function
floor function
algebra
functional equation
algebra proposed

[quote="pco"]@angelstt : since you seem interested in this functional equation (4 different problems about it), here is a fifth one :

Find all functions from $ \mathbb R^ \plus{} \to\mathbb R^ \plus{}$ (the set of all positive real numbers) such that : $ f(xf(y)) \equal{} f(x \plus{} y)$ $ \forall x,y > 0$[/quote]
Is my solution right?
$ f(x\plus{}y)\equal{}f(xf(y))\equal{}f(yf(x))$
[b]Case1[/b] if there aren't  $ y_1  \ne y_2$ such that $ f(y_1)\equal{}f(y_2)$ then xf(y)=yf(x)
Then, $ f(x)\equal{}ax$ . Then check it which is not suitable.
[b]Case2 [/b]if there are $ y_1  \ne y_2$ such that $ f(y_1)\equal{}f(y_2)$
      then $ f(x\plus{}y_1)\equal{}f(xf(y_1))\equal{}f(xf(y_2))\equal{}f(x\plus{}y_2)$ .
Then, $ f(x)\equal{}f(x\plus{}c)$ .($ c\geq 0$, c is the smallest frequency)
      [b] If[/b] $ c\equal{}0$then $ f(x)\equal{}const$.Check it suitable.
     [b]  If [/b]$ c>0$then, there exits $ a$ that $ f(a) \ne 1$
         +$ f(a)>1$
           $ f(xf(a))\equal{}f(x\plus{}a)$ so $ xf(a)\equal{}x\plus{}a\plus{}nc$( $ n \in N^*$ take x large enough)
so $ x(f(a)\minus{}1)\equal{}a\plus{}nc$
Then , take $ x_0\equal{}x\plus{}\frac{c}{2(f(a)\minus{}1)}$we got $ x_0.(f(a)\minus{}1)\equal{} a\plus{}nc\plus{}c/2\equal{}a\plus{}mc$ ($ n,m \in N^*$) Which can not true.
          Do the same for $ f(a)<1$
Then we obtain $ f(x) \equal{}const$

[quote="TRAN THAI HUNG"]
      ...
      [b] If[/b] $ c \equal{} 0$then $ f(x) \equal{} const$.Check it suitable.
      ...
[/quote]

sorry, i don't see that. If $ c\equal{}0$ we have $ f(x)\equal{}f(x)$. Can you explain? Thanks

[quote="TRAN THAI HUNG"][quote="pco"]@angelstt : since you seem interested in this functional equation (4 different problems about it), here is a fifth one :

Find all functions from $ \mathbb R^ \plus{} \to\mathbb R^ \plus{}$ (the set of all positive real numbers) such that : $ f(xf(y)) \equal{} f(x \plus{} y)$ $ \forall x,y > 0$[/quote]
Is my solution right?
$ f(x \plus{} y) \equal{} f(xf(y)) \equal{} f(yf(x))$
[b]Case1[/b] if there aren't  $ y_1 \ne y_2$ such that $ f(y_1) \equal{} f(y_2)$ then xf(y)=yf(x)
Then, $ f(x) \equal{} ax$ . Then check it which is not suitable.
[b]Case2 [/b]if there are $ y_1 \ne y_2$ such that $ f(y_1) \equal{} f(y_2)$
      then $ f(x \plus{} y_1) \equal{} f(xf(y_1)) \equal{} f(xf(y_2)) \equal{} f(x \plus{} y_2)$ .
Then, $ f(x) \equal{} f(x \plus{} c)$ .($ c\geq 0$, c is the smallest frequency)
      [b] If[/b] $ c \equal{} 0$then $ f(x) \equal{} const$.Check it suitable.
     [b]  If [/b]$ c > 0$then, there exits $ a$ that $ f(a) \ne 1$
         +$ f(a) > 1$
           $ f(xf(a)) \equal{} f(x \plus{} a)$ so $ xf(a) \equal{} x \plus{} a \plus{} nc$( $ n \in N^*$ take x large enough)
so $ x(f(a) \minus{} 1) \equal{} a \plus{} nc$
Then , take $ x_0 \equal{} x \plus{} \frac {c}{2(f(a) \minus{} 1)}$we got $ x_0.(f(a) \minus{} 1) \equal{} a \plus{} nc \plus{} c/2 \equal{} a \plus{} mc$ ($ n,m \in N^*$) Which can not true.
          Do the same for $ f(a) < 1$
Then we obtain $ f(x) \equal{} const$[/quote]

1) As prester said, $ \inf\{c$ such that $ f(x\plus{}c)\equal{}f(x) \forall x\}\equal{}0$ does not imply $ f(x)$ is constant (it would be true if $ f(x)$ was continuous, but this is not a constraint of the problem.

2) If $ f(x)\equal{}f(x\plus{}c)$ $ \forall x$, it does not allow you to conclude $ f(a)\equal{}f(b)$ $ \implies$ $ a\equal{}b\plus{}nc$ as you wrote ("$ f(xf(a)) \equal{} f(x \plus{} a)$ so $ xf(a) \equal{} x \plus{} a \plus{} nc$")

1) I agree I forgot Dirichlete function also have that condition.(@to prester:when I write c=0, I mean there is no minimum of c). :)
2) I agree If  $ f(x) \equal{} f(x \plus{} c)$,  it does not allow me to conclude   $ f(a) \equal{} f(b) \implies a \equal{} b \plus{} nc$ but if we also have $ f(y_1) \equal{} f(y_2)\implies\ y_1 \minus{} y_2$ is a frequency then we can conclude that since c is the smallest frequency.
Am I right? :) Thanks.
However, I still fell in solve this problem. Can you show your proof, please?

[quote="TRAN THAI HUNG"]1) I agree.
2) I agree If  $ f(x) \equal{} f(x \plus{} c)$,  it does not allow me to conclude   $ f(a) \equal{} f(b) \implies a \equal{} b \plus{} nc$ but if we also have $ f(y_1) \equal{} f(y_2)\implies\ y_1 \minus{} y_2$ is a frequency then we can conclude that since c is the smallest frequency.
Am I right? :) Thanks.[/quote]

 :blush:  You're right. $ f(a)\equal{}f(b)$ $ \implies$ $ |b\minus{}a|$ is a period and so $ c$ divides $ |b\minus{}a|$ if $ c\ne 0$
Sorry.

[hide="Solution"]
We claim that only constant functions work; it is clear that they do work.

[b]Observation:[/b] If $ f(a) \equal{} f(b)$ with $ b \minus{} a \equal{} s > 0$, then $ f(y \plus{} a) \equal{} f(yf(a)) \equal{} f(yf(b)) \equal{} f(y \plus{} b)$ for positive $ y$.  Set $ y \equal{} x \minus{} a$ to get that $ f(x) \equal{} f(x \plus{} s)$ for all $ x > a$.

If $ f$ is injective, then $ xf(y) \equal{} x \plus{} y$, so $ f(1) \equal{} 1 \plus{} \frac {1}{x}$.  However, this value is nonconstant, which is a contradiction, so $ f$ cannot be injective.  By the observation, there are reals $ t$ and $ u$ so that for all $ x > u$, $ f(x) \equal{} f(x \plus{} t)$.  Assume that $ f$ is nonconstant so that there is a $ z$ so that $ f(z) \ne 1$.

Let $ r$ be an arbitrary real greater than $ z$.  If $ f(z) > 1$, then set $ x \equal{} \frac {z \plus{} r}{f(z) \minus{} 1}$.  If $ f(z) < 1$, then set $ x \equal{} \frac {r \minus{} z}{1 \minus{} f(z)}$.  In the first case, we get $ xf(z) \equal{} x \plus{} z \plus{} r$, meaning that $ f(x \plus{} z) \equal{} f(x \plus{} z \plus{} r)$, and in the second case, we get that $ xf(z) \equal{} x \plus{} z \minus{} r$, so $ f(x \plus{} z \minus{} r) \equal{} f(x \plus{} z)$.  Either way, we have found a $ k$ and an $ l$ so that $ |k \minus{} l| \equal{} r$ and $ f(k) \equal{} f(l)$.  By the observation, there is a function $ g$ that maps reals to reals so that whenever $ x > g(r)$, $ f(x) \equal{} f(x \plus{} r)$.

Set $ h(r) \equal{} g(r) \minus{} t\lfloor {\frac {g(r)}{t} }\rfloor$ so that $ h(r) < t$ for all $ r$.  Let $ \max \{t, u \} \equal{} v$; if $ x > v$, then $ f(x) \equal{} f(x \plus{} g(r) \minus{} h(r)) \equal{} f(x \plus{} g(r) \minus{} h(r) \plus{} r) \equal{} f(x \plus{} r)$ for all $ r$.  Then, there is a constant $ c$ so that $ f(x) \equal{} c$ is constant whenever $ x > v$.  For any $ w$, put in $ x \equal{} \frac {w}{c}$ and set $ y > v$.  This gives that $ f(w) \equal{} f(xf(y)) \equal{} f(x \plus{} y) \equal{} c$, so $ f$ is constant everywhere.

Hence, only constant functions work.

[/hide]

[u][b]@QuattroMaster [/b][/u]:
I do agree with your solution with a little correction :
[quote="The QuattoMaster 6000"] ... then $ f(x) \equal{} f(x \plus{} g(r) \minus{} h(r)) \equal{} f(x \plus{} g(r) \minus{} h(r) \plus{} r) \equal{} f(x \plus{} r)$ for all $ r$.  [/quote]
In fact, this is not $ \forall r$ : it is $ \forall r > z$ but this does not destroy the proof.

Congrats.

[u][b]@Nixmtp[/b][/u] :
Here is my solution, using the same idea than QuattroMaster, but with a little bit different end :

Let $ P(x,y)$ be the equation $ f(xf(y)) \equal{} f(x \plus{} y)$

$ f(x) \equal{} 1$ $ \forall x$ is a solution. So we'll from now consider $ \exists u$ such that $ f(u)\ne 1$

1) If $ \exists a,T > 0$ such that $ f(a) \equal{} f(a \plus{} T)$ then $ f(x) \equal{} f(x \plus{} T)$ $ \forall x\ge a$
=================================================================================
Let $ x > a$ :
$ P(x \minus{} a,a)$ $ \implies$ $ f((x \minus{} a)f(a)) \equal{} f(x)$
$ P(x \minus{} a,a \plus{} T)$ $ \implies$ $ f((x \minus{} a)f(a \plus{} T)) \equal{} f(x \plus{} T)$ and so $ f(x) \equal{} f(x \plus{} T)$ $ \forall x > a$
And, since $ f(a) \equal{} f(a \plus{} T)$ : $ f(x) \equal{} f(x \plus{} T)$ $ \forall x\ge a$
Q.E.D.

2) $ \exists a,b > 0$ such that $ f(x) \equal{} f(x \plus{} T)$ $ \forall x > a,\forall T > b$
====================================================================
Let $ u$ such that $ f(u)\ne 1$
2.1) if $ f(u) > 1$
---------------
Let $ T > 0$ : $ P(\frac {u \plus{} T}{f(u) \minus{} 1},u)$ $ \implies$ $ f(\frac {uf(u) \plus{} Tf(u)}{f(u) \minus{} 1}) \equal{} f(\frac {u \plus{} T}{f(u) \minus{} 1} \plus{} u)$ $ \equal{} f(\frac {uf(u) \plus{} Tf(u)}{f(u) \minus{} 1} \minus{} T)$
And so, using the point 1) above : $ f(x) \equal{} f(x \plus{} T)$ $ \forall x\ge \frac {uf(u) \plus{} Tf(u)}{f(u) \minus{} 1} \minus{} T$
Q.E.D.

2.2) if $ f(u) < 1$
----------------
Let $ T > u$ : $ P(\frac {u \minus{} T}{f(u) \minus{} 1},u)$ $ \implies$ $ f(\frac {uf(u) \minus{} Tf(u)}{f(u) \minus{} 1}) \equal{} f(\frac {u \minus{} T}{f(u) \minus{} 1} \plus{} u)$ $ \equal{} f(\frac {uf(u) \minus{} Tf(u)}{f(u) \minus{} 1} \plus{} T)$
And so, using the point 1) above : $ f(x) \equal{} f(x \plus{} T)$ $ \forall x\ge \frac {uf(u) \minus{} Tf(u)}{f(u) \minus{} 1}$
Q.E.D.

