$ Find the smallest natural number $ k$ such that $ f^k(0) \equal{} 0.$

[quote="orl"]Let $ a, b \in \mathbb{N}$  with $ 1 \leq a \leq b,$ and $ M \equal{} \left[\frac {a \plus{} b}{2} \right].$ Define a function $ f: \mathbb{Z} \mapsto \mathbb{Z}$ by
\[ f(n) \equal{} \begin{cases} n \plus{} a, & \text{if } n \leq M, \\
n \minus{} b, & \text{if } n \geq M. \end{cases}
\]
Let $ f^1(n) \equal{} f(n),$ $ f_{i \plus{} 1}(n) \equal{} f(f^i(n)),$ $ i \equal{} 1, 2, \ldots$ Find the smallest natural number $ k$ such that $ f^k(0) \equal{} 0.$[/quote]

I suppose we must read "$ n \minus{} b$, if $ n > M$" (instead of "$ n\geq M$")

[hide="My solution"]

If $ k(x)$ is the least cycle for $ x$, we have $ f^{k(x)}(x) \equal{} x$, then we have $ p(x)a \minus{} q(x)b \equal{} 0$ where $ p(x)$ is the number of "$ \plus{} a$" moves and $ q(x)$ the number of "$ \minus{} b$" moves for the least cycle (and so $ p(x) \plus{} q(x) \equal{} k(x)$)

And so the least values possible for $ p(x)$ and $ q(x)$ are $ p(x) \equal{} \frac {b}{\gcd(a,b)}$ and $ q(x) \equal{} \frac {a}{\gcd(a,b)}$

So $ k(x)\geq \frac {a \plus{} b}{\gcd(a,b)}$

Now, if $ x \equal{} 0$, $ f^i(0)$ can only take values in $ ([\frac {a \plus{} b}{2}] \minus{} b,[\frac {a \plus{} b}{2}] \plus{} a]$, all multiples of $ \gcd(a,b)$ (since the starting value $ 0$ is such a multiple), so $ \frac {a \plus{} b}{\gcd(a,b)}$ values.

So a cycle is encountered in the first $ \frac {a \plus{} b}{\gcd(a,b)}$ moves. As a consequence, this cycle includes $ 0$ (else we would have a cycle whose length would be $ < \frac {a \plus{} b}{\gcd(a,b)}$

And so the result : $ k(0) \equal{} \frac {a \plus{} b}{\gcd(a,b)}$

(Remark : with some other departure points, the cycle may not involve the departure point and so $ k(x) > \frac {a \plus{} b}{\gcd(a,b)}$)

[/hide]


algebra polynomial quadratics geometry rectangle parallelogram trapezoid

The solution you wrote makes no sense to me. Spoiler: [hide]Notice that the polynomial has degree 28, so the coefficient of x^27 will be the negative sum of the roots. If a and b are the roots of the first quadratic, a and b are roots with multiplicity 5 in the big polynomial. Likewise, if c and d are roots of the the second quadratic then they each have multiplicity 9 in the big polynomial. So, the sum of the roots of the big polynomial is 5(a+b)+9(c+d) = 5(-5)+9(1)=-16. So the answer is the negative of that, i.e. 16.[/hide]

notice that sinA=sinB only if A=B, or A=(180-B)
if all the angles were equal, then A=B=C=D=90, and the quadrilateral would be a rectangle, which is clearly not the case.
for the A=(180-B) case, we must have 2 angles equal to A and 2 equal to (180-A), as any other combination would imply 3 angles of 90 degrees, and thus a rectangle (figure this out by adding the angles and setting them equal to 360).
so now we look at which orientation of the "2 angles equal to A and 2 equal to (180-A) case" works.  we could have the angles which are equal in measure oppose each other, or we could have the each angle oppose it's supplementary angle.
for the first case, we get a parallelogram, which requires two pairs of equal size, so that doesn't work.
fo the second case, we have an isosceles trapezoid, with one pair of equal sides (the legs).  so the two legs can have lengths of 5,6,or,7 (5,6,7 are the given side lengths, so we make x 5,6,or 7 to make a pair)

I'm inclined to believe that Zabelman's answer, rather than the book's answer, is correct.  Haven't really checked it yet, but his logic is solid.  It also appears that the authors of the book were trying to say exactly the same thing that he said, but that they said it worse.

