That's an excellent question.  What contest is this from?

By compare, do you mean find the ratio between?
You know that c= 1+1/6-1/20+1/42-1/72...
a= 1-1/6-1/20-1/42...
But that is an interesting problem if you're comparing the ratios...

I believe compare means put in order of magnitude i.e. largest to smallest or vice versa.

we also know that b<a because 2005/2<1003

amirhtlusa.. you have the wrong summations

Clearly, $c$ is by far the smallest. On the other hand, $\ln 2<a<\ln 2+\frac1{2005}$ and $\ln 2<b<\ln 2+\frac1{2005}$. Those estimates used calculus, and I'm not sure there's an elementary way to approach it. The interesting question here is how to compare $a$ and $b$.

Actually, that's a beautiful question- so beautiful that I won't spoil it.

[hide="answer"]$a=b$[/hide]
[hide="solution"]Write $a=\sum_{k\text{ is odd}}\frac1k\left(1-\sum_{n>1,2^nk<2005}\frac1{2^n}\right)$ The inner sum is $\frac1{2^mk}$, where $2^mk$ is the largest number of that form less than or equal to 2005. Rearranging the sum, it becomes the sum of the reciprocals of the integers between $\frac{2005}2$ and 2005, which is $b$.[/hide]

yea it is great
but during a contest i think that I would've just done
[hide] subtract b from a and find out if its < or > 0
so then you get all the odd terms minus the evens till 1002 minus 2*evens till 2004
then I noticed that the averages even out so it =0[/hide]

Thanks for your excellent solution! Nice job! :)

This suggests a nice problem.

Let $H_k = 1/1 + 1/2 + \dots + 1/k$ be the k'th harmonic number.

Let $a_n, b_n, c_n$ be the above sums with $2005$ replaced by any $n = 4m+1$.

Show that $a_n$ and $b_n$ can be expressed simply in terms of the $H_k$, and explain why this is not true for $c_n$.


calculus
derivative
calculus computations

y is it called that? i mean its not really partial is it?

Because you're taking the derivative with respect to one variable of many.

That method is used in Multivariable Calculus, when you have multiple independent variables and one dependent variable, in which when you do derivatives, you can only do it with one at a time, so that's why it's called partial differentiation.

well, well i didn't know you could do gauss1181 could do calculus!


advanced fields
advanced fields unsolved

Is it true that if we have a series of independent positive random variables $\sum \xi_{k}$ then the sufficient condition for convergence a.s. is convergence of $\sum E\xi_{k}$? I think it is because we can switch $E$ and $\sum$, but i also know that Kolmogorov theorem says that there must be also $\sum D\xi_{k}<\infty$  :?:

Kolmogorov's necessary and sufficient condition (also known as "the three series theorem") says that in general the following three series must converge: $\sum_{k}P(|\xi_{k}|>1), \quad \sum_{k}E\xi'_{k}, \quad \sum_{k}D\xi'_{k}$ where $\xi'_{k}=\begin{cases}\xi_{k}, |\xi_{k}|\le 1\\0, |\xi_{k}|>1\end{cases}$
I leave it to you to check that, for positive $\xi_{k}$, the condition $\sum_{k}E\xi_{k}<+\infty$ implies these three. The condition $\sum_{k}D\xi_{k}<+\infty$ is not necessary for convergence a.s.


function limit integration inequalities trigonometry calculus geometry

A recent attempt at proving the Riemann Hypothesis can be found here.

[url=http://arxiv.org/PS_cache/arxiv/pdf/0806/0806.0892v1.pdf]Zhang's Proof[/url]

The proof relies on showing that the function:
\[ \Phi(t)\equal{}4\pi\sum_{n\equal{}1}^{\infty}\left\{2n\pi^4 e^{9t/2}\minus{}3n^2e^{5t/2}\right\}\exp(\minus{}n^2 \pi e^{2t})\]
satisfies:
\[ \lim_{t\to\infty}\frac{\ln|\Phi^{(n)}(t)|}{t}\equal{}\minus{}\infty;\quad n\in\mathbb{Z}^{\plus{}}\]
and the entire function:
\[ \Xi(z)\equal{}\int_0^{\infty}\Phi(t)\cos(zt)dt\]
has genus at most of 1.  He doesn't explain this.  Can someone help explain how these are satisfied?

[edit] corrected statement

Thanks,
Shaw

$ \Xi$ has order 1, and hence has genus $ \leq$ 1 by Hadamard's factorization theorem

Ok, thanks Kalle.  The proof is at a level I believe I can follow (with some additional work).  May try to take it apart to see if I can find a flaw.