3) The only solutions are constant functions
============================================
Let $ a,b > 0$ such that $ f(x) \equal{} f(x \plus{} T)$ $ \forall x > a,\forall T > b$ (according to point 2) above)
Let $ x > y > a$

Let $ T_1 \equal{} b \plus{} 1 > b$. Using point 2, we get $ f(x) \equal{} f(x \plus{} b \plus{} 1)$
Let $ T_2 \equal{} b \plus{} 1 \plus{} x \minus{} y > b$. Using point 2, we get $ f(y) \equal{} f(y \plus{} b \plus{} 1 \plus{} x \minus{} y) \equal{} f(x \plus{} b \plus{} 1)$
And so $ f(x) \equal{} f(y) \equal{} c$ $ \forall x,y > a$ for some constant $ c$

Let then $ u > 0$ $ P(\frac u{f(a \plus{} 1)},a \plus{} 1)$ $ \implies$ $ f(u) \equal{} f(a \plus{} 1 \plus{} u{f(a \plus{} 1)}$ and, since $ a \plus{} 1 \plus{} u{f(a \plus{} 1)} > a$, we get $ f(u) \equal{} c$ $ \forall u > 0$
Q.E.D

And so (since the case $ f(x) \equal{} 1$ $ \forall x$ is also a constant solution and since obviously $ f(x) \equal{} c$ fits the requested equation :

The only solutions to this equation are $ \boxed{f(x) \equal{} c}$ $ \forall x > 0$ where $ c$ is a positive constant

Thank you very much , i really enjoyed your beautiful solutions.

[quote=horizon]my solution:
$x=1$,we have $f(f(y))=f(y=1)$,so $f(x(f(y+1))=f(x+f(y))=f(x+y+1)$
so $f(y)-y-1=T$ is period for any positive reals $y$
$f(xf(y))-xf(y)-1=T$,$f(x+y)-xf(y)-1=T$,$f(x+y)-(x+y)-1=T$,
so $T=x(1-f(y))+y$,if $f(x)$ is not constant,then we have $T$ can be all reals on interval $T>k$ or $T<k'$
so $f$ must be constant[/quote]
Where does $T=x(1-f(y))+y$ come from?
Shouldn't it be $0=x(1-f(y))+y$

Pardon me I try to proceed in a simple way . Can I substitute x = 0 ? In that case is f(y)=f(0) partially satisfy f(x) is a constant function? Partially because how can I determine f(0). Someone kindly help From the solutions above can I say f(0)=1 ?

[quote=pco]Find all functions $f: \mathbb R^\plus{}\to\mathbb R^\plus{}$ such that \[ f(xf(y))\equal{}f(x\plus{}y)\] for all positive reals $x$ and $y$.[/quote]

Assume $A= \{a | f(b+a)=f(b) , b>N\}$ .clearly $|xf(y)-x-y| \in A$. let $t \in A$. we have $f(y)^n t \in A$ for every integer $n$ aswell. now if $u+v \in A$ and also $v$ is not an element in $A$ we have :
$$f((x+u)f(y+v))=f(xf(y))$$
hence $x(f(y+v)-f(y))+uf(y+v)$ now since $f(v+y)$ and $f(y)$ are not equal we get that for some $a<b$ we have $(a,b) \in A$ which gives that big enough $x$ have a constant $f(x)$ now just put $y$ to be big enough and we are done

It is routine to check that $f$ is not injective. Suppose there exist $x_1>x_2$ such that $f(x_1)=f(x_2)$, then the equation implies, for all $x>x_2$, $f(x)=f(x+p)$ for some $p>0$.

Since $f\equiv 1$ works, let $f(n)\neq 1$ for some $n$. If $f(n)>1$, then in the equation setting $x=\tfrac{y+n}{f(n)-1}$ and $y=n$ yields $f(c)=f(c-y)$ for some $c$ and any $y>0$. If $f(n)<1$, then for any $y>n$, setting $x=\tfrac{y-n}{1-f(n)}$ yields similar. And so from above, $f(x)=f(x+p)$ for all $x>c_1$ and for all $p>c_2$, where $c_1$, $c_2$ fixed.

For any $y>x$, setting $p=y-x+p$ gives $f(x)=f(y+p)=f(y)$, hence $f$ is constant over $(c_1,\infty)$. Finally, original equation for fixed $y>c_1$ yields $f(xf(y))$ constant for all $x>0$. Hence $f$ is constant, which works.


Vote NOW!!!
Please state the month u were born =]
(i couldnt do all 12 months cuz when i tried to post it said there was too many choices)

[quote=".Ana."]December huh[/quote]

ME2!!!! Which day?

My birthday is posted on another thread, guess it!

Someone born February 29th?

I've posted it in another thread.

[hide="Hint"][hide="Dont want to guess?"][hide="Are you sure?"]It is in my username[/hide][/hide][/hide]

Posted in another thread about this already.
I was born in February.


Let $p(x)$ be a $1999$-degree polynomial with integer coeffficients that is equal to $\pm 1$ for $1999$ different integer values of $x$. Show that $p(x)$ cannot be factored into the product of two plynomials with integer coefficients.

[hide]
suppose the polynomial can be factored into a product of polynomials with integer coefficients, $A(x)B(x)$
wolog, let $A(x)$ have the lower degree. Then the degree of $A(x)$ is at most $999$. For each of the $1999$ values, $A(x)$ and $B(x)$ both give integer results, and the product of these results is $1$ or $-1$. This means that for each of the $1999$ values, $A(x)$ gives $1$ or $-1$. Then $A(x)$ gives $1$ at least $1000$ times, or $A(x)$ gives $-1$ at least $1000$ times, which is impossible for a polynomial of degree $999$.
[/hide]


geometry
parallelogram
similar triangles
geometry solved

let ABCD be convex quadrilateral .AC cuts BD at O .X,Y,Z,T are the projections of O on AB,BC,CD,DA.Prove that ABCD is parallelogram iff OX=OZ,OY=OT

If ABCD is a paralelogram, then obviously OX=OZ and OY=OT.

The converse: if OX=OZ and OY=OT.
If AB||CD (or AD||BC) then OX=OZ (or OY=OT) means that the similar triangles OAB and OCD (or OBC and OAD) have congruent corresponding heights, so they are congruent so ABCD is a paralelogram.

Suppose no opposite sides of ABCD are paralel. Then there exists M=AB \cap CD and N=AD \cap BC.
OX=OZ and OY=OT means MO and NO are bisectors for AMC and CNA.

Lets look at AMC and ANC. MO and No are bisectors of AMC and ANC, so CO/OA=CM/MA=CN/NA. Let O' be the point where the exterior bisector of AMC hits AC. We have O'C/O'A=OC/OA. It is easy to see that O' is also the point where the exterior bisector of ANC hits AC.

If we do the same looking at the triangles BMD and BND we get that the exterior bisectors of BMD and BND meet BD at a point O".

From these, we get O=O'=O" which is impossible because for example O lies in [AC] while O' does not.

So ABCD has 2 paralel opposite sides, so from what we saw earlier, it is a paralelogram.
----------------------------------------------------------------------------------
Sorry about my earlier mistake. I've put C instead of D someplaces.

I think it is a Television Network :D.
Replace C by D where you see NBC or MBC.


Two small identical balls lying on a horizontal plane are connected by a weightless spring. One ball is fixed at the origin and the other is free. The balls are charged identically as a result of which the spring length increases n times. Determine the ratio of new frequency (when balls are charged) to old frequency (when balls were uncharged) when the free ball is displaced slightly from its mean position.

[hide="Ans"] $ \sqrt{\frac{3n\minus{}2}{n}}$ [/hide]

$ \minus{} kx \equal{} ma$

$ \omega^2 \equal{} \frac {k}{m}$

Now force at eq. pos is given by
$ k(n \minus{} 1)x \equal{} \frac {KQ^2}{(nx)^2}$

When the ball is displaced by dx , the net force is$ \minus{} (k(n \minus{} 1)(x \plus{} dx) \minus{} \frac {KQ^2}{(nx \plus{} dx)^2}) \equal{} \minus{} [\frac {KQ^2}{(nx)^2} (1 \minus{} \frac {1}{(1 \plus{} \frac {2dx}{nx})}) \plus{} k(n \minus{} 1)dx] \equal{} \minus{} [ k(n \minus{} 1)x(\frac {2dx}{nx}) \plus{} k(n \minus{} 1)dx] \equal{} \minus{} k(n \minus{} 1)dx(\frac {2}{n} \plus{} 1) \equal{} ma$

This gives $ \omega^2 \equal{} k(n \minus{} 1)(\frac {2}{n} \plus{} 1)/m$

Wrong again :( But where is the glitch

Cool work Conjurer , u made the same mistake i made initially but still appreciable effort  :coolspeak:

Here is my solution  :wink:
[hide][img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/2/c/d/2cd322fc424a562d303edf981271e041ce5424f2.gif[/img][/hide]

Yesterday I solved that problem, interesting, but I was waiting others tries.

In slogger's reply, when I click to reveal hidden content, I get only the word Image...  Do you know how can I get his solution? Or, maybe you could "transfer" it to me using Latex?

Should I put my solution?

Of course u can post ur solution

Geniuses should not stop from posting a solution just because a lesser solution has been posted before :D

Also , u can right click on the place where Image is written and click view image

Slogger's solution, made easier to view :): [hide][img]http://www.mathlinks.ro/latexrender/pictures/2/c/d/2cd322fc424a562d303edf981271e041ce5424f2.gif[/img][/hide]


limit
function
complex numbers
calculus
calculus computations

You should first consider a more basic question: $(-1)^{x}=?$

I think sometimes it's not real, and when it's real, I think it's $-1$. So the answer is $-1$?

The point I tried to make is that one cannot define $(-1)^{x}$ for fractional values of $x$ in a way that is consistent with algebraic rules such as $(a^{x})^{y}=a^{xy}$. For instance, is $(-1)^{2/3}=((-1)^{2})^{1/3}=1^{1/3}=1$? Of course not, because $1^{3/2}$ is not equal to $-1$. And so on.

So $1$ is not the answer, right?
What is it?

[quote="Jimmy"][quote="mlok"]Suppose we agree that $(-1)^{x}$ means $e^{\pi ix}$, where the exponential function is defined by its Taylor expansion $e^{z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}$. If so, then the answer is $1$, as PTynan89 wrote.

But do we all agree with this interpretation of $(-1)^{x}$? Some people think that $(-1)^{1/3}=-1$, which is not the same as $e^{\pi i/3}=\frac{\sqrt{3}}{2}+\frac{i}{2}$.[/quote]

Well in that case we can define $-1$ as $e^{\pi i+2\pi i n}, \ n\in\mathbb{Z}$, then we have $(-1)^{x}= e^{\pi i x+2\pi i n x}= e^{(\pi i x)(1+2n)}$, and thus $\lim_{x\to 0}(-1)^{x}= \lim_{x\to 0}e^{(\pi i x)(1+2n)}= e^{0}= 1$.[/quote]
This is correct if n doesn't depend on x. But if we take n=[1/(2x)]

In my post $n$ was meant to be an arbitrary integer, although it could coincidentally be $\frac{1}{2x}$.