As for why this is true, in any monic polynomial (leading coefficient = 1) of degree n, the coefficient of the term of degree (n - 1) is the additive inverse of the sum of the roots of the polynomial.  To show this, write it out as a product of linear factors and then look at what would happen in the term with (n - 1) x's of that product.

Also, for this particular question, you can simply write out the product and figure out how you could get the x^27 power.  Since, as Zabelman said, the largest possible exponent is x^28, we need to "lose" exactly one power of x in every term.  We could do this in any of the 5 (x^2 + 5x + 2) factors by taking one (5x) term and taking all the rest (x^2) terms, giving you a total of 25x^27, or take it in the 9 (x^2 - x + 3) terms by taking the x term in one place and the x^2 terms everywhere else, giving you a total of (-9)x^27, which together give a coefficient of 16, as Zabelman said.

[hide]

First of all, when x<1, its negative.

Secondly, as x goes from 1 to 2, and as x goes from 2 to 3, and as x goes from 4 onwards, each of those parts has the LHS of the inequality going from infinity to minus infinity (as x is just bigger than 1, the 1/(x-1) is extremely high, as x gets close to 2 from below, 1/(x-2) is extremely low, etc). Of course the function is decreasing in each interval.

So our intervals look like (1,a), (2,b), (3,c), and they don't overlap. So the sum of the lengths is a-1+b-2+c-3 = a+b+c-6. Now we have to find a,b,c.

When we substitute a,b,c into the inequality, the LHS must come out to equal 1. So if we multiply both sides by (x-1)(x-2)(x-3), we get a cubic polynomial with roots a, b, c. Now the coefficient of x^3 is 1, and the polynomial also is (x-a)(x-b)(x-c), which expands to x^3 - (a+b+c)x^2 + something. So all we want is the coefficient of x^2.

The coefficient of x^2 in (x-1)(x-2)(x-3) - (x-1)(x-2) - (x-1)(x-3) - (x-2)(x-3) is (-1-2-3-1-1-1) = -9. So a+b+c=9.

Finally, the sum of the lengths is 9-6 = 3.

[/hide]

Tell me if theres any parts of that which you don't understand.

[quote="takuanitromars36"]Ok, I got the answer that TripleM posted. Here are some more I don't get:

1. What are all ordered triples of numbers (x,y,z) which satisfy the system of equations

(x+y)(z+1) = 24
(x+z)(y+1) = 24
(y+z)(x+1) = 24?

I thought to subtract the first equation from the second equation, and the second equation from the third equation. Then, I factored these two resulting equations. They are the following:

(x-1)(z-y) = 0
(z-1)(y-x) = 0
[/quote]

Missed one: (y-1)(z-x) = 0.

I'll take the first equation and set the first factor to zero. Hence, x = 1. Sub that into the original equations, you get

(1 + y)(1 + z) = 24
y + z = 12

Rewrite the second equation as z = 12 - y and sub it back into the first:

(1 + y)(1 + 12 - y) = 24
(1 + y)(13 - y) = 24
13 + 12y - y2 = 24
y2 - 12y + 11 = 0
(y - 1)(y - 11) = 0
y = 1 or y = 11.

When y = 1, z = 11. When y = 11, z = 1. We can deduece that x = 11, y = 1, z = 1 would also be valid.

If instead of setting x - 1 = 0, we set y = z, then we'd have (by the original equations)

(x + z)(z + 1) = 24 --> z2 + zx + z + x = 24
2z (x + 1) = 24 --> zx + z = 12

Hence z2 + x = 12, so x = 12 - z2. Sub this into zx + z = 12:

z(12 - z2) + z = 12
12z - z3 + z = 12
z3 - 13z + 12 = 0

The only real root is z = 1, so y = 1, and x = 11. We've found them before, so we can conclude that the solutions are

(x, y, z) = (11, 1, 1), (1, 11, 1) and (1, 1, 11).