I find the zeta function interesting to study.  Some background is needed first to understand what Zhang is attempting to show above.  We start with Jensen's Theorem that each of the following inequalities is  necessary and sufficient for the entire function $ f(z)\neq 0$ with genus $ 0$ or $ 1$ to have only real zeros:
\[ y\frac {\partial}{\partial y}\big|f(x + iy)\big|^2\geq 0\quad\quad \frac {\partial^2}{\partial y^2}\big|f(x + iy)\big|^2\geq 0;\quad - \infty < x,y < \infty
\]
Take for example $ f(z) = sin(x + iy)$:
\begin{align*} y\frac {\partial}{\partial y}\big|\sin(x + iy)\big|^2 & = y\frac {\partial}{\partial y}\big(\sin^2 x + \sinh^2 y\big) = y\sinh 2y\geq 0 \\
\frac {\partial^2}{\partial y^2}\big|\sin(x + iy)\big|^2 & = \frac {\partial}{\partial y}\sinh 2y = 2\cosh2y > 0 \end{align*}
and thus by Jensen's Theorem, $ f(z) = \sin(z)$ has only real zeros.
Now consider the function defined as $ \displaystyle\xi(z) = \frac {z(z - 1)}{2\pi^{z/2}}\Gamma(z/2)\zeta(z)$ and a variation of this function given by $ \Xi(z) = \xi(1/2 + iz)$. The Riemann Hypothesis can be formulated in terms of the $ \Xi$ function:  The Zeta function has non-trivial zeros only along the critical line if the $ \Xi$ function has only real zeros.  Now the $ \Xi$ function can be written as:
\[ \Xi(z) = \int_{ - \infty}^{\infty}\Phi(t)\exp(izt)dt
\]
with
\[ \Phi(t) = 2\pi\sum_{n = 1}^{\infty}\left\{2n\pi^4 e^{9t/2} - 3n^2e^{5t/2}\right\}\exp( - n^2 \pi e^{2t})
\]
Now, assuming $ \Xi(z)$ satisfies all the criteria of Jensen's Theorem, then $ \Xi(z)$ will only have real zeros if the following are true:
\begin{align*} y\frac {\partial}{\partial y}\left|\int_{ - \infty}^{\infty} \Phi(t)\exp(izt)dt\right|^2 & \geq 0 \\
\frac {\partial^2}{\partial y^2}\left|\int_{ - \infty}^{\infty}\Phi(t)\exp(izt)dt\right|^2 & \geq 0 \end{align*}
I haven't yet obtained Jensen's reference on expressing these partials as double integrals but I think I can if $ \displaystyle \overline{\Xi(x + iy)} = \int_{ - \infty}^{\infty}\Phi(t)e^{i(x - iy)t}dt$ is correct?   I'm not sure but if so then since $ |u|^2 = u\overline{u}$, by Leibniz and the product rule:
\begin{align*}\frac {\partial}{\partial y}\left\{\int_{ - \infty}^{\infty}\Phi(t)e^{i(x + iy)t}dt\int_{ - \infty}^{\infty}\Phi(s)e^{i(x - iy)s}ds\right\} & = \\
& \hspace{ - 30pt}\int_{ - \infty}^{\infty}\Phi(t)e^{i(x + iy)t}dt\left(s\int_{ - \infty}^{\infty}\Phi(s)e^{i(x - iy)s}ds\right) \\
& \hspace{ - 30pt} + \int_{ - \infty}^{\infty}\Phi(s)e^{i(x - iy)s}ds\left( - t\int_{ - \infty}^{\infty}\Phi(t)e^{i(x + iy)t}dt\right) \\
& \hspace{ - 40pt} = \int_{ - \infty}^{\infty}\int_{ - \infty}^{\infty}\Phi(t)\Phi(s)e^{i(s + t)x}e^{(s - t)y}(s - t)dsdt \end{align*}
I'd need to justify combining two single integrals into a double one though.  And if I take another partial:
\[ \frac {\partial^2}{\partial y^2}\big|\Xi(x + iy)\big|^2 = \int_{ - \infty}^{\infty}\int_{ - \infty}^{\infty}\Phi(t)\Phi(s)e^{i(s + t)x}e^{(s - t)y}(s - t)^2dsdt
\]
we can then state one of Jensen's necessary and sufficient condition for the Riemann Hypothesis to hold:
\[ \int_{ - \infty}^{\infty}\int_{ - \infty}^{\infty}\Phi(t)\Phi(s)e^{i(s + t)x}e^{(s - t)y}(s - t)^2dsdt\geq 0;\quad - \infty < x,y < \infty
\]
Jensen give another necessary and sufficient condition based on a variation of this double integral which I do not currently know how to derive:
\[ \int_{ - \infty}^{\infty}\int_{ - \infty}^{\infty}\Phi(t)\Phi(s)e^{i(s + t)x} (s - t)^{2n}dsdt\geq 0;\quad - \infty < x < \infty,\quad n=0,1,2,\cdots
\]
It is this last expression that Zhang is attempting to prove.

[edit] corrected formulas

Wait a minute.  Guess that's basic Complex Analysis:

Consider $ \displaystyle\Xi(x)\equal{}\int_{\minus{}\infty}^{\infty} \Phi(t)e^{ixt}dt;\quad x\in\mathbb{R}$.  Then $ \displaystyle\Xi(x)\equal{}\int_{\minus{}\infty}^{\infty}\Phi(t)\cos(xt)\plus{}i\int_{\minus{}\infty}^{\infty} \Phi(t)\sin(xt)dt$.

Now, it is know that $ \Phi(t)$ is an even function and therefore $ \displaystyle \int_{\minus{}\infty}^{\infty}\Phi(t)\sin(xt)dt\equal{}0$ and thus by the Reflection Principle, $ \displaystyle\overline{\Xi(z)}\equal{}\Xi(\overline{z})\equal{}\int_{\minus{}\infty}^{\infty}\Phi(t)e^{i(x\minus{}iy)t}dt$

Seems like a nice approach. I have never seen Jensen's theorem, looks very interesting.

I'm having some difficulty finding an accessible reference for Jensen's work.  He died before publishing the results stated here; G. Polya reviewed his work and latter published it but in German. So for now, I'll have to leave some holes in the preliminary lead in to Zhang's work.  How about this:  Suppose independent of any Riemann Hypothesis, we had the following assignment.  Given that the function $ \Psi(x)$ is a rapidly decreasing, positive, and even function, is:
\begin{align*} (47) \quad f_n(x) & = \int_{ - \infty}^{\infty}\int_{ - \infty}^{\infty}\Psi(s)\Psi(u)e^{i(s + t)x} (s - t)^{2n}dsdt \\
    (49) \hspace{3.5em} & \overset{?}{=} 2\int_0^{\infty}\cos(ux)\left\{\int_0^{\infty}\Psi\left(\frac {u + v}{2}\right)\Psi\left(\frac {u - v}{2}\right)v^{2n}dv\right\}du;\quad x\in\mathbb{R}, n = 0,1,2,\cdots \end{align*} where $ u=s+t$ and $ v=s-t$ and I'm using the same equation numbers as Zhang.  This part looks ok to me.  Note I can write $ (47)$ as:\begin{align*}
f_n(x)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Psi(s)\Psi(u)\cos\big[x(s+t)\big](s-t)^{2n}dsdt \\ &+i\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Psi(s)\Psi(u)\sin\big[x(s+t)\big](s-t)^{2n}dsdt
\end{align*} and since the sine integral is odd across the line $ s=-t$, I can rotate the axes by $ \pi/4$ and make a change of variable in the second integral:
$ \displaystyle u=\frac{s}{\sqrt{2}}+\frac{t}{\sqrt{2}}$ and
$ \displaystyle v=-\frac{s}{\sqrt{2}}+\frac{t}{\sqrt{2}}$
then:
$ \displaystyle s=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ and

$ \displaystyle t=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$

Which is then: \[ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(1)\Psi(s)\Psi(t)
\sin\big[xu\sqrt{2}\big](-\sqrt{2}v)^{2n}dudv\] which looks odd to me across the $ u$-axis and thus the integral would be zero although I'm not sure. The second change of variable is then used by Zhang to arrive at $ (49)$. Anybody see anything wrong with just this much?