[quote="OHO"]So what's the conclusion?[/quote]

$\lim_{x\to 0}(-1)^{x}= 1$

[quote="Jimmy"]Well in that case we can define $-1$ as $e^{\pi i+2\pi i n}, \ n \in \mathbb Z$, then we have $ (-1)^{x}= e^{\pi i x+2\pi i n x}= e^{(\pi i x)(1+2n)}$, and thus $\lim_{x \to 0}(-1)^{x}= \lim_{x \to 0}e^{(\pi i x)(1+2n)}= e^{0}= 1$.[/quote]
You're still not answering mlok's objection :!:

Furthermore, if one assumes that "the usual rules" work with complex exponents, then one can find another proof of the fact that $0 = 1$. It's somewhere on the forum :huh:

[b]perfect_radio wrote:[/b][quote]
You're still not answering mlok's objection  :!:

Furthermore, if one assumes that "the usual rules" work with complex exponents, then one can find another proof of the fact that $0 = 1.$ It's somewhere on the forum  :huh: [/quote]
As Jimmy wrote we can write -1 as $e^{\pi i+2\pi i n},\n\in \mathbb Z$
so,in general whenever we write $(-1)^{1/3}$,we should express it in a more general form as $e^{(\pi+2n\pi)/3},$ which for different values of n will assume 3 values $-1,\frac{\sqrt{3}}{2}+\frac{i}{2},\frac{\sqrt{3}}{2}-\frac{i}{2}$

Well, is $(-1)^{x}$ is multivalued: $(-1)^{x}=\{e^{\pi i x(1+2n)}\colon n\in\mathbb Z\}$, then $\lim_{x\to 0}(-1)^{x}$ is multivalued too: it is the set $\{z\in\mathbb C\colon |z|=1\}$.  :ninja:

And what is the definition of the limit of a multivalued function?  :huh:

[b]mlok:3rd post[/b]
[quote]
The point I tried to make is that one cannot define $(-1)^{x}$ for fractional values of $x$ in a way that is consistent with algebraic rules such as $(a^{x})^{y}=a^{xy}$. For instance, is $(-1)^{2/3}=((-1)^{2})^{1/3}=1^{1/3}=1?$ Of course not, because $1^{3/2}$ is not equal to $-1$. And so on.[/quote]

In previous post the point I wanted to clearify was that the above rule of algebra is valid even for complex numbers,
and $1^{3/2}=e^{3n\pi}=1 and-1$

[b]mlok:5th post[/b]
[quote]Well, is $(-1)^{x}$ is multivalued: $(-1)^{x}=\{e^{\pi i x(1+2n)}\colon n\in\mathbb Z\}$, then $\lim_{x\to 0}(-1)^{x}$ is multivalued too: it is the set $\{z\in\mathbb C\colon |z|=1\}.$

And what is the definition of the limit of a multivalued function? [/quote]
I definitely agree with your above argument and would say that as you pointed out,that the given limit does not exit.But you should also note that as $x\to 0$ $n$ can take infinite values,and choosing a large $n$ from the set will prove that [b]Jimmy's[/b] conclusion was wrong and the limit will not tend to 1.

[b]perfect_radio wrote:[/b]
[quote]
You're still not answering mlok's objection :!:

Furthermore, if one assumes that "the usual rules" work with complex exponents, then one can find another proof of the fact that $0=1$ . It's somewhere on the forum :huh: [/quote]

I could not find it anywhere in the forum :maybe: can you help with it?
Also,do you agree with my above arguments?


inequalities
geometry
perimeter
function
inequalities proposed

Given an arbitrary triangle prove the inequality:

$P \geq \frac{3\sqrt{3}}{8} R^2$

i think its /4

fix R, we want to maximize the area of the triangle.  It can be written as $R^2/2(sinX + sinY + sinZ)$ where X,Y,Z are the angles OAB OBC OCA.  The sinX + .. is well known, has maximum sqrt(3)/2.  So minimum is R^2 3sqrt(3)/4

ps. if you mean P for perimeter, please state that.. you should state it if P is area anyway.

Singular,i think u found the maximum and not the minimum :

P=(abc)/4R. a=2RsinA,etc so P=2R 2 sinAsinBsinC. The function ln(sinx) is concave and so sinAsinBsinC <= 3sqrt(3)/8 and thus P <= 3R 2 sqrt(3)/4

As far as the initial ineq is concerned, if P denotes the area, i think it is wrong : take A=2π/3, Β=C=π/6. Then sinAsinBsinC=sqrt(3)/8 an the whole expression is 2R 2 sqrt(3)/8.

You're right galeos, I've must have miscalculated when counting minimum.

However $P\leq \frac{3\sqrt{3}}{4}R^2$ is true, well-known and easy  :D


calculus
derivative
parameterization
LaTeX
real analysis
real analysis unsolved

please if anyone has an idea how to solve p(x,t)=q(x,t)-q(x,t)q(x,t)'.  here ' means the derivative with respect to t. I am sure that is no explicit solution but there must be some conditions such that I can write q in terms of p. One of the conditions is that p must be continuous, otherwise this has no sense. But can anyone give me an advise how to approach the problem?

thanks a lot
mikiduta

What is $ x$ doing there? :? Is it just a parameter and the equation you are interested in is merely $ q(t)(1\minus{}q'(t))\equal{}p(t)$ or, equivalently, $ q'(t)\equal{}1\minus{}\frac{p(t)}{q(t)}$?

Also, what does it mean "to write $ q$ in terms of $ p$"? Are you just looking for the conditions for a solution to exist? One (rather trivial) statement of this sort is that if $ p$ is continuous and $ p\le 0$, and $ q(t_0)>0$, then the solution exists for all $ t>t_0$. I doubt it is of much use for you but it is hard to tell more without having a more clear idea of what you are really after. :)

Thank you very much for looking at the problem that I posted.
Indeed this equation has the form(after you rewrite it)[code]q_{'}(t)=1-frac{p(t)}{q(t)}. The x , indeed is not in the equation.
I want to show that the equation is invertible (that I can take out q in terms of p), i.e has a solution, even though this solution is not an explicit one. The things that I didn't say is that I have no certain conditions about how p and g must be.I just need conditions (as generally as possible) such that this equation will have solutions.Can i have p in W^{1,1}? I found this problem while doing my research and no colleague was able to help me out with any idea.
May you please explain why it is trivial that the equation has solutions when [/code] p\leq 0[code]. I am slow in understanding things but the nice thing is that once i get them I am fast:))?

Thank you soooooooo much for your help!

Sincerely,
Mikiduta

No mathematical help, but to use LaTeX on the forums, see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 or [[LaTeX]].

Thank you very much for looking at the problem that I posted.
Indeed this equation has the form(after you rewrite it) .
Code:
$ q_{'}(t)\equal{}1\minus{}\frac{p(t)}{q(t)}$.
The Code: $ x$ , indeed is not in the equation.
I want to show that the equation is invertible (that I can take out Code:$ q$ in terms of Code $ p$), i.e has a solution, even though this solution is not an explicit one. The things that I didn't say is that I have no certain conditions about how $ p$ and $ g$ must be.I just need conditions (as generally as possible) such that this equation will have solutions.Can i have $ p$ in $ W^{1,1}$?I found this problem while doing my research and none of my colleague was able to help me out with any idea.
May you please explain why it is trivial that the equation has solutions when Code: $ p\leq 0$. I am slow in understanding things but the nice thing is that once i get them I am fast:))?

Thank you soooooooo much for your help!

Sincerely,
Mikiduta

Can any of you help me?
I really need some ideas?

Thanks a lot!
Mikiduta

Hi I am new on this site.
I saw your problem and as weird as it seams I am thinking at the same problem....I will suggest to use Peano theorem-theorem of existence of the solution.  But I do not see why you need $ p\leq 0$. The continuity is important because otherwise you cannot apply the theorem above. Do you need the solutions to be unique?

thanks a lot!
I don not get either why the $ p$ must be less or equal then $ 0$....
I have no other idea...I tried to expand in power series to see if I can work by hand this problem...but it is not working.

I hope we will find an idea!

Thanks,
mikiduta

Ah, of course, continuity of $ p$ is enough to ensure the [b]local[/b] existence and uniqueness. Just recall your classical existence and uniqueness theorem for general ODE. The problem is that if $ p$ and $ q$ are positive, the solution may easily develop a singularity in finite time when $ t$ goes up and $ q$ hits zero. The negativity of $ p$ together with the positivity of the initial data were there to prevent this and for nothing else.

thank you very much for your details.
This explains a lot.

Thanks,
mikiduta

Do you think there can be any other general constraint that I can ask suck that this equation will have solution?for example if i have $ q_{'}$ bounded by a large constant, $ p$ continuous....etc
Thanks,
miki


Let
\[ f_n(x)\equal{}\frac{x}{nx \plus{} 1}\]
be a sequence $ (f_n)_1^\infty$.
Investigate uniform convergence on

   (i) $ [0,1]$

   (ii) $ [1,\infty[$

we have uniform convergence on both intervals right  :?:

For $ x\ge 0,\ 0 \le \frac {x}{nx \plus{} 1} < \frac {x \plus{} \frac1n}{nx \plus{} 1} \equal{} \frac1n.$

Yes, that's uniform convergence.


what is it and how do i start it?

seems like many acceptees to the so-called "prestigious colleges" have had some sort of experience with this.

[url=http://www.nmun.org/]http://www.nmun.org/[/url]

This looks like it, I could be wrong, though.

I think I'm doing it now (sounds funny) for my AP World History class, but I don't know what about it goes on a resume or anything. All it means for me is more work for my already hw-flooded world history class.

Model UN could be run as a club (like in my school) or a class (as in paladin8's school- this is called Global Classrooms). it is run through the UN Assosiation of the USA (UNA-USA) and has more than one conference every year (look at [url=http://www.mun.org/]www.mun.org/[/url] for the dates and other info on the program)

it's a club in my school.  most of the smart and top students in my school are in it. i'm not, i think it's boring.
it's a serious club

Well, I don't know much about it. I worked a few years as High School teacher back in Costa Rica, and they were running that same project. It seems like a debate club or something like that, and somehow several of the kids on those highschools meet once a year. I think it is part of (or related to) the IB, because, at least in CR, just schools with the IB implemented participated in the project. Usually each "team" is assigned to a country, and they should read about the country and present debate as if they were defending such country.

I think it's easier to be enthusiastic about when run as a club.  If you're not sure about to join, your school probably doesn't have a group.  Therefore, I would suggest that you find at least  3 people who are interested, a conference somewhere in your area (many colleges will hold high school conferences, we have 3 major ones in SC that I know of), and a teacher who is willing to sponsor you.  Usually history teachers or those interested in some form of debate are more than willing to coach.  The focus of the club is primarily debate on international issues using a set of rules similar to that of the UN.  Our team at Summerville is made up of 9 members, 3 people per member-state, and is run by an AP history teacher.  We meet after school one day a week, usually for about an hour, though we typically are flexible and often cancel practice.  It's a great way to show colleges a wider variety of talents, demonstrating your ability to deal with people and interest in world affairs.  It especially helps in those that seek well-rounded students.  If you have any more questions, talk to me and I'll try to answer.

my school's model un club is a complete joke.  i've been a part of it this past year (9th grade, my first year in high school) and although there are a few smart people in it, the majority of people don't do any work.  we go to the harvard model un conference and the ivy league model un concerence at upenn, and the kids just want a vacation.  that's really all that anyone cares about the club...they just want the vacation from school where there's a dance.  in the committee sessions, they go outside and talk with other kids and don't try to participate it all.  it's really a shame, so i probably won't be doing it next year.


probability
ratio
Putnam
calculus
real analysis
function
algebra

A fair die bearing number 1 through 6 on its faces is thrown repeatedly until the
running total first exceeds 12. What is the most likely total that will be obtained?

[hide]If exceeding $ 12$ means at least $ 13$, then I think it's $ 13$.

Before first exceeding $ 12$, we may have a running total of $ 12,11,10,9,8,$ or $ 7$.

For each of the above running totals, there exists a number between $ 1$ and $ 6$ (inclusive) which , when added, can help it reach $ 13$.

But a running total of 7 before the last throw makes reaching $ 14$ impossible.

Similarly a running total of 7 or 8 before the last throw makes reaching $ 14$ and $ 15$ impossible.

And a total of 18, the least likely of all, requires a running total of 12 before the last throw.[/hide]

You've neglected to consider the fact that arriving at those previous numbers might have different probabilities to begin with.

No: you have a weighted average of uniform distributions on $ \{13\}$, $ [13, 14]$, $ [13, 15]$, ..., $ [13, 18]$.  This gives a greatest likelyhood of arriving at 13, regardless of the weights.

(The actual probabilities are very ugly -- roughly speaking, they work out to 28%, 23.7%, 19.2%, 14.6%, 9.7% and 4.8%.