[quote="takuanitromars364. A scientific supply house sells every integral gram weight from 1 gram to 100 grams inclusive. What is the least number of these weights a pharmacist would need to use, together with a two-pan balance scale, to measure correctly any integral gram-weight from 1 gram to 100 grams? [Note: Weights may be placed on either or both pans of the balance scale.][/quote]

Not exactly sure but I believe the weights needed are 1, 3, 9, 27 and 81, so 5 weights.

Here's where the "not sure" comes in. I haven't got a proof for it (or at least I think my proof is incomplete), but the answer of 5 weights is correct. It's just like a rule that if you need least number of weights and can place them on both sides it's the powers of 3.

BTW You can swap that 81g weight with any weight between 60 and 80g inclusive and it will probably work just fine. :lol:

If anyone has a formal proof I'd like to hear about it.

[quote="Templar"]If anyone has a formal proof I'd like to hear about it.[/quote]

Represent the weights that are on the scale in balanced ternary form.  This is similar to ternary (base-3), but we use the digits, [tex]\{-1, 0, +1\}[/tex], in lieu of the conventional ternary digits, [tex]\{0, 1, 2\}[/tex].

As with any base-3 number, the number represented by the n-digit balanced ternary number is [tex]B=b_{n-1}3^{n-1} + b_{n-2} 3^{n-2} + \ldots +  b_2 3^2 + b_1 3^1 + b_0 3^0[/tex].

The weights are available in sizes of [tex]3^i[/tex], where [tex]0\le i < n[/tex].  If a particular weight is placed in the left pan, we represent that digit-position with [tex]b_i=-1[/tex].  If a particular weight is not on the balance, we represent that digit-position with [tex]b_i=0[/tex].  If a particular weight is placed in the right-hand pan, we represent that digit-position with [tex]b_i=+1[/tex].

The largest (and smallest) possible values of [tex]B[/tex] are [tex]\pm\sum_{i=0}^{n-1}3^i=\pm\frac{3^n-1}{2}}[/tex].

Let [tex]c_i=b_i+1[/tex], [tex]0\le i < n[/tex].  Now [tex]c_i\in\{0, 1, 2\}[/tex].  So, we consider the conventional ternary number, [tex]C=c_{n-1}3^{n-1} + c_{n-2} 3^{n-2} + \ldots +  c_2 3^2 + c_1 3^1 + c_0 3^0[/tex].  Then [tex]C=B+\frac{3^n-1}{2}[/tex].  There is a one-to-one mapping between B and C.  Since C is known to contain [tex]3^n[/tex] distinct integer values, B must also contain [tex]3^n[/tex] distinct integer values.

Therefore, B will contain all integers, [tex]\{-\frac{3^n-1}{2}, -\frac{3^n-1}{2}+1, \ldots, -2, -1, 0, +1, +2, \ldots, \frac{3^n-1}{2}-1, \frac{3^n-1}{2}\}[/tex]

The maximum weight we can place in either pan is [tex]\frac{3^n-1}{2}[/tex].  So, that's the maximum amount we can weigh.  And we can weigh every non-negative integer weight up to that maximum amount.

[quote="takuanitromars36"]I'm not too familiar with base 3. Can you explain in more detail how this works?[/quote]

If you need general information on number bases, let me refer you to:  [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12354[/url].

If you are at the "Getting Started" level, please understand that "balanced ternary" numbers have been studied enough to give them a name, but they are *rarely* used.  Learn the conventional number bases.  [My use of balanced ternary numbers was an unusual tool to prove the weighing problem.]