I've reviewed equations $ (49)$ through $ (57)$ and although the author is a bit sloppy with notation, and equation cross-referencing, I believe the math is OK and illustrates some interesting problems in basic Calculus.  Anyone disagree?  However,  I detected a flaw in Zhang's work at equation $ (58)$:

Consider the integral:
\[ I_1(x) = \int_0^{\frac {1}{x}} f(ux)\int_0^{\infty}\Psi\left(\frac {u + v}{2}\right)\Psi\left(\frac {u - v}{2}\right)dvdu;\quad x > 0
\]
where $ \Psi$ is even, positive, and rapidly decreasing and $ f$ well-behaved to make everything converge.  Letting:
\[ u = \frac {\cos(\theta)}{x},\quad\quad v = v;\quad 0 < \theta < \pi/2
\]
the Jacobian is:
\[ \frac {\partial(u,v)}{\partial(\theta,v)} = - \frac {\sin(\theta)}{x}
\]
making the other substitutions, I obtain:
\begin{align*} \int_0^{\frac {1}{x}} f(ux)\int_0^{\infty}\Psi\left(\frac {u + v}{2} \right)\Psi\left(\frac {u - v}{2}\right)dvdu & \\
& \hspace{ - 155pt} = \int_{0}^{\pi/2}\frac {\sin(\theta)}{x}f\big(\cos(\theta)\big)\int_0^{\infty}\Psi\left(\frac {x^{ - 1}\cos(\theta) + v)}{2}\right)\Psi\left(\frac {x^{ - 1}\cos(\theta) - v}{2}\right)v^{2n}dvd\theta \end{align*}
I'm pretty sure that's right.  I checked it numerically with $ f(x) = \cos(x)$ and $ \Psi(x) = e^{ - x^2}$.   I realize numerical estimates are not enough but it does give some credence I think.  Zhang does not have the $ x$ in the denominator and this I think makes his estimates in equation $ (59)$ no longer valid.  I could be wrong of course: if this was water, I'd be standing on tip-toes to breathe.
Anyone here interested in this is willing to look at it?  I'll contact the author and give him a chance to respond.


AMC
USA(J)MO
USAMO
geometry
similar triangles

USAMO Winners:  The 12 scores in the range 38-27, inclusive
Winners names, schools and states will be posted on the AMC website later today.

Honorable Mention:  The 13 scores in the range 26-23, inclusive.

Blue MOSP:  The 16 non-12th grade students with scores in the range 26 - 18, inclusive.

Red MOSP: The 21 9th grade students with scores in the range 8-15, inclusive.

Training for CGMO: 8 girls with scores in the range 8-16, inclusive.

Official invitations will  be going out today (Mon, May 11)  and tomorrow (Tues, May 12)

Steve Dunbar
MAA Director, American Mathematics Competitions

Hmmmm... are CGMO and Red MOSP people mutually exclusive?

Don't think so... In fact, I think the CGMO group usually trains as part of Red? (maybe? I have no idea now that I think about it, but it makes sense since no girl got above 16...)

Hmm, I think my grandma who speaks only chinese might have picked up the phone while everyone else was away. That or what I described will happen tomorrow. Any insight on how this will affect my MOP invitation?

Just tell you grandma not to pick any phone call, I guess? It probably won't affect anything though, as they should e-mail you information later.

HOLY GUANO I GOT A 15!!!
sorry for the trolling, i solved one problem and i though each problem was worth 7 points. ...

yeah, i solved one problem o.0 it was 4
i misread 2 and 6, couldn't think of any observations to put on 3, observed unhelpful similar triangles on #1 and #5
EDIT: shoot firefox automatically took me to 2005. gah
when do you find this year's scores?
at least now i have a goal :)

he clicked on the wrong link, so he saw his "score" in 2005 of that number

I got an 8, but I'm a sophomore.  I have to study so hard to make blue mop next year... Anyone mind giving me mop notes? :)

I got an 20..  but i am canadian. SO does it mean i am done with this year's math fun?.. I want to go to MOP!

University applications are not the reason to go to MOP. It's about the fun and the contest training- which in your case would apply toward potentially making the Canadian IMO team.


LaTeX
calculus
advanced fields
advanced fields unsolved

Fo each $ a>0$, let $ x_a$ denote the sequence $ \left(1/n^a \right)_{n\equal{}1}^{\infty}$.
For which $ a$ do we have $ x_a \in l^\infty$?
For which $ a$ do we have $ x_a \in l^1$ ?
For which $ a$ do we have $ x_a \in l^2$?

$ a\geq 0 \iff x_a\in l^\infty$
$ a>1/2 \iff  x_a\in l^2$
$ a>1 \iff x_a\in l^1$

$ \text{\LaTeX}$ tip: I prefer \ell to l. $ \ell^1,\ell^2,\ell^{\infty}$ rather than $ l^1,l^2,l^{\infty}.$

Of course, once you strip off the abstract language, this is pretty much a standard calculus problem.


I have Linux and WIN XP dual system. i have internet connection through modem in WIN XP. but recently i wanted to make available it in linux as well. but on pack of modem it is said that: supported drivers for windows 98/ME/00/XP. that means it is not possible to connect internet at Linux?

It means it's a windmodem. You need to download specific drivers for you modem(if they are available). Try using linuxant drivers and slmodem drivers. Also, try to google for drivers for modem chipset, not the modem itself.

I installed Fedora Core the other day and failed to use the Internet. The weird thing is that I can access the modem through an IP (192.168.100.1), which is used for initialization. On the screen I get, it is said to me that all tasks are completed, so it should work (at least, it works when I get such a message under Windows). When I tell it to search proper bandwidth again, the modem is reset (so the connection works, til there at least), but do I need any drivers for the Internet to work?

I'd be grateful if anyone can answer my question. I'm a total newbie in this Linux stuff.