Notice that those probabilities are close to $ \frac{6}{21}$, $ \frac{5}{21}$, $ \frac{4}{21}$, $ \frac{3}{21}$, $ \frac{2}{21}$, and $ \frac{1}{21}$.  Is that a lucky coincidence? :wink:

Makes sense, thanks.

I take it that the probability of the most likely first roll exceeding $ n$ being $ n \plus{} 1 \ldots n \plus{} 6$ would tend to $ \frac {6}{21} \ldots \frac {1}{21}$, respectively, as $ n \to \infty$, or something like that?

Yes, that's right.  Would anyone like to prove it?

See also this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=20848]old thread[/url] on a continuous version.

The nicest possible answer for the excess comes with a geometric distribution in the discrete case and an exponential distribution in the continuous case.

[quote="Ravi B"]Notice that those probabilities are close to $ \frac {6}{21}$, $ \frac {5}{21}$, $ \frac {4}{21}$, $ \frac {3}{21}$, $ \frac {2}{21}$, and $ \frac {1}{21}$.  Is that a lucky coincidence? :wink:[/quote]

This result is very intuitive.
We may take "exceeding $ 1000$ as an example

For $ 1001$,
obviously $ 995$, $ 996$, $ 997$, $ 998$ and $ 999$ ,$ 1000$ are possible running totals before the last throw.

For $ 1002$
$ 996$, $ 997$, $ 998$, $ 999$ and $ 1000$ are possible running totals before the last throw.

For $ 1003$
$ 997$, $ 998$, $ 999$ and $ 1000$ are possible running totals before the last throw.
....
For 1006,
$ 1000$ is the only possible running totals before the last throw.
And the probabilities of arriving these running totals are very close. So the probabilities should be in the ratio of $ 6: 5: 4: 3: 2: 1$

We can expect the probability of a running total of $ n$ to be close to that of a running total of $ n \plus{} 1$ for large $ n$, hence the result.

No idea how to prove it though  :blush: .

Yes, you've got the basic idea.  To make it rigorous, imagine that we keep rolling forever.  It's well known that the probability of ever hitting a total of $ k$ approaches $ \frac{2}{7}$ as $ k$ approaches infinity.  (That was a Putnam problem from 1960.)  Thus the probability of ever using a particular transition, say from $ k$ to $ k \plus{} 2$, approaches $ \frac{2}{7} \cdot \frac{1}{6}$, which is $ \frac{1}{21}$.

For the first total exceeding $ n$ to be $ n \plus{} 1$, we must have used one of the 6 transitions ending in $ n \plus{} 1$.  Hence the probability of doing so approaches $ \frac{6}{21}$.  In the same way, we get the other probabilities.

Now, actually proving that that probability of ever hitting $ k$ approaches $ \frac27$ is harder. Several approaches follow:

[hide="Special case, quantitative, minimal calculus"]We think in terms of generating functions. Let $ f(n)$ be the probability of hitting $ n$ with one die roll, and $ F(n)\equal{}\frac16(x\plus{}x^2\plus{}x^3\plus{}x^4\plus{}x^5\plus{}x^6)$ the corresponding generating function. Let $ h(n)$ be the expected number of times we hit $ n$ when rolling repeatedly and adding terms (which is also equal to the probability of ever hitting $ n$, since we can only hit once). We can easily see that $ h(0)\equal{}1$ and $ h(n)\equal{}\sum_{m\le n}f(m)h(n\minus{}m)$; we reach $ n$ by first reaching $ n\minus{}m$ and then rolling $ m$ with probability $ f(m)$.
That convolution is multiplication for generating functions, and the equations become $ H(x)\equal{}1\plus{}H(x)F(x)$. Solving for $ H$, $ H(x)\equal{}\frac1{1\minus{}F(x)}\equal{}\frac{6}{6\minus{}x\minus{}x^2\minus{}x^3\minus{}x^4\minus{}x^5\minus{}x^6}$.
Now, we decompose this with partial fractions. $ \frac{6}{6\minus{}x\minus{}x^2\minus{}x^3\minus{}x^4\minus{}x^5\minus{}x^6}\equal{}\frac{2}{7}\left(\frac1{1\minus{}x}\plus{}\frac{15\plus{}10x\plus{}6x^2\plus{}3x^3\plus{}x^4}{6\plus{}5x\plus{}4x^3\plus{}3x^3\plus{}2x^4\plus{}x^5}\right)$

That $ \frac{2}{7}\frac1{1\minus{}x}$ is the limiting average of $ h(n)$; we want to estimate how fast the rest goes to zero.
Since all zeros of $ 6\plus{}5x\plus{}4x^3\plus{}3x^3\plus{}2x^4\plus{}x^5$ are strictly outside the unit circle, the coefficients of the power series of $ h$ go to zero geometrically fast. How fast?
The zeros of $ 6\plus{}5x\plus{}4x^3\plus{}3x^3\plus{}2x^4\plus{}x^5$ are approximately $ \minus{}1.4918$, $ \minus{}0.8058\pm 1.2229i$, and $ 0.5517\pm 1.2533i$. The latter pair have the smallest absolute value, at approximately $ 1.369$. Expanding $ \frac{15\plus{}10x\plus{}6x^2\plus{}3x^3\plus{}x^4}{6\plus{}5x\plus{}4x^3\plus{}3x^3\plus{}2x^4\plus{}x^5}$ in complete partial fractions, we get the pair $ \frac{\minus{}0.2758\mp 0.6267i}{x\minus{}(0.5517\pm 1.2533i)}$, the pair $ \frac{0.4029\mp 0.6115i}{x\minus{}(\minus{}0.8058\pm 1.2229i)}$, and the singleton $ \frac{0.7459}{x\plus{}1.4918}$.
Estimating by absolute values,
$ \left|\frac27\cdot\frac{15\plus{}10x\plus{}6x^2\plus{}3x^3\plus{}x^4}{6\plus{}5x\plus{}4x^3\plus{}3x^3\plus{}2x^4\plus{}x^5}\right|<\frac27\left(\frac{2\cdot 0.7}{x\minus{}\frac43}\plus{}\frac{3\cdot 0.75}{x\minus{}\sqrt{2}}\right)$
(termwise in the power series)
We then have that $ \left|h(n)\minus{}\frac27\right|<0.3\cdot \left(\frac34\right)^n\plus{} \frac12\cdot 2^{\minus{}\frac n2}$.

From here, the rest is as Ravi B wrote. The excess $ e_n(k)$ (the probability we get $ n\plus{}k$ when we first get $ \ge n$) has formula $ e_n(k)\equal{}\sum_{j>k}f(j)h(n\plus{}k\minus{}j)$ for $ k\ge 0$, which converges to $ \frac27\sum_{j>k}f(j)$.[/hide]

[hide="General case, includes complex analysis. Very long."]
Here let $ f(n)$ be a nice enough probability distribution on the nonnegative integers with mean $ \mu>0$ and variance $ \sigma^2$. As before, let $ F(x)$ be the corresponding generating function, $ h(n)$ the expected number of times we hit $ n$, and $ H(x)$ the generating function for $ x$. We have $ H(x)\equal{}\frac1{1\minus{}F(x)}$.
Now, we want to estimate $ H(n)$ for large $ n$, but we can't just use partial fractions because we're not working with polynomials anymore. We use complex analysis instead. For $ 0<r<1$, $ h(n)\equal{}\frac1{2\pi i}\int_{|z|\equal{}r}\frac{H(z)}{z^{n\plus{}1}}\,dz\equal{}\frac1{2\pi}\int_0^{2\pi}\frac{H(re^{i\theta})}{r^n}e^{\minus{}in\theta}\,d\theta$ by the Cauchy integral formula, since $ F$ is analytic on the unit disk. Unfortunately, dividing by $ r^n$ blows up; we really want to use $ r>1$.

For $ r>1$, we must also account for poles; $ 1\minus{}F(1)\equal{}0$ since $ F$ comes from a probability distribution, and there may be other zeros of $ 1\minus{}F$ or even barriers to analytically extending it. We draw on $ f$ being "nice" here; suppose that there is some $ R>1$ such that $ F(z)$ is analytic on the disk $ |z|<R$ and $ F(z)\equal{}1$ only at $ z\equal{}1$ inside the disk. The former requires that the probabilities $ f(n)$ decay at least geometrically as $ n\to\infty$, and the latter requires that the possible values do not have a common factor (if $ f(n)\equal{}0$ whenever $ n$ is odd, odd integers are unreachable and $ F(\minus{}1)\equal{}1$).
Now, choose $ r$ between $ 1$ and $ R$. By the residue theorem,
$ h(n)\equal{}\frac1{2\pi i}\int_{|z|\equal{}r}\frac{H(z)}{z^{n\plus{}1}}\,dz \minus{} \text{Res}_{z\equal{}1}\frac{H(z)}{z^{n\plus{}1}}$
$ \equal{}\frac1{2\pi}\int_0^{2\pi}\frac{H(re^{i\theta})}{r^n}e^{\minus{}in\theta}\,d\theta \minus{} \text{Res}_{z\equal{}1}\frac{1}{z^{n\plus{}1}(1\minus{}F(z))}$
$ \equal{}O(r^{\minus{}n})\plus{}\frac1{F'(1)}\equal{}O(r^{\minus{}n})\plus{}\frac1{\mu}$, using that we have a simple pole $ \frac1{1\minus{}F(z)}\equal{}\frac1{1\minus{}(1\plus{}\mu(z\minus{}1)\plus{}\frac{\mu^2\minus{}\mu\plus{}\sigma^2}{2}(z\minus{}1)^2\plus{}O(z\minus{}1)^3)}.$

Now define $ e_n(k)$ to be the probability we hit $ n\plus{}k$ when first matching or exceeding $ n$. We have $ e_n(k)\equal{}\sum_{j>k}f(j)h(n\plus{}k\minus{}j)\equal{}\sum_{j>k}f(j)\left(\frac1{\mu}\plus{}O(r^{\minus{}n\minus{}k\plus{}j})\right)$
$ \equal{}\frac1{\mu}\sum_{j\equal{}k\plus{}1}^{n\plus{}k}f(j)\plus{}\sum_{j\equal{}k\plus{}1}^{n\plus{}k}O(r^{\minus{}j})O(r^{\minus{}n\minus{}k\plus{}j})\equal{}O(nr^{\minus{}n\minus{}k})\plus{}\frac1{\mu}\sum_{j\equal{}k\plus{}1}^{n\plus{}k}f(j)$
As $ n\to\infty$, this converges to the tail end distribution $ e(k)\equal{}\frac1{\mu}P(X>k)$, where $ X$ is the base random variable with distribution $ f$. We can easily calculate its mean
$ \sum_k k\cdot e_k\equal{}\frac1{\mu}\sum_k\sum_{j>k}kf(j)\equal{}\sum_j\sum_{k<j}kf(j)\equal{}\frac1{\mu}\sum_j \frac{j^2\minus{}j}{2}f(j)$
$ \equal{}\frac1{2\mu}\left(E(X^2)\minus{}E(X)\right)\equal{}\frac{\mu^2\minus{}\mu\plus{}\sigma^2}{2\mu}.$

From the excess, we can also extract the expected number of terms it takes to reach $ n$; that's $ \frac1{\mu}\left(n\plus{}\frac{\mu^2\minus{}\mu\plus{}\sigma^2}{2\mu}\right)\equal{}\frac{n\minus{}\frac12}{\mu}\plus{}\frac12\plus{}\frac{\sigma^2}{2\mu^2}$ plus something that goes to zero at least geometrically.

There is another path as well; we can focus on the expected number of terms from the start. That's actually what I worked out first, and put in my notes.