The numbers $a$, $b$, $c$, and $d$ must be the distinct solutions to the equation

\[x^4+x^2+kx+64=0\]

Because it must be possible to factor the above as

\[(x-a)(x-b)(x-c)(x-d)=0\]

we know that $a+b+c+d=0$ and $ab+ac+ad+bc+bd+cd=1$.  So, we know that the following is true:

\[(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)=(0)^2-2\]

If we expand the above, we get

\[a^2+b^2+c^2+d^2=-2\]

which was what we wanted.


inequalities
algebra
polynomial
real analysis
real analysis unsolved

Hi!

I need to show the following:

\[ \begin{gathered}
  {\max _{x \in \left[ {a,b} \right]}}\frac{{{T_k}\left( {\frac{{b \plus{} 2a \minus{} 2x}}
{{b \minus{} a}}} \right)}}
{{{T_k}\left( {\frac{{b \plus{} a}}
{{b \minus{} a}}} \right)}} \leqslant 2{\left( {\frac{{\sqrt {\frac{b}
{a}}  \minus{} 1}}
{{\sqrt {\frac{b}
{a}}  \plus{} 1}}} \right)^k},0 < a < b \hfill \\
  {T_k}(x) \equal{} \frac{1}
{2}\left( {x \plus{} \sqrt {{x^2} \minus{} 1} } \right) \plus{} \frac{1}
{2}\left( {x \minus{} \sqrt {{x^2} \minus{} 1} } \right){\text{ on }}\left[ { \minus{} 1,1} \right] \hfill \\
\end{gathered} \]

T_k stands for the Chebyshev Polynomials of the first kind

I expanded the fraction:

\[ \frac{{{T_k}\left( {\frac{{b \plus{} 2a \minus{} 2x}}
{{b \minus{} a}}} \right)}}
{{{T_k}\left( {\frac{{b \plus{} a}}
{{b \minus{} a}}} \right)}} \equal{} ... \equal{} \frac{{{{\left( {b \plus{} a \minus{} 2x \plus{} 2\sqrt {\left( {x \minus{} b} \right)\left( {x \minus{} a} \right)} } \right)}^k} \plus{} {{\left( {b \plus{} a \minus{} 2x \minus{} 2\sqrt {\left( {x \minus{} b} \right)\left( {x \minus{} a} \right)} } \right)}^k}}}
{{{{\left( {\sqrt a  \plus{} \sqrt b } \right)}^{2k}} \plus{} {{\left( {\sqrt a  \minus{} \sqrt b } \right)}^{2k}}}}\]

How do I show now:

\[ \frac{{{{\left( {b \plus{} a \minus{} 2x \plus{} 2\sqrt {\left( {x \minus{} b} \right)\left( {x \minus{} a} \right)} } \right)}^k} \plus{} {{\left( {b \plus{} a \minus{} 2x \minus{} 2\sqrt {\left( {x \minus{} b} \right)\left( {x \minus{} a} \right)} } \right)}^k}}}
{{{{\left( {\sqrt a  \plus{} \sqrt b } \right)}^{2k}} \plus{} {{\left( {\sqrt a  \minus{} \sqrt b } \right)}^{2k}}}} \leqslant 2{\left( {\frac{{\sqrt {\frac{b}
{a}}  \minus{} 1}}
{{\sqrt {\frac{b}
{a}}  \plus{} 1}}} \right)^k}\]

Thanks in advance for your help!

PiDeltaPhi

Are you sure you didn't mean $ T_k\left(\frac{b\plus{}a\minus{}2x}{b\minus{}a}\right)$ instead of  $ T_k\left(\frac{b\plus{}\color{red}2\color{black}a\minus{}2x}{b\minus{}a}\right)$ ? :?

Hi!

Ouch, you are absolutely correct, there should be a instead of 2a...I'll redo all calculations...