Could you provide some information about your hardware(especially modem).


Trivial by Mihailescu's Theorem  :rotfl:

[quote="t0rajir0u"] Mihailescu's Theorem[/quote]

What is it?

[quote="ZetaX"]Mihailescu is the one who proved Catalan's conjecture.[/quote]

But what is Mihailescu's Theorem ?  :huh:

Well, proving a conjecture leads often to them be named after you... ;)

Oh, and there is also google ;)

Ok! It here http://mathworld.wolfram.com/CatalansConjecture.html   :D

But above problem has level is Olympiad?

i smel here short listed problem 2000. find all natural numbers such:$a^{m}+1=(a+1)^{n}$. and also there is generalisation for this $a^{m}+1|(a^{k}+1)^{n}$. i will  post my solution using elementry number theory. :wink:

Both are trivial after using Zsygmoni  :P (and it is elementary)

[quote="ZetaX"] Zsygmoni  [/quote]

What is it? again :D

Sorry, it's spelled "Zsigmondy", see http://mathworld.wolfram.com/ZsigmondyTheorem.html .

[quote=N.T.TUAN]Find all triples $(x,y,z)$ of positive integers such that $(x+1)^{y+1}+1=(x+2)^{z+1}$.[/quote]
By Mihailescu's Theorem:
$\Rightarrow (x,y,z)=(1,2,1)$
By Zsigmondy's Theorem:
$\exists$ prime $p/ p|(x+1)^{y+1}+1, p\nmid x+2$
$\Rightarrow 0\equiv (x+2)^{z+1} \pmod{p} (\Rightarrow \Leftarrow)_\blacksquare$
But there are exceptions:
$\Rightarrow (x,y,z)=(1,2,1)$


[quote="Algadin"]Let $ a.b.c$ be ral numbers such that $ 4(a \plus{} b \plus{} c) \minus{} 9 \equal{} 0$. Find max
$ S \equal{} (a \plus{} \sqrt {a^2 \plus{} 1})^b(b \plus{} \sqrt {b^2 \plus{} 1})^c(c \plus{} \sqrt {c^2 \plus{} 1})^a$[/quote]
Lock it,mod,please,see here http://www.mathlinks.ro/viewtopic.php?t=168836


This problem is from Vol II of "Introduction from Calculus and Analysis" by R. Courant and F. John. (this book is my calculus textbook)
I think some problems in the book are interesting  (this does not imply that they are difficult)

Find the values of $a,b,c,d$, such that the equation $Af_{xx}+2Bf_{xy}+Cf_{yy} =0$ becomes
(a) $f_{\xi\xi}+f_{\eta\eta} = 0$
(b) $f_{\xi\eta} = 0$
after the transfrom of $\xi = ax+by$ and $\eta = cx+dy$ ($ad-bc\neq 0$)

liyi, is there supposed to be a condition $AC-B^2>0$?

Okay, your are required to discuss yourself.
For (a) we need $AC>B^2$;
For (b) we need $AC<B^2$.


In Zuminglish, all words consist of only the letters $ M$, $ O$, and $ P$. As in english, $ O$ is said to be a vowel and $ M$ and $ P$ are consonants.  A string of $ M$'s, $ O$'s, and $ P$'s is a word in Zuminglish if and only if between any two $ O$'s there appear at least two consonants.  Let $ N$ denote the number of 10-letter Zuminglish words.  Determine the remainder when $ N$ is divided by $ 1000$.

Can there be a word like MMMMMMM or PPPPPPPPP? or does each word need all three letters?

yes all m's and all p's are possible.

O does not have to be used

[hide="solution"]We find the maximum number of O's:

O--O--O--O

4. Now there are five cases:

no O's. Then there are $ 2^{10}\equal{}1024$ possibilities.

one O. There are 10 spaces for the O to go and $ 2^9\equal{}512$ arrangements of the other 9 consonants. $ 5120$

two O's. We have O--O, and we have 6 more consonants. We can make a 1-1 correspondence to putting 6 consonants in three spaces to solving the equation $ a\plus{}b\plus{}c\equal{}6$ for non-negative integers $ a$, $ b$, and $ c$. There are $ \binom{6\plus{}3\minus{}1}{3\minus{}1}\equal{}28$. And there are $ 2^8\equal{}256$ ways to arrange the consonants, $ 256\cdot 28\equal{}1024\cdot 7\equal{}7168$.

three O's. We now have O--O--O, and there are three consonants to put into 4 spaces. a+b+c+d=3, $ \binom{3\plus{}4\minus{}1}{4\minus{}1}\equal{}35$, and there are $ 2^7\equal{}128$ ways to arrange the consonants, $ 128\cdot 35\equal{}4480$

four O's. There is only one configuration, and that is O--O--O--O. There are $ 2^6\equal{}64$ ways to arrange the consonants.

$ 1024\plus{}5120\plus{}7168\plus{}4480\plus{}64\equal{}\boxed{17856}$[/hide]

That's not right, since all AIME answers are from 000-999 :P

He said determine the remainder, so it would be 856.

Hmm

I get 12C0*2^10+10C1*2^9+8C2*2^8+6C3*2^7+4C4*2^6=15936

so $ 936$

Just rephrase the question as how many ways are there so no more than 1 O can appear in every block of three.  Then for the rest its 2^(10-x) where x is the number of Os.

the future is right and 1=2 is wrong, although I haven't tried to find an error in 1=2's proof.  One can also do this recursively:  [hide]Let $ f(n)$ be the number of words that end in $ O$, $ g(n)$ be the number of words that end in $ O*$ for some consonant $ *$ and let $ h(n)$ be the number of words that end in $ **$ for consonants $ *$ (not necessarily the same).  Then $ f(n) \equal{} h(n \minus{} 1)$, $ g(n) \equal{} 2f(n \minus{} 1)$ and $ h(n) \equal{} 2g(n \minus{} 1) \plus{} 2h(n \minus{} 1)$.  We also have initial conditions $ f(0) \equal{} g(0) \equal{} 0, h(0) \equal{} 1$ (or, if you're uncomfortable with that, $ f(2) \equal{} g(2) \equal{} 2, h(2) \equal{} 4$), and it's relatively simple to calculate $ f(10) \plus{} g(10) \plus{} h(10) \equal{} 15936$.[/hide]

the only little mistake 1=2 made is that he miscalculated $ \binom{3 \plus{} 4 \minus{} 1}{4 \minus{} 1}$ to be $ 35$ .. but it is $ \binom{3 \plus{} 4 \minus{} 1}{4 \minus{} 1} \equal{} \binom{6}{3} \equal{} 20$

the rest is ok and his solution seems to be quite (or even exactly) the same as the one from the-future.

wayne


I heard somewhere that for any polynomial...