For this method, let $ g(n)$ be the expected number of terms required to reach or exceed $ n$. Conditioning on the first term, we get that $ g(0)\equal{}0$ and $ g(n)\equal{}1\plus{}\sum_{m\equal{}0}^n f(m)g(n\minus{}m)$ for $ n\ge 1$. In terms of the generating functions, this becomes $ G(x)\equal{}\frac{x}{1\minus{}x}\plus{}F(x)G(x)$, or $ G(x)\equal{}\frac{x}{(1\minus{}x)(1\minus{}F(x))}$. As before, we want to estimate $ g(n)$, and we use the same integral on a circle of radius $ r>1$:
$ g(n)\equal{}\frac1{2\pi i}\int_{|z|\equal{}r}\frac{G(z)}{z^{n\plus{}1}}\,dz \minus{} \text{Res}_{z\equal{}1}\frac{G(z)}{z^{n\plus{}1}}$
$ \equal{}\frac1{2\pi}\int_0^{2\pi}\frac{G(re^{i\theta})}{r^n}e^{\minus{}in\theta}\,d\theta \minus{} \text{Res}_{z\equal{}1}\frac{1}{z^n(1\minus{}z)(1\minus{}F(z))}$
Substituting $ t\equal{}z\minus{}1$ and expanding $ F$ in its power series centered at $ 1$, that residue becomes $ \text{Res}_{t\equal{}0}\frac{1}{(1\plus{}t)^n(\minus{}t)(\minus{}\mu t\minus{}(\mu^2\minus{}\mu\plus{}\sigma^2)\frac{t^2}{2}\plus{}O(t^3))}$
$ \equal{}\text{Res}_{t\equal{}0}\frac1{\mu t^2}(1\plus{}t)^{\minus{}n}\left(1\plus{}\frac{\mu^2\minus{}\mu\plus{}\sigma^2}{2\mu}t\plus{}O(t^2)\right)^{\minus{}1}$
$ \equal{}\text{Res}_{t\equal{}0}\frac1{\mu t^2}(1\minus{}nt\plus{}O(t^2))(1\minus{}\frac{\mu^2\minus{}\mu\plus{}\sigma^2}{2\mu}t\plus{}O(t^2))$
$ \equal{}\text{Res}_{t\equal{}0}\frac1{\mu}t^{\minus{}2}\minus{}\frac{n}{\mu}t^{\minus{}1}\minus{}\frac{\mu^2\minus{}\mu\plus{}\sigma^2}{2\mu^2}t^{\minus{}1}\plus{}O(1)$

The desired residue is the coefficient $ \minus{}\frac{n}{\mu}\minus{}\frac{\mu^2\minus{}\mu\plus{}\sigma^2}{2\mu^2}$ of $ \frac1t$, and thus $ g(n)\approx \frac{n}{\mu}\plus{}\frac{\mu^2\minus{}\mu\plus{}\sigma^2}{2\mu^2}$ for large $ n$, with geometrically decaying error.
This method doesn't get the distribution of the excess directly, but it does get the expected value, and we can easily take differences of $ g$ to find $ h$ and get the distribution from there.

The whole argument here was constructed by analogy with Kent's argument in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=20848]old thread[/url] I linked; Laplace transforms in the continuous case become $ z$-transforms in the discrete case, and an inversion formula integrating over a line becomes an inversion formula integrating over a circle. It's quite likely that I assumed more niceness than necessary, but the residue arguments of I leave the disk are a lot nicer than anything I could do on the unit circle.[/hide]

In your first solution, just knowing that the zeros of $ 6 \plus{} 5x \plus{} 4x^2 \plus{} 3x^3 \plus{} 2x^4 \plus{} x^5$ are outside the unit circle was enough to get the limit $ \frac{2}{7}$, right?  I guess your extra work was to get an explicit bound on the error.

Another solution is to apply the standard convergence result about finite Markov chains (stochastic matrices) to a particular 6-state diagram (matrix of order 6).  But I guess that is relying on another result.

The Putnam book I own has an elementary proof of $ \frac{2}{7}$.  Using your notation, let $ h(n)$ be the probability that we ever hit $ n$.  Let $ a(n)$ be the minimum of $ h(n \minus{} 6)$, $ h(n \minus{} 5)$, $ h(n \minus{} 4)$, $ h(n \minus{} 3)$, $ h(n \minus{} 2)$, and $ h(n \minus{} 1)$.  Let $ b(n)$ be the corresponding maximum.  Then we can show pretty easily that $ a$ is (weakly) increasing and $ b$ is (weakly) decreasing.  With a little more work, we get the limiting probability squeezed in between.

I know another nice proof of the $ \frac{2}{7}$ limit using a coupling argument, but since that's going beyond "Intermediate Topics", I will omit it here.  The idea is that if we start our running total artificially to be 1 with probability $ \frac{6}{21}$, 2 with probability $ \frac{5}{21}$, 3 with probability $ \frac{4}{21}$, 4 with probability $ \frac{3}{21}$, 5 with probability $ \frac{2}{21}$, and 6 with probability $ \frac{1}{21}$, then the probability of hitting $ n$ is [i]exactly[/i] $ \frac{2}{7}$ for all $ n$.  In fact, that same coupling argument can establish the results about Markov chains that I alluded to earlier.

Yes, the extra work in the first solution was to get an explicit bound.

That other argument looks like it still needs a version of "the other roots are all strictly outside the unit disk". We have some errors (which live in the invariant subspace of net probability zero), and we need to show that their contribution decays. I like the partial fractions for getting that out in the open.
The Markov chain version is annoyingly specific; it only works for bounded random variables. The elementary solution bounding between the last six values has the same restriction.
As for going far beyond "Intermediate"- that's why I put everything in spoiler tags.
[Added in edit][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=185160]Here[/url] is another version of the argument for the discrete uniform case.[/edit]

The $ \frac27$ is really $ \frac {1}{\text{mean}}$. If we don't worry about proving convergence, that's pretty easy to see intuitively.
One other note: my definition of excess is shifted by one from the original version in this thread, and starts at zero. It's an arbitrary choice, and easily adjusted between versions.

I see that you guys liked my problem huh?

Say this was on the Putnam. Approximately where would it be?

Take a look at JBL's post #4, and re-consider whether it would be on the Putnam in the first place  :P

That, and the Putnam has changed a lot in 50 years. The old Putnam question wasn't the original question of the thread, either.

Essentially this question (I think they sent you out to 700 instead of just 12) was also on the Princeton University Math Competition (they asked it as an expected value question, I think, but it ended up boiling down to the same type of reasoning) and it's appeared on the forum in at least one place, but I can't find it :(

I posted the general version [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=242473]here[/url].  It's an extremely poor putnam problem because it's pretty straightforward if you know some markov chain theory, because the details are annoying to work out, and because there is almost nothing to do from a problemsolving perspective -- I figured out what the generalization was immediately upon reading the original PUMaC problem.

e: this refers to my generalization.  your problem is really just a textbook exercise and hence unsuitable for the putnam.

[quote="Ravi B"]I know another nice proof of the $ \frac {2}{7}$ limit using a coupling argument, but since that's going beyond "Intermediate Topics", I will omit it here.  The idea is that if we start our running total artificially to be 1 with probability $ \frac {6}{21}$, 2 with probability $ \frac {5}{21}$, 3 with probability $ \frac {4}{21}$, 4 with probability $ \frac {3}{21}$, 5 with probability $ \frac {2}{21}$, and 6 with probability $ \frac {1}{21}$, then the probability of hitting $ n$ is [i]exactly[/i] $ \frac {2}{7}$ for all $ n$.  In fact, that same coupling argument can establish the results about Markov chains that I alluded to earlier.[/quote]

I would call this 'starting the chain at stationarity', but I guess calling it a coupling works just as well.


If $ \log_2 \Big(\log_3 (\log_2 x) \Big) \equal{} 0$, then $ x^{\minus{}\frac{1}{2}}$ equals

This means that $ \log_{3}(\log_{2}x)\equal{}1$, which in turn means that $ \log_{2}x\equal{}3$, or that $ x\equal{}8$.  Then $ x^{\minus{}1/2}\equal{}\frac{1}{2\sqrt{2}}$


This is a topic that has been on the news a lot in the recent years, illegal immigrants. What do you think we should do about it?

[quote="1=2"]Go to war with Iran and make a draft. :wink:

come on.[/quote]

I was going more on the line of mexican immigrantion...because thats what the majority of illegal immigration is mexican...

Give them a chance to be a legal US citizen. I thought that's what the US was supposed to be about during the mid-1800s - a land of hope.

[quote="i_like_pie"]To possibly attract more responses, maybe this should go to The Round Table?[/quote]

It might get chucked back here so this is fine.

@ZhangPejin: Sure they need to pass a test, but that test is only a background check I believe. You can't have the entirety of illegal immigrants be registered as a terrorist...

[quote="ZhangPeijin"]Oh in order to get into the US, you just need a Visa.
But there's like a minor US history test for citizenship.[/quote]

AAAAAHHHH! *runs away*

[rhetorical]If this was a land of the free, why are people being shipped away?[/rhetorical]

It took me, an American-raised, legal immigrant, a guy who is more patriotic towards the US than many white Americans, someone who seeks to improve this country (and I think I can), 12 years to get my greencard, and I am not yet allowed to apply for citizenship. I've tutored American kids, gotten American peers off drugs, served the American elderly in nursing homes, and picked squash for American food kitchens.

I don't think someone who is here illegally and who contributes less to the community than I do should have it easier. That's just not fair.

And please, don't say it's that they have it much harder, I've lived without a bed...

You notice that new story awhile back?
About the woman who had a son born in the US while she was here illegally?
They make it sound like the American Government is so cruel that they would separate the boy's mother from her etc, etc.
Though I was pretty sure that the boy would've been able to went back to Mexico with her, but idk.

[quote="miyomiyo"][quote="ZhangPeijin"]No but basically, in order to become an american citizen after you're 18, you need to pass a test, and most illegal immigrants would be unable to pass that test.
And they also need funding.
Most illegal immigrants are just people who were too poor to live in Mexico, which must be pretty poor seeing as Mexico is almost a third world country.[/quote]Actually, 60% of home-born Americans can't pass the test.[/quote]

we took that test in History Class, and we got past the requirement... and i'm not a US citizen...

I'm too lazy to take it again and become a US citizen.


What integer is closets to the fourth power of $ \sqrt{2 \plus{} \sqrt{2 \plus{} \sqrt{2}}}$

We are given $ x=\sqrt{2+\sqrt{2+\sqrt2}}\implies x^2=2+\sqrt{2+\sqrt2}\implies x^4=6+\sqrt2+4\sqrt{2+\sqrt2}$. Since $ 1.4<\sqrt2\<1.5$ and thus

$ \sqrt{3.24}=1.8<\sqrt{2+\sqrt2}<\sqrt{3.61}=1.9$, we have $ 6+1.4+4(1.8)=14.6<x^4<6+1.5+4(1.9)=15.1\implies\boxed{15}$.

Since this is a countdown, I feel like I'm being stupid and missing some trick...

[quote=math154]We are given $ x=\sqrt{2+\sqrt{2+\sqrt2}}\implies x^2=2+\sqrt{2+\sqrt2}\implies x^4=6+\sqrt2+4\sqrt{2+\sqrt2}$. Since $ 1.4<\sqrt2<1.5$ and thus

$ \sqrt{3.24}=1.8<\sqrt{2+\sqrt2}<\sqrt{3.61}=1.9$, we have $ 6+1.4+4(1.8)=14.6<x^4<6+1.5+4(1.9)=15.1\implies\boxed{15}$.

Since this is a countdown, I feel like I'm being stupid and missing some trick...[/quote]

[quote=nowunkie]I have a different way. It may be quicker, too. This way could easily be done in the brain, and does not involve any decimals.

[hide] Starting from the third step of math154's solution, we have $ x=\sqrt{2+\sqrt{2+\sqrt2}}\implies x^2=2+\sqrt{2+\sqrt2}\implies x^4=6+\sqrt2+4\sqrt{2+\sqrt2} $
We know $ 2+\sqrt{2}\approx 4 $ Sort of, not really, but close enough. You will see why. Now we have a perfect square, yay! $ x^4\approx 6+\sqrt{2}+4\sqrt{4}=6+\sqrt{2}+8 $ We know $ \sqrt{2} $ is approximately 1 point something close to $ 5 $, but since we already compensated before for approximating $ \sqrt{2+\sqrt{2}} $, our final answer is $ 6+1+8\approx \boxed{15} $[/hide][/quote]

I love this solution!