PiDeltaPhi

Edit: It seems that my calculations were done, using a instead of 2a. The inequality remains the same:

\[ \frac{{{{\left({b\plus{}a\minus{}2x\plus{}2\sqrt{\left({x\minus{}b}\right)\left({x\minus{}a}\right)}}\right)}^{k}}\plus{}{{\left({b\plus{}a\minus{}2x\minus{}2\sqrt{\left({x\minus{}b}\right)\left({x\minus{}a}\right)}}\right)}^{k}}}}{{{{\left({\sqrt a\plus{}\sqrt b }\right)}^{2k}}\plus{}{{\left({\sqrt a\minus{}\sqrt b }\right)}^{2k}}}}\leqslant 2{\left({\frac{{\sqrt{\frac{b}{a}}\minus{}1}}{{\sqrt{\frac{b}{a}}\plus{}1}}}\right)^{k}} \]

Yes, but then all you need to do is to recall that $ |T_k|\le 1$ on $ [\minus{}1,1]$ and use the inequality from your other post. So, where is the difficulty? :?

Hi fedja!

\[ \begin{gathered}
  {\text{Ok, }}\left| {{T_k}} \right| \leqslant 1: \hfill \\
  1 \geqslant {T_k}\left( {\frac{{b \plus{} a}}
{{b \minus{} a}}} \right) \geqslant \frac{1}
{2}{\left( {\frac{{\sqrt {\frac{b}
{a}}  \plus{} 1}}
{{\sqrt {\frac{b}
{a}}  \minus{} 1}}} \right)^k} \hfill \\
   \Leftrightarrow 2{\left( {\frac{{\sqrt {\frac{b}
{a}}  \minus{} 1}}
{{\sqrt {\frac{b}
{a}}  \plus{} 1}}} \right)^k} \geqslant 2 \cdot {T_k}\left( {\frac{{b \plus{} a}}
{{b \minus{} a}}} \right) \cdot {\left( {\frac{{\sqrt {\frac{b}
{a}}  \minus{} 1}}
{{\sqrt {\frac{b}
{a}}  \plus{} 1}}} \right)^k} \geqslant 1 \hfill \\
  {\text{Now we have the right side of the inequality}}{\text{. But: }} \hfill \\
  2 \cdot {T_k}\left( {\frac{{b \plus{} a}}
{{b \minus{} a}}} \right) \cdot {\left( {\frac{{\sqrt {\frac{b}
{a}}  \minus{} 1}}
{{\sqrt {\frac{b}
{a}}  \plus{} 1}}} \right)^k} \equal{} 2 \cdot \frac{{{{\left( {\sqrt a  \plus{} \sqrt b } \right)}^{4k}} \plus{} {{\left( {a \minus{} b} \right)}^{2k}}}}
{{{{\left( {b \minus{} a} \right)}^{2k}}}} \hfill \\
  {\text{Now I}}{\text{ need to show that: }} \hfill \\
  2 \cdot \frac{{{{\left( {\sqrt a  \plus{} \sqrt b } \right)}^{4k}} \plus{} {{\left( {a \minus{} b} \right)}^{2k}}}}
{{{{\left( {b \minus{} a} \right)}^{2k}}}} \geqslant \frac{{{{\left( {b \plus{} a \minus{} 2x \plus{} 2\sqrt {\left( {x \minus{} b} \right)\left( {x \minus{} a} \right)} } \right)}^k} \plus{} {{\left( {b \plus{} a \minus{} 2x \minus{} 2\sqrt {\left( {x \minus{} b} \right)\left( {x \minus{} a} \right)} } \right)}^k}}}
{{{{\left( {\sqrt a  \plus{} \sqrt b } \right)}^{2k}} \plus{} {{\left( {\sqrt a  \minus{} \sqrt b } \right)}^{2k}}}} \hfill \\
\end{gathered} \]

Now, here is my difficulty: How do I show the last inequality?

You seem utterly confused. :(

All you need to do is to denote $ \frac ba$ by $ t$, estimate the numerator on the left hand side by $ 1$ (because there $ T_k$ is evaluated at some point between $ \minus{}1$ and $ 1$), express the denominator in terms of $ t$ (here the formula with the square root is useful), and use the (trivial) inequality from your other post.