$ m \minus{} n \mid P(m) \minus{} P(n)$

the proof relies on the fact that $ m^q \minus{} n^q$ is divisible by $ m \minus{} n$ and so is pretty simple, right?

[b]Are there any restrictions for these values m, n, P(m), P(n)?[/b]

Also:

Let let P(x) be a polynomial with integer coefficients such that $ P(2) \equal{} 3$ and $ P(7) \equal{} \minus{}5$

 But by the property above, $ 7\minus{}2 \equal{} 5 \mid \minus{}5 \minus{} 3 \equal{} \minus{}8 \Longrightarrow 5 \mid \minus{}8$, which is obviously NOT true!

I thank you in advance.

[b]How can this be?[/b]

[b]Does it also mean something about how there are no integer roots or something like that and how does it follow from $ 5 \nmid \minus{}8$?[/b]

There are two ways to interpret the identity: it holds as a polyomial identity in $ m$ and $ n$, or it holds with $ m, n$ integers and $ P(x)$ having integer coefficients.

What exactly is your second question?  Which part of the proof confuses you?

[quote="theax"]Let let P(x) be a polynomial with integer coefficients such that $ P(2) \equal{} 3$ and $ P(7) \equal{} \minus{} 5$

 But by the property above, $ 7 \minus{} 2 \equal{} 5 \mid \minus{} 5 \minus{} 3 \equal{} \minus{} 8 \Longrightarrow 5 \mid \minus{} 8$, which is obviously NOT true![/quote]If you (or someone) can find a polynomial with INTEGER coefficients to fit $ P(2)\equal{}3,P(7)\equal{}\minus{}5$, I'll buy you an ice cream  :icecream:

Also:

Let let P(x) be a polynomial with integer coefficients such that P(2) = 3 and P(7) = -5

But by the property above, 7-2 = 5 \mid -5 - 3 = -8 \Longrightarrow 5 \mid -8, which is obviously NOT true!

I thank you in advance.

How can this be?

Does it also mean something about how there are no integer roots or something like that and how does it follow from 5 \nmid -8?

Question 1: Well, it's confusing that the property is quite easily proven, but yet it doesn't make sense in the question above (taken from intermediate algebra by aops).

Question 2: does the fact that it doesn't make sense with the property have anything to do with the fact that there are no integer roots and how does it follow from the fact that it doesn't make sense with the property? (I know it has something to do with it, but I just can't remember how it follows...)

You're using a lot of pronouns.  Please be more specific about what's confusing you.  What, exactly, doesn't make sense?

If it helps, look at what goes wrong when you try to find a [b]linear[/b] polynomial satisfying the problem constraints.

Here is a NON-INTEGER linear polynomial fitting your information:

[hide]
P(x) = ax + b
2a + b = 3
7a + b = -5

a = -8/5
b = 31/5

$ P(x) \equal{} \frac { \minus{} 8}{5}x \plus{} \frac {31}{5}$[/hide]

where
P(2) = 3
P(7) = -5

First off - I would like to thank all the people who so promptly replied!
 :D

[i]Also:

Let let P(x) be a polynomial with integer coefficients such that $ P(2) \equal{} 3$ and $ P(7) \equal{} \minus{}5$

But by the property above, $ 7\minus{}2 \equal{} 5 \mid \minus{}5 \minus{} 3 \equal{} \minus{}8 \Longrightarrow 5 \mid \minus{}8$, which is obviously NOT true![/i]

But then, the property was proved for polynomials in general! A contradiction!

[b]Question 1: Can someone explain the contradiction (because I'm probably overlooking something here)?[/b]

[b]Question 2:]Does the fact that there is a contradiction imply that there are no integer roots (or something like that) and how does it follow from $ 5 \nmid \minus{}8$ - I remember some property like that I learned before but now cannot re-derive it, or maybe my memory is faulty (if that is the case, can you provide a counterexample so that I can clearly understand it)?[/b]

I greatly appreciate any help or insight into this matter!

[quote="theax"][b]Question 1: Can someone explain the contradiction (because I'm probably overlooking something here)?[/b][/quote]
Any polynomial with integer coefficients $ P(x)$ has the property that $ 7 \minus{} 2 | P(7) \minus{} P(2)$.  Suppose $ P(x)$ is an integer polynomial such that $ P(7) \equal{} \minus{} 5, P(2) \equal{} 3$.  Then $ 7 \minus{} 2 | \minus{} 5 \minus{} 3$; contradiction.  Hence no such polynomial exists.
[quote="theax"][b]Question 2: Does the fact that there is a contradiction imply that there are no integer roots (or something like that) and how does it follow from $ 5 \nmid \minus{} 8$ - I remember some property like that I learned before but now cannot re-derive it, or maybe my memory is faulty (if that is the case, can you provide a counterexample so that I can clearly understand it)?[/b][/quote]
No.  What it means is that for any fixed $ a$ the polynomial $ P(x) \minus{} P(a)$ is always divisible by $ x \minus{} a$; this is another way to state the factor theorem.

This is perhaps clearest when $ a \equal{} 0$; then $ P(x) \minus{} P(0)$ is always divisible by $ x$, and that's essentially what makes this problem work.  If we translate the points $ 2$ and $ 7$ to $ 0$ and $ 5$, then it's obvious what the problem is: $ P(5) \minus{} P(0)$ needs to be divisible by $ 5$, and we're given that it's not.

[quote="theax"]THANKS!
 :D

So just to confirm:
So the fact that this is a "contradiction" does NOT imply that there are no integer roots???[/quote]There is major confusion as to what "contradiction" you are referring to because:

Contradiction #1: The contradiction you are talking about is the claim that $ m\minus{}n\nmid P(m)\minus{}P(n)$.