Let $ a,b,c>0$. Show that:

\[ \frac{1}{a^2\plus{}bc}\plus{}\frac{1}{b^2\plus{}ca}\plus{}\frac{1}{c^2\plus{}ab}\le \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}\left(\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\right)\]

[quote="Pain rinnegan"]Let $ a,b,c > 0$. Show that:
\[ \frac {1}{a^2 \plus{} bc} \plus{} \frac {1}{b^2 \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ab}\le \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}\left(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\right)\]
[/quote]
The inequality is equivalent to
\[ \sum \frac{a\plus{}b\plus{}c}{b\plus{}c}\geq \sum \frac{ab\plus{}bc\plus{}ca}{a^{2}\plus{}bc},\]
or
\[ \sum \frac{a}{b\plus{}c}\plus{}\sum \frac{a^{2}}{a^{2}\plus{}bc}\geq \sum \frac{a(b\plus{}c)}{a^{2}\plus{}bc}.\]
By the Cauchy-Schwarz Inequality, we have
\[ \begin{aligned}
\frac{1}{a^{2}\plus{}bc} &\equal{}\frac{b\plus{}c}{(a^{2}\plus{}bc)(b\plus{}c)}\equal{}\frac{b}{(a^{2}\plus{}bc)(b\plus{}c)}\plus{}\frac{c}{(a^{2}\plus{}bc)(c\plus{}b)} \\
&\leq \frac{b}{\left( a\sqrt{b}\plus{}\sqrt{bc}\sqrt{c}\right) ^{2}}\plus{}\frac{c}{\left( a\sqrt{c}\plus{}\sqrt{bc}\sqrt{b}\right) ^{2}}\equal{}\frac{1}{(a\plus{}c)^{2}}\plus{}\frac{1}{(a\plus{}b)^{2}}.
\end{aligned}\]
Therefore,
\[ \begin{aligned}
\sum \frac{a(b\plus{}c)}{a^{2}\plus{}bc} &\leq \sum \left[ \frac{a(b\plus{}c)}{(a\plus{}b)^{2}}\plus{}\frac{a(b\plus{}c)}{(c\plus{}a)^{2}}\right] \equal{}\sum \left[ \frac{b(c\plus{}a)}{(b\plus{}c)^{2}}\plus{}\frac{c(a\plus{}b)}{(b\plus{}c)^{2}}\right]  \\
&\equal{}\sum \frac{a}{b\plus{}c}\plus{}\sum \frac{2bc}{(b\plus{}c)^{2}}.
\end{aligned}\]
It suffices to prove that
\[ \frac{a^{2}}{a^{2}\plus{}bc}\plus{}\frac{b^{2}}{b^{2}\plus{}ca}\plus{}\frac{c^{2}}{c^{2}\plus{}ab}\geq \frac{2bc}{(b\plus{}c)^{2}}\plus{}\frac{2ca}{(c\plus{}a)^{2}}\plus{}\frac{2ab}{(a\plus{}b)^{2}}.\]
Since $ (b\plus{}c)^{2}<2(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca),$ we have
\[ \frac{2bc}{(b\plus{}c)^{2}}\equal{}\frac{1}{2}\minus{}\frac{(b\minus{}c)^{2}}{2(b\plus{}c)^{2}}\leq \frac{1}{2}\minus{}\frac{(b\minus{}c)^{2}}{4(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca)}.\]
It follows that
\[ \begin{aligned}
\sum \frac{2bc}{(b\plus{}c)^{2}} &\leq \frac{3}{2}\minus{}\frac{(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2}\plus{}(a\minus{}b)^{2}}{4(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca)} \\
&\equal{}\frac{(a\plus{}b\plus{}c)^{2}}{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca}.
\end{aligned}\]
Therefore, it is enough to check that
\[ \frac{a^{2}}{a^{2}\plus{}bc}\plus{}\frac{b^{2}}{b^{2}\plus{}ca}\plus{}\frac{c^{2}}{c^{2}\plus{}ab}\geq \frac{(a\plus{}b\plus{}c)^{2}}{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca},\]
which is true according to the Cauchy-Schwarz Inequality. Equality holds if and only if $ a\equal{}b\equal{}c.$

The following is true too:

Let $ a,b,c>0$. Prove that:

\[ \frac{1}{a^2\plus{}bc}\plus{}\frac{1}{b^2\plus{}ca}\plus{}\frac{1}{c^2\plus{}ab}\le \frac{1}{\sqrt[3]{abc}}\left(\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\right)\]

[quote="Pain rinnegan"]The following is true too:

Let $ a,b,c > 0$. Prove that:
\[ \frac {1}{a^2 \plus{} bc} \plus{} \frac {1}{b^2 \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ab}\le \frac {1}{\sqrt [3]{abc}}\left(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\right)\]
[/quote]
Yes, it's a hard problem  :maybe:
My solution is based on the following ineq:
$ 2\sum\ ab(a^4 \plus{} b^4) \plus{} 5\sum\ a^3b^3 \plus{}$ $ 7abc\sum\ a^3 \plus{} 12a^2b^2c^2 \ge 2\sum a^2b^2(a^2 \plus{} b^2) \plus{} 8abc \sum ab(a \plus{} b)$
It's obviously trues because
$ 2\sum ab(a^4 \plus{} b^4) \ge 2\sum\ a^2b^2(a^2 \plus{} b^2)$
$ 5\sum\ a^3b^3 \ge \frac {5}{2}abc\sum ab(a \plus{} b)$ (by AM-GM)
and $ 7abc \sum a^3 \plus{} 12a^2b^2c^2 \ge \frac {11}{2}abc \sum ab(a \plus{} b)$
Indeed, this proof is rather long. I'll post it soon. Wait for another proofs  :)


Find the largest integral value of $ x$ which solves: $ \frac{1}{3}<{\frac{x}{5}}<{\frac{5}{8}}$

In reality only an upper bound is necessary:
x/5<5/8 => x<25/8=3.125, so x=3 is the largest integer.

Please use LaTeX!!!

Because each $ \frac{1}{5}$ is $ 0.2$, We find multiples of $ 0.2$ less than $ \frac{5}{8}\equal{}0.625$, finding that $ \frac{3}{5}\equal{}0.6$ is the largest, since another $ 0.2$ added would bring the solution above the decimal value of $ \frac{5}{8}$. In general, it is much easier to compare fractions when converted to a decimal.

If $ \frac{x}{5} < \frac{5}{8}$, then $ 8x < 25$. Therefore, $ x < \frac{25}{8}$, so the answer is 3.


I was on Emory University's website checking out their early admission acceptance rate

http://www.emory.edu/ADMISSIONS/admission-aid/apply.htm

and the website showed that 32% acceptance rate for Regular, and 34% for early decision.

That's almost like no differences. could anyone explain this?

thanks

Why should applying ED make you more likely to get in? There's really only one factor, which is the commitment factor. Applying ED suggests that the school is your top choice, and that makes you slightly more desirable. But otherwise, there's nothing. When a school has a significantly higher ED acceptance rate, it's probably because the ED pool was self-selected into being higher quality.

Personally, I don't see the point of ED and I'm glad that Harvard precipitated its decline in the Ivy League (well I don't know if any other top university has followed yet besides Princeton, but presumably after those two did, other schools are considering it).

Virginia has also followed the lead of Harvard and Princeton in going to a single-deadline system, if I remember correctly.

http://www.virginia.edu/undergradadmission/admission.html

MIT uses [b]non-binding[/b] early action :D .


Hi everyone,
does anyone know how to do these two problems??? i have been working on them for hours and am unable to get an answer. please if you could help me i would appreciate it!!! :)

sin[Arccos(-square root 2/2)-pie/4]

        my attempt: with using the unit circle i got 90 degrees or 1 or 0 or 180 degrees. i dont know if the anwswer is to be in degrees or not as the instructions for both of these problems was just to find each value.
cos(2Arctan squareroot 3/3)
        my attempt was getting cos(2 times the square root of 12) by using the unit circle

PLEASE if you could help me work out these problems i would very much appreciate it!

Thank you very much

[quote="chocolatelover"]Hi everyone,
does anyone know how to do these two problems???
cos(2Arctan squareroot 3/3)...
[/quote]

Step by step evaluation... That's the Key:
First off,
$ \arctan \frac {\sqrt {3}}{3} \equal{} \arctan \frac {1}{\sqrt {3}}.$

Draw an equilateral triangle of side 2. It is straightforward now that

$ \arctan \frac {1}{\sqrt {3}} \equal{} \frac {\pi}{6}.$

Hence, $ \cos \left(2 \arctan \frac {\sqrt {3}}{3}\right) \equal{} \cos \left(\frac {\pi}{3}\right) \equal{} \frac {1}{2}.$

Hope that helps...

Thought there were 2 problems there. All I did was sketch the resolution process of the second one [8th line of your original post].

no you were right there were two problems... but i ment in the problem that you did it stated 2Arctan what happened to the 2 in the process that you did? i did not follow what happened to the 2

[quote="chocolatelover"]thank you so much for taking your time to help me!!
i have one question though...what happend to the 2???
thanks[/quote]

Just edited that part, man... Sorry.

its ok thanks for the help...i thought that was what probably happened just forgot the 2...but the answer still works out the same so its ok. but thank you very much for the help

[quote="chocolatelover"]

sin[Arccos(-square root 2/2)-pie/4]...

[/quote]

Real problem is the precomputation of $ \arccos \left( \minus{} \frac {\sqrt {2}}{2}\right)$. The rest would follow from the [b]sin-of-a-sum[/b] identity. Gimme a sec to work out the sol. in some detail.

Your idea of introducing the [color=red]unit circle[/color] is OK. Though, you have to be careful of the way you pick the value for $ \arccos \left( \minus{} \frac {\sqrt {2}}{2}\right)$. It turns out to be $ \frac {3\pi}{4}$ in this case. Therefore,

$ \arccos \left( \minus{} \frac {\sqrt {2}}{2}\right) \minus{} \frac {\pi}{4} \equal{} \frac {\pi}{2},$

and the mentioned identity would'nt be necessary after all. You know, $ \frac {\pi}{2}$ is one of them notable angles in Trigs.

ok thanks i would really appreciate that :)

[quote="coquitao"][quote="chocolatelover"]

sin[Arccos(-square root 2/2)-pie/4]...

[/quote]

Your idea of introducing the [color=red]unit circle[/color] is OK. Though, you have to be careful of the way you pick the value for $ \arccos \left( \minus{} \frac {\sqrt {2}}{2}\right)$. It turns out to be $ \frac {3\pi}{4}$ in this case. [/quote]

Why? The näive approach dictates to our minds that there are 2 possibilities for an x to satisfy $ \cos x \equal{} \minus{} \frac {\sqrt {2}}{2}.$ Yet, we must always pick the value for x that belongs to $ [0,\pi]$.

[quote="chocolatelover"]ok i understand that but then dont you have to take the sin of pie/2??[/quote]

You are right... Didn't write it, though. In any case, here it is:

$ \sin \frac{\pi}{2} \equal{} 1$.

See you!


geometry
ratio
geometric transformation
reflection
parallelogram

[quote="Andreas"]1. Does there exist a positive integer $N$ such that the number formed by the last two digits of the sum
$1 + 2 + 3 + \cdot\cdot\cdot + N$ is $98$?

2. Let $ABCD$ be a rectangle. If $P$ and $Q$ are points respectively on $AD$ and $DC$ such that the areas of the triangles $BAP, PDQ$ and $QBC$ are all equal, find the ratios $DP/PA$ and $DQ/QC$.

3. Let $P$ be an interior point of an equilateral triangle $ABC$ such that $AP^2 = BP^2 + CP^2$.
Prove that $\angle BPC = 150^o$.

PS: I didn't solve the last. Any idea for it?[/quote]

[hide="1."]1. Consider $n(n+1) \mod 10$. In order for $n(n+1)$ to end in $6$, and $\frac{n(n+1)}{2}$ to possibly end in $8$, $n \equiv 7 \mod 10$
So let $n$ be of the form $7 + 10a_1 + 100a_2 + ..... 10^{n}a_n$.
So $n+1$ is in the form  $8 + 10a_1 + 100a_2 + ..... 10^{n}a_n$

To see if $\frac{n(n+1)}{2}$ ends in $98$, it suffices to consider multiplying the last two digits of $n$ and $n+1$, then dividing by $2$.