$ x\plus{}\frac{4}{x}\plus{}a\equal{}\sqrt{x}\minus{}\frac{2}{\sqrt{x}}\plus{}4$
Prove that solution of equation must not exceed 1

[quote="nguyensilva"]$ x \plus{} \frac {4}{x} \plus{} a \equal{} \sqrt {x} \minus{} \frac {2}{\sqrt {x}} \plus{} 4$
Prove that solution of equation must not exceed 1[/quote]

That's wrong : For $ a\equal{}0$, for example, there is a solution $ x\equal{}4>1$

[quote="nguyensilva"][quote="pco"][quote="nguyensilva"]$ x \plus{} \frac {4}{x} \plus{} a \equal{} \sqrt {x} \minus{} \frac {2}{\sqrt {x}} \plus{} 4$
Prove that solution of equation must not exceed 1[/quote]

That's wrong : For $ a \equal{} 0$, for example, there is a solution $ x \equal{} 4 > 1$[/quote]
No, i said number of solution ^^[/quote]

That was not very clear.

But it's wrong again : For $ a\equal{}0$, forexample, we have at least two solutions : $ 2$ and $ 4$, and so the number of solutions is $ > 1$

Denote $ y \equal{} \sqrt{x}\minus{}\frac{2}{\sqrt{x}}$; notice $ y^2 \equal{} x\plus{}\frac{4}{x}\minus{}4$, so the given equation becomes $ y^2 \minus{} y \plus{} a \equal{} 0$, with real roots $ y \equal{} \frac {1 \pm \sqrt{1\minus{}4a}} {2}$ if $ a \leq \frac {1} {4}$.

On the other hand, $ (\sqrt{x})^2 \minus{} y\sqrt{x} \minus{} 2 \equal{} 0$ has one and one only real positive root $ \sqrt{x} \equal{} \frac {y \plus{} \sqrt{y^2 \plus{} 8}} {2}$, no matter the sign or value of real number $ y$.

Therefore, for $ a \equal{} \frac {1} {4}$ we have $ y \equal{} \frac {1} {2}$ and $ \sqrt{x} \equal{} \frac {1 \plus{} \sqrt{33}} {4}$, while for $ a < \frac {1} {4}$ we have two distinct real $ y$ values, hence two positive real values for $ \sqrt{x}$ (if $ t^2 \minus{}y_1t \minus{} 2 \equal{} t^2 \minus{} y_2t \minus{} 2 \equal{} 0$ for $ y_1 \neq y_2$, then $ (y_1 \minus{} y_2)t \equal{} 0$, so $ t\equal{}0$, but then $ 0\equal{}2$, absurd). For example, for the given example $ a\equal{}0$, we have $ y_1\equal{}0$ and $ y_2\equal{}1$, leading to $ \sqrt{x} \equal{} \sqrt{2}$, respectively $ \sqrt{x} \equal{} 2$.


p1,p2....p10 be prime numbers(pi>2).for each i<>j we have pi^16-pj^16:pk(k<>i,k<>j)Can be?

It is Given 10 odd prime numbers. Could it so happen that difference of the sixteenth degrees any two divided by any of remaining?

No, it cannot happen. The argument below is far from optimal, and it wouldn't surprise me if there's a better bound on the number $ m$ of such primes. However, the argument is fairly simple and works well for $ m\equal{}10$ so presumably it is more or less the intended solution.

[hide="proof"]
 Let $ p_1, p_2, \ldots, p_{10}$ be distinct odd primes and assume $ p_1 < p_2 < \ldots <p_{10}$. We will prove a stronger statement: it's impossible that $ p_{10}$ divides $ p_i^{16}\minus{}p_j^{16}$ for all $ i,j < 10$.