Contradiction #2: The contradiction that t0rajir0u is talking about is part of a proof and saying that no such polynomial exists.

Here is what you need to understand. Contradiction #1 is not a valid counterexample because contradiction #2 (t0rajir0u's argument) proves that there is no such polynomial given by your example.

Are there integer roots? There is no basis to support that. However, what the second part of t0rajir0u's argument is that the statement $ m\minus{}n|P(m)\minus{}P(n)$ is related to the Factor Theorem.

Try to get your facts straight and you're good to go.

IMPORTANT PS: We hope you know that the condition that the polynomial must have INTEGER coefficients is relevant, which t0rajir0u argues why your "contradiction" does not work.

:blush:

thanks!

notice that I used quotes on the contradiction (which implied that it was invalid) but I was merely using the same terms that I had established before.

But thanks for the clarification!  :D

[quote][i]Also:

Let let $ P(x)$ be a polynomial with integer coefficients such that $ P(2) \equal{} 3$ and $ P(7) \equal{} \minus{}5$

But by the property above, $ 7\minus{}2 \equal{} 5 \mid \minus{}5 \minus{} 3 \equal{} \minus{}8 \Longrightarrow 5 \mid \minus{}8$, which is obviously NOT true!

But then, the property was proved for polynomials in general! A contradiction![/i]

Question 1: Can someone explain the contradiction (because I'm probably overlooking something here)?

Question 2:]Does the fact that there is a contradiction imply that there are no integer roots (or something like that) and how does it follow from 5 \nmid -8 - I remember some property like that I learned before but now cannot re-derive it, or maybe my memory is faulty (if that is the case, can you provide a counterexample so that I can clearly understand it)?

I greatly appreciate any help or insight into this matter![/quote]

[b]So the answer to number 2 is NO, right?[/b]

Thanks

[quote="theax"]:blush:

thanks!

...

[b]So the answer to number 2 is NO, right?[/b]

Thanks[/quote]t0rajir0u and I say NO


(A.Myakishev, 9--10) Given triangle $ ABC$. One of its excircles is tangent to the side $ BC$ at point $ A_1$ and to the extensions of two other sides. Another excircle is tangent to side $ AC$ at point $ B_1$. Segments $ AA_1$ and $ BB_1$ meet at point $ N$. Point $ P$ is chosen on the ray $ AA_1$ so that $ AP\equal{}NA_1$. Prove that $ P$ lies on the incircle.

Denote  AA1∩⊙I = Q , where ⊙I is the incircle of △ABC
O1,O2,O3 are midpoint of BC,CA,AB respectively   I’ is the incenter of △O1O2O3
AA1∩O2O3=M
It is well-known N is called Nagel point  so we have I’I=I’N   I’M⊥O2O3
IQ/IaA1=r/Ra=AI/AIa   =〉IQ∥IaA1   =〉 IQ⊥O2O3  =〉IQ∥I’M  =〉
QM=MN  =〉 QA=NA1   =〉 P=Q
Done!

You can see here http://www.mathlinks.ro/viewtopic.php?t=337716

Sorry to revive.But I have another solution :)
Let the incircle be tangent to $BC,CA$ at $D E$.
Draw the diameter $DD'$ of $(I)$.
Let $O_a$ be $A$-excircle of triangle $ABC$,this circle is tangent to the extension of $AC$ at $A_2$.
The homothety center $A$ sends $(I)\rightarrow (O_a)$ will send $D'\rightarrow A_1, E\rightarrow A_2$.
This means $A,D',A_1$ are collinear and $\frac{D'A}{D'A_1}=\frac{AE}{EA_2}$
Note that $EA_2=EC+CA_2=EC+CA_1=CD+CA_1=CD+BD=BC$.
Applyinh Menelaus theorem for triangle $AA_1C$ with line $BNB_1$:
$\frac{NS}{NA}.\frac{B_1A}{B_1C}.\frac{BC}{BA_1}=1$
Note that $B_1A=EC=DC=BA_1$.
Hence:
$\frac{NA_1}{NA}=\frac{B_1C}{BC}=\frac{AE}{BC}=\frac{D'A}{D'A_1}$
But $NA_1+NA=AA_1=D'A+D'A_1$ so it implies that $NA_1=AD'$, or we conclude that $D'$ is exactly $P$.
Done!

Dear Mathlinkers,
for a nice synthetic proof...

http://jl.ayme.pagesperso-orange.fr/Docs/26.%200.%20Segments.pdf   p. 6-7.

Sincerely
Jean-Louis

[quote=jayme]Dear Mathlinkers,
for a nice synthetic proof...

[url=http://jl.ayme.pagesperso-orange.fr/Docs/26.%200.%20Segments.pdf] see here [/url]   p. 6-7.

Sincerely
Jean-Louis[/quote]


For any non-negative integer n,
f(n) = 1, if n = 0,
f(n) = (f(n/2))^2, if n>0 and n is even
f(n) = 2f(n-1), if n >0 and n is odd.
The "recursion depth" of f(n) is the number of other integers m such that the value of f(m) is calculated whilst computing f(n). [The recursion depth of f(4) is 3.]
For any non-negative integer n,
g(n) = 0, if n = 0,
g(n) =1+ g(n/2), if n>0 and n is even,
g(n) =1+ g(n-1), if n>0 and n is odd.
(i) What is g(2^k) where k is a non-negative integer?
(ii) What is g(2^l + 2^k) when l>k>=0 are integers?
(iii) Why is the value of g(n) = the recursion depth of f(n)?

(i)-
 $ g(2^k) \equal{} 1 \plus{} g(2^{k \minus{} 1}) \equal{} 2 \plus{} g(2^{k \minus{} 2}) \equal{} ... \equal{} k \plus{} g(1) \equal{} k \plus{} 1$
(ii)-
$ g(2^k \plus{} 2^l) \equal{} g(2^k(2^{l \minus{} k} \plus{} 1)) \equal{} k \plus{} g(2^{l \minus{} k} \plus{} 1) \equal{} k \plus{} 1 \plus{} g(2^{l \minus{} k}) \equal{} k \plus{} 1 \plus{} l \minus{} k \plus{} 1 \equal{} l \plus{} 2$

Extension: Find a simple way to describe $ f(n)$ for all $ n$.