$\frac{(7 + 10a_1)(8+10a_1)}{2} = \frac{56 + 150a_1 +100{a_1}^2}{2} = 28 + 75a_1 + 50{a_1}^2.$

So for $\frac{n(n+1)}{2}$ to end in $98$, $28 + 75a_1$ must end in $98$.

But $28 + 75a_1$ ends in only $28, 03, 53, 78$.

So $\frac{n(n+1)}{2}$ never ends in $98$.[/hide]

Let $R$ be the reflection of the point $P$ w.r.t. the midpoint $M$ of the side $[BC].$

$PA^2=PB^2+PC^2,\ 4AM^2=3a^2,\ 4AM^2=2(AR^2+PA^2)-4MP^2$ $,\ 4MP^2=2(PB^2+PC^2)-a^2;$

$2AR^2=4AM^2-2PA^2+4MP^2=3a^2-2(PB^2+PC^2)$ $+[2(PB^2+PC^2)-a^2]=2a^2\Longrightarrow AR=a.$

Therefore $R\in C(A,a)$, the quadrilateral $BPCR$ is a parallelogram and $A=60^{\circ}.$

Thus, $m(\widehat {BPC})=m(\widehat {BRC})=\frac 12(360^{\circ} -A)=150^{\circ}.$

Last one is cute...

Let Q be the point such that $\angle QBP = 60$ and QB = PB.  Since we then have QBP equilateral and QBA and PBC congruent, $PB^2 + PC^2 = PQ^2 + QA^2$ which equals $AP^2$ only when $\angle PQA = 90 \implies \angle BQA = 150 \implies \angle BPC = 150$.

For 1, if $ \frac {n(n \plus{} 1)}{2} \equal{} 98 \plus{} 100k$, then $ (2n \plus{} 1)^2 \equal{} 100q \plus{} 85$, but a square can't be 10 mod 25.

For 2, WLOG let AB=1, AP=m, PD=n, DQ=p, then $ m\equal{}np\equal{}(m\plus{}n)(1\minus{}p)\equal{}n\minus{}pm$
Thus $ np\equal{}n\minus{}p(pn)$, $ p\equal{}\frac{\sqrt{5}\minus{}1}{2}$, the rest follows easily.


Galois Theory
superior algebra
superior algebra unsolved

Let $ p$ be a prime and let $ F$ be a field. Let $ K$ be a Galois extension of $ F$ whose Galois group is a p-group.
Suppoe that $ L$ is a Galois extension of $ K$ whose Galois group is a p-group.
Prove that the Galois closure of $ L$ over $ F$ is a Galois extension of $ F$ whose Galois group is a p-group.

[hide="Hint"]Every $ \sigma\in Gal(L/F)$ induces an automorphism $ \sigma'\in Gal(K/F)$ by restriction to $ K$.[/hide]

[quote="Third Edition"]Every $ \sigma\in Gal(L/F)$ induces an automorphism $ \sigma'\in Gal(K/F)$ by restriction to $ K$.[/quote]
$ L/F$ may not be Galois extension.
What is $ Gal(L/F)$?

It is notation for the automorphisms of $ L$ that fix $ F$. The extension doesn't have to be Galois to talk about that (maybe some places use $ Aut(L/F)$ when the extension is not Galois?).


limit
calculus
symmetry
function
calculus computations

What is the maximum value of (1/2)^x + (1/2)^(1/x) for x > 0

hello, we have $ f(x) \le 1$ for all $ x \in \mathbb{R}$ and $ x>0$. Further we have $ \lim_{x \to \infty}f(x)\equal{}1$.
Sonnhard.

Ugh we can't even apply Jensen's to this. I can't think of anything but calculus here.

If it does end up going to calculus, I would suggest the substitution $ x \equal{} 2^y \implies \frac{1}{x} \equal{} 2^{\minus{}y}$ to gain some degree of symmetry: \[ f(y) \equal{} \left( \frac{1}{2} \right) ^{2^y} \plus{} \left( \frac{1}{2} \right) ^{2^{\minus{}y}}\; ,\; \; y \in \mathbb{R} \]
[hide="Calculus"]Differentiate, set to 0, and rearrange to get $ \frac{2^y \minus{} 2^{\minus{}y}}{2} \equal{} y$. We know a lot about this function because it is closely related to hyperbolic sin! In fact, we can easily show that there must be exactly 3 solutions to this: $ y\equal{}0$ and $ y\equal{} \pm a$ for some $ a$.

From there, we can use the fact that $ f$ is eventually increasing and approaches $ 1$ as $ y$ increases without bound, to argue that the the graph must have one of two possible shapes:
1. A relative minimum at $ y\equal{}0$, non-extremal stationary points at $ \pm a$, and increases asymptotically to $ 1$ (symmetric for negative $ y$)
2. A relative maximum at $ y\equal{}0$, relative minima at $ \pm a$, and then increases asymptotically to $ 1$ (symmetric for negative $ y$)

Case (1), we have a global minimum at $ y\equal{}0$, but we know that $ f(0)\equal{}1 \equal{} \lim_{y \to \infty} f(y)$, which is invalid in our situation. This means our graph assumes the shape outlined in case (2). Now it is obvious that the maximum must either be the attainable value at $ y\equal{}0$ or the unattainable limit as $ y$ increases (or decreases) without bound. We find that the maximum is the attainable value $ f(0)\equal{}1$.[/hide]


The number 3003 is a palindrome because its digits are the same forward or backward.  The number 1 is also a palindrome.  How many whole numbers between 1 and 3003 are palindromes?

[hide]  Lets find all the plaindromes including 1 and 3003.

x= 9

xx=9

xxx=90

xxxx (maximum of 2)=20

3003= 1

9 + 9 + 90 + 20 + 1 = 129, 129-2=[b]127[/b] [/hide]

[hide="Can"] [hide="any"] [hide="of"]       [/hide] [/hide] [/hide]  It requires some math.....

1 digits:
9

2 digits:
9

3 digits:

9*10 = 90

4 digits:

2*10 + 1

Total: 129?

G-UNIT is correct, because it asks for palindromes [i]between[/i] 1 adn 3003.

[quote="math92"][quote="G-UNIT"]delete this[/quote]
what was the point of that?[/quote]

My guess is that G-UNIT posted something, then edited it. You wouldn't see the "edited" line then. Although, then G-UNIT could just delete it then.

To delete your own post click on the x next to the quote button.


ratio
geometry
symmetry
geometric transformation
reflection

Consider $ \triangle ABC$ with vertices $ A(4,8),B( \minus{} 1,2),$ and $ C(0, \minus{} 3).$ Find the point $ D$ such that $ \triangle ABD,\triangle ACD$ and $ \triangle BCD$ all have the same area.

Let $ G$ be the centroid; let $ A'\in AG\cap BC$. Then $ A'$ is the midpoint of $ BC$. It is well known that $ \frac{AA'}{GA'}\equal{}\frac{1}{3}\equal{}\frac{[ABC]}{[BCG]}$ since we can drop altitudes from $ A$; $ G$ to $ BC$ the the ratio of the heights is $ 1: 3$ and the bases are the same so the area ratio is also $ 1: 3$. By symmetry of argument, all these triangles have an area that is a third of the original triangle so they are equal.

I didn't know that, I used matrices instead. But I didn't know whether the absolute values of the expressions were positive or negative so I had to try out possibilities.

I didn't prove that I found all solutions. In fact, there are three other solutions!

In fact, one can show that the triple ratio $ u: v: w\equal{}[PBC]: [PCA]: [PAB]$ uniquely defines a point $ P$ in the plane of $ ABC$. (The areas are signed, if $ P$ is in the same half plane as $ A$ wrt line $ BC$, $ [PBC]$ is positive. If $ P$ and $ A$ are on opposite sides of $ BC$, then $ [PBC]$ is negative.)

Noting that $ [PBC]\plus{}[PCA]\plus{}[PAB]\equal{}[ABC]$ (using directed areas), the problem postulates that

$ |[PBC]|\equal{}|[PCA]|\equal{}|[PAB]|$

We can normalize to $ u\plus{}v\plus{}w\equal{}1$. So we either have the solutions $ (u,v,w)\equal{}\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right),(1,1,\minus{}1)$ and permutations.

The first solution corresponds to the centroid. The second corresponds to the reflection of a vertex in the midpoint of the opposite side.


Given a non negative integer $ n$
Proof: there [b]exists[/b]  a [b]unique[/b]  polynomial $ p(x)$ which satisfies
(1)$ P(0)\equal{}1$
(2)$ degP(x)\equal{}n$
(3)$ (1\plus{}x)P(x)\plus{}(1\minus{}x)P(\minus{}x)\equal{}2$


[quote="BanishedTraitor"]$x,y,z$ are positive numbers. Given that $x^{2}+y^{2}+z^{2}=1$. Find the minimum value of $S=\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}$.

[hide="Spoiler"]Consider $S^{2}$[/hide][/quote]
We have $S^{2}\geq3 ({x}^{2}+{y}^{2}+{z}^{2})=3$ by using AM-GM inequality
So $minS=\sqrt{3}$ when $x=y=z=\frac{1}{\sqrt{3}}$


articles
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This article was published this summer, but only recently came to my attention.  What do you all think of it?
[url=http://www.theglobeandmail.com/servlet/ArticleNews/TPStory/LAC/20030816/MATHMAIN16/TPFocus/]Strength in numbers[/url]

I'm somewhat annoyed.  Granted, it's not a major newspaper, but they could have done much better.  For instance, they call it an "arithmetic program"; they make fun of the campers ("Those campers who have made it out of bed shovel scrambled eggs down the hatch -- one, seeking optimal efficiency, lowers his mouth to the same horizontal plane as his plate -- and another day at Mathcamp has begun."), they ridicule the idea of spending time doing math instead of 'normal' things (they quote someone, probably very loosely: "there's a lot of freedom and no curfew . . . because if we had to go to bed at a certain time, that might interrupt some important mathematical ideas."), etc.

And don't forget all the bad puns.  "Strength in numbers," indeed.  This is a fact of being a math person -- the New York City Math Team makes in into the New York Post from time to time when we win ARML or NYSML, but it's always in this tiny column at the bottom of the page that usually has pedophilia-type stories above it.  (I'm actually serious -- in 2002, we were on a page with a story about R. Kelly's sex video with underage girls and a story about an overly-active priest.)  Anyhow, this guy seems particularly set on degrading the mathcampers, which is a real shame.  But there will always be the point that they "could be running around playing sports," or some other nonsense like that.  It's life as a math person to be so regarded.

while at the ROSS program, we were interviewed by a newspaper, and in it they referred to "number theory" as "numbers theory" and got mod incorrectly (i think they said somehting like 25 is 3 mod 24) or something like that

Oddly enough, Mathcamp's website, http://www.mathcamp.org, links to the article.  I suppose comparatively ignorant press is better than no press at all.   Some people might recall an anti-drug ad that had someone say "I'd rather go to math camp than smoke a joint," which rather annoyed us, but became a joke (indeed, I learned about it through such jocular references).

I am myself a huge fan of Mathcamp, having gone in 2001 and 2002 (and I'm intending to return again this year).  It is definitely much more than what any such article could make it to be.

On the subject of bad math puns:
Decimals have their point.
Calculus has its limits.
Algebra is full of unknowns.
Trig lets you see things from many different angles.
And many more...

[quote]Oddly enough, Mathcamp's website, http://www.mathcamp.org, links to the article.[/quote]

That's how I found the article.  I'm somewhat surprised that they linked to it (and said it "gives a good overall feel for life at Mathcamp"), but I suppose anyone who is serious about math would just laugh at that article and ignore the insults anyway.

I think an online friend (it may even have been Alison's mom, although I'm not remembering for sure) first told me about the link through an email message to an email discussion list. I agree that the article conveys an attitude of the reporter that it's weird to like math, but that's what made the story newsworthy: finding a whole camp for unexpected people, namely likers of math. I printed out the whole article to show my son someday, but I haven't actually shown it to him yet (again, if I remember right). I think math-likers do need to meet one another in person to get over the many occasions on which they will be regarded as weird.