Note that $ a^{16}\minus{}b^{16} \equal{} (a\minus{}b)(a\plus{}b)(a^2\plus{}b^2)(a^4\plus{}b^4)(a^8\plus{}b^8)$. Since $ p_{10}$ is the largest, it certainly cannot divide any $ p_i\minus{}p_j$. It also cannot divide any $ p_i\plus{}p_j$ since then it would divide $ \frac{p_i\plus{}p_j}{2}$ as well, which is impossible since it's too large. Therefore $ p_{10}$ does not divide $ p_i^2\minus{}p_j^2$ for any $ i,j < 10$.

Suppose that $ p_{10}$ divides $ p_i^2\plus{}p_j^2$ as well as $ p_i^2\plus{}p_k^2$. Note that $ p_i$ is common to both pairs. This is clearly impossible, since now $ p_{10}$ divides the difference $ p_k^2\minus{}p_j^2$, and we just proved that this never happens. Thus the pairs $ \{i,j\}$ such that $ p_{10}$ divides $ p_i^2\plus{}p_j^2$ are disjoint, so there can be at most 4 such pairs.

By removing one of the $ p_i$'s from each such pair, we can leave 5 primes < $ p_{10}$ such that $ p_{10}$ doesn't divide any $ p_i^2\plus{}p_j^2$. We already know it doesn't divide $ p_i^2\minus{}p_j^2$ either, so it doesn't divide $ p_i^4\minus{}p_j^4$ for any $ i,j$ among the 5 remaining primes. Now repeat the argument: if $ p_{10}$ divides both $ p_i^4\plus{}p_j^4$ and $ p_i^4\plus{}p_k^4$, we get a contradiction, so the pairs $ \{i,j\}$ such that $ p_{10}$ divides $ p_i^4\plus{}p_j^4$ are disjoint, and so there can be at most 2 such pairs.

By removing one of the $ p_i$'s from each such pair, we are left with 3 primes < $ p_{10}$ such that $ p_{10}$ doesn't divide either the square sum or the 4th power sum of any pair. It must therefore divide $ p_i^8\plus{}p_j^8$ for all pairs $ i,j$ among those three. But then it divides some $ p_k^8\minus{}p_j^8 \equal{} (p_k^4\minus{}p_j^4)(p_k^4\plus{}p_j^4)$, which is impossible.

Observe that, without modifications, this argument does not apply if we only have 9 primes to begin with, since at the first step we'd be left with only 4 primes less than the largest one, and after the second step we may have just 2, and this leads to no contradiction. That's why I expect this to be the intended solution. However, I'm fairly sure that a much better bound can be derived by using more of the information provided.

[/hide]

Incidentally, $ (3,5,17)$ are three primes satisfying $ p_k | p_i^{16}\minus{}p_j^{16}$ for distinct $ i,j,k$. There are many other such triples.


algebra
function
domain
absolute value
complex numbers
calculus
calculus computations

I need to find the domains on FoG, F+G, and F/G

f(x)=x/2(x^2)-x-3
      Domain of f=(-infinity to -1)(-1,3/2)(3/2 to infinity)
      Range of f=??

g(x)=abs(2x-5)
      Domain of g= all reals
      Range of g= [zero to infinity)

FoG= abs(2x-5)/2(abs(2x-5)^2)-abs(2x-5)-3
      How do you square an absolute value?

I suspect you're not using enough parentheses. Please clarify what functions you're actually talking about. For example, your first reads as $ \frac{x}{2}\cdot x^2\minus{}x\plus{}3$, and I don't think you intended that.

The general principle here as that the domain is assumed to be everywhere the formula makes sense. As long as you're not dividing by zero or taking the square root of something negative*, these make sense.

* What do you mean, complex numbers? They don't exist.

Sorry about the lack of parentheses

f(x)= x[size=200]/[/size](2(x^2)-x-3)

Fog= abs(2x-5)[size=200]/[/size](2(abs(2x-5)^2)-(abs(2x-5))-3)


The spectral radius of Hilbert matrix is increasing with the size $ n$.

Increasing, but bounded. The infinite Hilbert matrix does define a bounded operator from $ \ell^2$ to $ \ell^2.$

I don't know how to compute the radius of a Hilbert matrix in terms of n.