EDIT: Oops, I meant $ g(n)$.

Yes, obviously 2^n. But what about (iii)? Is there a connection?

Oh, I meant to write $ g(n)$ above.

For (iii), there is a connection: if it takes $ n$ steps to find $ f(n/2)$ where $ n$ is even, then we can find $ f(n)$ in just one more step, so $ g(n)\equal{}g(n/2)\plus{}1$ if $ n$ is even.  The logic for $ n$ odd is exactly the same.

The fact that this recursion for $ g(n)$ gives the actual "recursion depth" of $ n$, i.e. the minimal number of steps to find $ n$ is less obvious - it helps to have an explicit form for $ g(n)$ if you want to prove this.


[quote="chien than"]Let  $a,b,c\geq 0,a+b+c=1$
Find min $A=ab+2bc+3ca$[/quote]
$\min A=0$ and holds when $a=1,$ $b=c=0.$


geometry
Euler
3D geometry
prism
geometry solved

Can you prove the incircle theorem of Darij's avatar.

Bomb

I didn't get your meaning clearly

I  think you meaned:
we have two circle with centers $O\ ,\ I$ if we have $OI^2=R^2-2Rr$ ($R,r$ are radius of bigger and smaller circles in order) prove that for any point on bigger circle we can construct a triangle that another circle be it's incircle.

if you meaned this this not hard at all.

to prove it use Euler theorem

[quote="bomb"]Can you prove the incircle theorem of Darij's avatar.[/quote]

In fact, the theorem is called Poncelet's porism for triangles (I am saying this since I may change my avatar sometime in future...).

We had some topics about it:

http://www.mathlinks.ro/Forum/viewtopic.php?t=42330
http://www.mathlinks.ro/Forum/viewtopic.php?t=31633
http://www.mathlinks.ro/Forum/viewtopic.php?t=41614

  Darij


Gauss
geometry
geometric transformation
search
calculus

You may have to speak a foreign language to read some of it.  It is possible that someone has just typed some of that up and put it online, so you might try searching for that.  Giving Amazon a shot is always worthwhile.  Otherwise, your only good chance I think is specialized publishers, and I don't have any idea how you should best go about that.  I warn you that it is almost certainly going to be difficult going, even if you can find them in English.  Hope this was a little helpful.

[quote="ComplexZeta"]Disquisitiones Arithmeticae by Gauss is for sale on Amazon for about $40.[/quote]

Yes, it's an [url=http://www.amazon.com/exec/obidos/ASIN/0300094736/learninfreed]English translation[/url] by Arthur
A. Clarke.

There are a few English translations of part of Euler's vast output, but it's still worthwhile to learn Latin to follow Abel's advice to learn from the masters.

[quote="walrus"]I'd really like to read some classic math texts written by the masters (as per Abel's advice) e.g. those written by Gauss, Galois, Legendre, etc, but I don't know where I could find them or even what exact titles to look for.  Any help anyone can give me would be greatly appreciated.[/quote]

Round Table is a good forum for this question, but I would personally expect this kind of question in the Other Problem-Solving Topics forum. A good first place to look for info on your quest for books by the masters would be the book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0321016181/learninfreed]A History of Mathematics: An Introduction[/url] by Victor J. Katz, which would tell you who the masters are, and probably provide citations to the English translations available for their works. Definitely a lot of the good stuff is still in the original Latin (or French, after a certain period). I also like John Stillwell's book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0387953361/learninfreed]Mathematics and Its History[/url] 2nd edition, which has some interesting citations to works by the masters, and is just plain a good book to read about mathematics.

Hope this helps!

There's almost certainly German and Russian stuff available as well.  I'm sure things were written and translated among those four languages -- Latin, French, German, and Russian -- but I'm not sure how much of it would ever have been translated to English.

Reading Newton might be fun.

[quote]Reading Newton might be fun.[/quote]

Yeah, you have to read [i]Principia[/i].

Many of the "classic" texts of mathematics have been translated into several languages. In particular, I know that Dover has published several of them in English translation - including Euclid's [u]Elements[/u] and Descartes' [u]Geometry[/u]. They're not, however, necessarily easy to read. The notation is sometimes bad, and the names of things are different than what we call them today. My advice is to read a good textbook on a subject, and supplement that with the classic texts later on. It's also good to just read History of Mathematics books like E.T. Bell's [u]Men of Mathematics[/u] and Katz's [u]A History of Mathematics: An Introduction[/u].

Some translations can also be found online. An online version of Euclid's [u]Elements[/u] can be found here: http://www.math.princeton.edu/~cojocaru/art-research-rm.html

And a lot of translations can also be found in bounded collections of author's works, or as parts of other books. References are usually found in books on the History of Mathematics, or you can use these sites to search for them:

http://www.amazon.com/
http://www.ericweisstein.com/encyclopedias/books/MathematicsClassics.html
http://www.ericweisstein.com/encyclopedias/books/GeneralMathematicsHistory.html
http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html

I don't understand why a person would want to read one of these classic books for educational purposes (although the reading may be enjoyable). Why would reading a book by Newton on calculus be more informative or enlightening than reading a good modern calculus textbook?

It becomes more important when you become a graduate student or a researcher. But for students of you guys ages I don't think it is a good way to "learn" a subject.

[quote="gauss202"]
Some translations can also be found online. An online version of Euclid's [u]Elements[/u] can be found here: http://www.math.princeton.edu/~cojocaru/art-research-rm.html
[/quote]
Did you perhaps copy the wrong link? That isn't Euclid's Elements. It's "the Art of Research"

Sorry, I can't resist.

[quote="gauss202"]And a lot of translations can also be found in bounded collections of author's works.[/quote]

'bounded collection' -- would that be a collection which is contained entirely in a ball of some radius R, centered at a point x?