All the rest that I have heard about MathCamp, and I definitely have heard reports that Alison's mom has passed on from other people, has been uniformly favorable. It's on our list here as a program to apply to when my son is the right age.

"Mathcamp is the negative reciprocal of all that."

???

"Mathcamp was not created to provide a summer wonderland in the woods, but to be an incubator where gifted students aged 15 to 18 could pursue their beloved science beyond the bounds of outdated and unimaginative curricula, not to mention burnt-out teachers."

Outdated curricula?  I agree that the math curricula in school isn't the best, but I don't like the sound of "retro-math" at all.

"He has four adult daughters in his native England, and on this side of the Channel, two teenaged boys as well as a two-year old son he watches two days a week, each grouping from a subsequent marriage.)"

Wouldn't "this side of the English Channel" be referring to France, since the implication is that it's on the side of the English channel opposite of England?

But I like the second half of the article better.  I was amused at the fact that there is a cheerleader camp at the same place...

[quote="mathfanatic"]"Mathcamp is the negative reciprocal of all that."

"He has four adult daughters in his native England, and on this side of the Channel, two teenaged boys as well as a two-year old son he watches two days a week, each grouping from a subsequent marriage.)"[/quote]

It looks like you could teach the editor of that newspaper a thing or two about English, as well as about math.

Gollly Gee TokenAdult you stay up late I think its 12 olock rightnow.

My son has attended Mathcamp for each of the past 3 years, and hopes to go again. I'm a big Mathcamp fan. But I wasn't offended at all by the article. If was written by a non-math person as entertainment for an audience of non-math persons. It was meant to be amusing, not repectful. Taken in that light, I thought the description of events gave a pretty positive picture of why someone who loves math might want to go to Mathcamp.

(some of the language peculiarities like "arithmetic camp" and "other side of the channel" may have to do with the author being Canadian. For instance, maybe they use the word arithmetic in a more general sense than we do, to mean mathematics. Canadians, what do you think?)

A lot of mathematicians say arithmetic to mean number theory.

I didn't really like the way the writer did this article.....


Let AB be circle O's diameter, CD is a chord, AE is perpendicular to CD, AE intersects CD at E, BF is perpendicular to CD, BF intersects CD at F. Prove EC=DF. :ninja:

[hide]If $M$ is the midpoint of $EF$ , then in trapez $AEFB$ we have that $OM // AE //BF$. But then  $OM \perp EF$ , so $M$ is also the midpoint of $CD$. Finally

    $CE = ME - MC= MF - MD = DF$[/hide]
 Babis

Anyone else with a different solution. :D  ;)

I don't understand your last step, it seems like you are assuming it, can you explain why?


[quote="ckck"][quote="The QuattoMaster 6000"][quote="b-flat"]Simplify the following:

$\sqrt[4]{284+48\sqrt{35}}$[/quote]
[hide] $\sqrt[4]{284+48\sqrt{35}}=\sqrt{12+2\sqrt{35}}=\sqrt{5}+\sqrt{7}$, which is our answer. [/hide][/quote]
I don't get your steps. Can you expand your solution a bit longer?[/quote]

Quatto means that he found the fourth root of 284 and 84, simplifing it to
$\sqrt{12+2\sqrt{35}}$

wait noo that doesn't make sense....quatto explain urself  :huh:

[quote="7h3.D3m0n.117"][/quote][quote="ckck"][quote="The QuattoMaster 6000"][quote="b-flat"]Simplify the following:

$\sqrt[4]{284+48\sqrt{35}}$[/quote]
[hide] $\sqrt[4]{284+48\sqrt{35}}=\sqrt{12+2\sqrt{35}}=\sqrt{5}+\sqrt{7}$, which is our answer. [/hide][/quote]
I don't get your steps. Can you expand your solution a bit longer?[/quote][quote="7h3.D3m0n.117"]

Quatto means that he found the fourth root of 284 and 84, simplifing it to
$\sqrt{12+2\sqrt{35}}$

wait noo that doesn't make sense....quatto explain urself  :huh:[/quote]

Usually with problems like these, you can assume a clean and tidy denesting, i.e. that $\sqrt{a \pm b \sqrt{c}}=m \pm n \sqrt c$. Solve by squaring, then setting rational and irrational parts equal. If this doesn't work, you can try solving other systems in a similar manner by assuming it denests into the form $\sqrt{m}\pm \sqrt{n}$, or other combinations of integers and radicals.

Just remember that the $\pm$ should be determined by what your nested expression looks like; and keep all radicals and denested radicals as real numbers (i.e. don't accidentally make something negative when it cannot be.)

[quote="ckck"][quote="jli"][quote="b-flat"]Simplify the following:

$\sqrt[4]{284+48\sqrt{35}}$[/quote]

[hide="solution/2"]Let $\sqrt[4]{284+48\sqrt{35}}=\sqrt{a+b\sqrt{c}}$

$\sqrt{284+48\sqrt{35}}=a+b\sqrt{c}$

$284+48\sqrt{35}=a^{2}+2ab\sqrt{c}+b^{2}c$

We know that $c=35$

$2ab=48\Rightarrow ab=24$

$a^{2}+35b^{2}=284$

Solving the 2 equasions we get $a=12$ and $b=2$.

So $\sqrt[4]{284+48\sqrt{35}}=\sqrt{12+2\sqrt{35}}$.

To simplify this, do something similar.  Im tired :| [/hide][/quote]
So then $\sqrt{12+2\sqrt{35}}$ simplified would be in the form of $a+b\sqrt{c}$ right?[/quote]

[b]IF[/b] it leads to a more simple expression, then yes. But if it does not, or if the system is inconsistent, then try other types of expressions, e.g. $\sqrt{m}+\sqrt{n}$ or others.

Well, actually, the way I did it way guessing and checking, kind of.
[hide] I assumed that the equation could be simplified.  Now, say that $\sqrt[4]{284+48\sqrt{35}}=\sqrt{a+b\sqrt{35}}\Rightarrow a^{2}=35b^{2}+2ab\sqrt{35}=284+48\sqrt{35}.$  This means that $a^{2}+35b^{2}=284$ and $2ab=48\sqrt{35}\Rightarrow ab=24\sqrt{35}$.  Guessing and checking from here gives that $a=12, b=2\Rightarrow \sqrt[4]{284+48\sqrt{35}}=\sqrt{12+2\sqrt{35}}$.  Now do a similar method for our new squareroot. [/hide]

\begin{eqnarray*}\sqrt{12+2\sqrt{35}}&=& \sqrt{5+2\sqrt{35}+7}\hfill \\ &=& \sqrt{(\sqrt 5+\sqrt 7 )^{2}}\hfill \\ &=& \sqrt 5+\sqrt 7 \hfill \\ \end{eqnarray*}


Let $s$ be the limiting sum of the geometric series

\[ 4 - \frac{8}{3} + \frac{16}{9} - \cdots, \]
as the number of terms increases without bound.  Then $s$ equals:

(A) a number between 0 and 1
(B) 2.4
(C) 2.5
(D) 3.6
(E) 12

Hello, MCrawford!

[quote]Let $S$ be the limiting sum of the geometric series: $ 4 - \frac{8}{3} + \frac{16}{9} - \cdots, $
as the number of terms increases without bound.  Then $S$ equals:

(A) a number between 0 and 1
(B) 2.4
(C) 2.5
(D) 3.6
(E) 12[/quote]

The sum of a convergent infinite geometric series is: $S \;= \;\frac{a}{1 - r}$
for first term $a$ and common ratio $r$.

We have: $a = 4,\;r = -\frac{2}{3}.$

Therefore: $\displaystyle{S \;= \;\frac{4}{1 - (-\frac{2}{3})} \;= \;\frac{12}{5} \;= \;2.4}$ . . . answer choice [b](B)[/b]

hi
it is a geometic series
with a=4 & r=-2/3:
     on calculating using infinite geometric series (S =a/(1-r))
it is 12/5


[hide="click to tell me your secrets"]1st - Brendan Fletcher & Bryce Taylor - 72
3rd - Christine Hong - 62
4th - David Lucia - 60
5th - Elizabeth someone (8th grade girl)
6th / 7th - Kevin Lang & someone else
8th - Shannon Taylor - 58
9th - 4? people tied with 51[/hide]

There were 18 questions worth 4 points each (1 point for a blank) on the contest.

Overall, the contest did a fairly good job of separating the state Qualifiers from the non-qualifiers.

As a related note, Forsyth Country Day school won Algebra II and I don't remember any other results, but I'll post them once they are on the web site.

[quote="b-flat"]

I don't remember any other results, but I'll post them once they are on the web site.[/quote]

XD

I don't think any words are necessary...

[url=http://www.math.uncc.edu/~hbreiter/problems/Writeup08.htm]Here are the results.[/url]

[hide="just comprehensive"]
[b]Individual Results[/b]
1st - Brendan Fletcher & Bryce Taylor - 72
3rd - Christine Hong - 62
4th - David Lucia - 60
5th - Alexander Mauney(8th grade girl)
6th - Elizabeth Shen
7th - Kevin Lang
8th - Shannon Taylor - 58
9th - 4? people tied with 51

[b]in-county schools:[/b]
1st - Providence Day School
2nd - Myers Park  High School
3rd - Charlotte Latin School

[b]out of county schools:[/b]
1st - C. E.  Jordan High School
2nd - Piedmont High School
3rd - North Lincoln High School[/hide]


Find all positive integer solutions to $abc-2=a+b+c$.

[hide] WLOG, assume $a \ge b \ge c$. We have $a = \frac{b+c+2}{bc-1}$ and $a \ge 1$, so $bc-1 \le b+c+2 \Rightarrow (b-1)(c-1) \le 4$. This gives us several cases:

$c = 1$: $ab-2 = a+b+1 \Rightarrow (a-1)(b-1) = 4$, which gives us solutions $(5, 2, 1); (3, 3, 1)$.

$c = 2$: $b = 2, 3, 4, 5$ which gives us the solution $(2, 2, 2)$.

$c = 3$: $b = 3$, no solutions.

So our solutions are $(2,2,2); (3,3,1); (5,2,1)$ and all permutations. [/hide]

[hide="easy"]If all of $a,b,c \ge 2$ then $abc\ge 4c$ Permutation and addition gives $abc \ge \frac{4}{3}(a+b+c)$ and so from the equation we obtain $a+b+c \le 6$. This means that $a=b=c=2$. Let $c=1$ then ...
:oops_sign:  - it's nearly the same with [b]paladin8[/b][/hide]
P.S. I offen observe "WLOG" here. Does it mean "easy", "brutal force" or somewhat else?

O, it was so easy to guess but I failed!


It's the same argument as proving that the product of limits is the limit of the products. If you make both terms close enough to the bound, you can make the product as close as you want to the product of the bounds.
There's also the step of showing that the product is an upper bound; that's what the positivity conditions are needed for.


[hide]
All I know is that the top reduces down to $14\sqrt{2}$.
[/hide]

And I can find nothing better than $7(2+\sqrt{2})\sqrt{\sqrt{5}-1}$


analytic geometry
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trig identities
Law of Cosines
calculus
calculus computations

Find the shortest distance between two points in polar coordinates.
(hint: use $ ds^2 \equal{} dr^2 \plus{} r^2d\theta^2$)

There's no need for differential notation. The distance formula in polar coordinates is the Law of Cosines:

The distance from $ (r_1,\theta_1)$ to $ (r_2,\theta_2)$ is
\[ \sqrt{r_1^2\plus{}r_2^2\minus{}2r_1r_2\cos(\theta_1\minus{}\theta_2)}\]

(I think I've assumed that $ r_1$ and $ r_2$ are nonnegative; I'll let someone else worry about what happens if you allow negative $ r.$)

Note that if our points are written as vectors $ \mathbf{v}_1, \mathbf{v}_2$, then $ |\mathbf{v}_1 \minus{} \mathbf{v}_2| \equal{} \sqrt { |\mathbf{v}_1|^2 \plus{} | \mathbf{v}_2|^2 \minus{} 2 \mathbf{v}_1 \cdot \mathbf{v}_2 }$, which is coordinate-free (and depends only on the inner product we specify) and fairly easy to compute in polar coordinates besides.  As I understand it, the form of the differential is only important if we specify a different metric.