[quote="confuted"]
Did you perhaps copy the wrong link? That isn't Euclid's Elements. It's "the Art of Research"[/quote]

Oops, yeah. I meant: http://babbage.clarku.edu/~djoyce/java/elements/Euclid.html


inequalities
combinatorics unsolved
combinatorics

Let $n$ be an odd positive integer and let $x_1, \ldots, x_n, y_1, \ldots, y_n$ be nonnegative real numbers satisfying $x_1 + \cdots + x_n = y_1 + \cdots + y_n$.  Show that there exists a proper, nonempty subset $J \subseteq \{1,\ldots,n\}$ such that \[\frac{n-1}{n+1}\sum_{j\in J}x_j \leq \sum_{j\in J} y_j \leq \frac{n+1}{n-1}\sum_{j\in J} x_j.\]

wlog we assume $n>1$ and that $\sum_i x_i=\sum_i y_i=1$.if the result were not true then for each $i$, $y_i<\frac{n-1}{n+1}x_i$ or $x_i<\frac{n-1}{n+1}y_i$.now if $y_i<\frac{n-1}{n+1}x_i$ then it must follow that for the complement set with sum $1-x_i,1-y_i$ the otherway inequality must hold(since the sums are the same),i.e.
 $y_i<\frac{n-1}{n+1}x_i\Rightarrow 1-y_i>\frac{n+1}{n-1}(1-x_i)$.since $n$ is odd one of these inequalities occurs atleast $\frac{n+1}{2}$ times,i.e. if $I_0=\{i|y_i<\frac{n-1}{n+1}x_i\},I_1=\{i|x_i<\frac{n-1}{n+1}y_i\}$ then either $|I_0|\geq\frac{n+1}{2}$ or $|I_1|\geq \frac{n+1}{2}$. say $|I_0|\ge \frac{n+1}{2}$. then since we have
 $y_i<\frac{n-1}{n+1}x_i$ and $1-x_i<\frac{n-1}{n+1}(1-y_i)$ it follows that(solve for the intersection of the lines)
$x_i>\frac{2}{n+1}$.but then $\sum_{i\in I_0} x_i > 1$ and that is a contradiction.


algebra
polynomial
algebra unsolved
function

Suppose that $a,b,c,d\in\mathbb{R}$ and $f(x)=ax^3+bx^2+cx+d$ such that $f(2)+f(5)<7<f(3)+f(4)$. Prove that there exists $u,v\in\mathbb{R}$ such that $u+v=7 , f(u)+f(v)=7$

Surely this is trivial?  $f(x)$ is a real polynomial, so $g(x)=f(x)+f(7-x)$ is a real polynomial, and in particular continuous, so by the intermediate value theorem (since $g(5)>7>g(4)$) there exists $u\in(4,5)$ such that $g(u)=7$.

Oh ,now I find a new solution:

I think this solution has a little different to Diarmuid's solution

We must prove the equality of $f(x)+f(7-x)=7$ has a root in real number. Suppose that $g(x)=f(x)=f(7-x)-7$ we know that $g(x)$ is a polynomial which its degree at most $2$ , $g(2)<0$ and $g(3)>0$ so that the equation $g(x)=0$, then it has at least one root.  :)

[color=red]Excuse me for my bad question [/color]


is possible to train calculating in human memory to calculate any problem in 1 second??? 5687,25 * 22 and others numbers? sorry for my bad english.

[quote="jiri.otevrel"]calculate any problem in 1 second???[/quote]

Er, no, not really.  All you have to do is make the problem arbitrarily long; just keep adding digits and eventually the capacity of the problem will be greater than the capacity of the human brain to store data, and then obviously such a calculation is not possible by the human brain.

check out vedic arithmetic

vedic math has lots of different methods of doing basic arithmetic (and even some higher level stuff) that are conducive to faster, mental calculations

Vedic methods involve different, multi-step techniques, but they allow a person to say.. multiply 4 digit numbers in their head.

it requires a lot of memorizing though. There's no real point when you have calculators... or even paper.

the only one that I know that I actually use is a squaring algorithm

$a^{2}= (a-b)(a+b)+b^{2}$
so something like 28^2 = (20)(36) + 8^2 = 784

Sure it is! These guys are true human calculators. They can get the answers faster than you can type problems into the calculator.

[youtube]EueFhYZ4HxI[/youtube]

[quote="hunter34"]But that isn't really a human calculator, it is a abacus which is a way to store numbers.[/quote]

did u watch all of the video?

Art Benjamin is amazing at mental calculation. Here's one video of one of his performances.

[url]http://video.google.com/videoplay?docid=6687235905416745311&q=art+benjamin&total=3198&start=0&num=10&so=0&type=search&plindex=0[/url]

[quote="Kyron"][quote="chickendude"]the only one that I know that I actually use is a squaring algorithm

$ a^{2}\equal{} (a\minus{}b)(a\plus{}b)\plus{}b^{2}$
so something like 28^2 = (20)(36) + 8^2 = 784[/quote]
How do you find $ b$?[/quote]
$ b$ is any number that he picks that makes the calculation convenient, because for any number $ b$ the equation will work.


factorial
superior algebra
superior algebra unsolved

Hi,

I'm looking for pointers to a problem I'm trying to slove. How do I show that there is an isomorphism between an R-Algebra and an factorial ring? Is there always a standard approach to these kind of problems?

Thanks.

Mike

Please be more explicit.
What kind of a ring is $R$?
Isomorphism with respect to which structure?


conics
hyperbola
quadratics
analytic geometry
graphing lines
slope
trigonometry

[color=darkblue] [u][b]Please (if it is possibly), an algebraical proof ![/b][/u] [/color]

[quote="Virgil Nicula"]$ \{\begin{array}{c}
a\ge 0\ ,\ b\ge 0\\\\
a^2+b^2=1\end{array}\ \ \implies\ \ 2\le (1+a)(1+b)\le \frac {3+2\sqrt 2}{2}$ .[/quote]

[color=darkblue]Solution 1[/color]

Suppose that $ (1 \plus{} a)(1 \plus{} b)$ takes some value $ k$, there are more than one intersection points between the graphs of $ a^2 \plus{} b^2 \equal{} 1\ (a\geq 0,\ b\geq 0)$ and $ (a \plus{} 1)(b \plus{} 1) \equal{} k.$ Let $ T\left(\frac {1}{\sqrt {2}},\ \frac {1}{\sqrt {2}}\right)$ be the point of tangency. Thus when the hyperbola $ (a \plus{} 1)(b \plus{} 1) \equal{} k$ passes through the point $ (a,\ b) \equal{} (1,\ 0),\ (0,\ 1)$, we have $ k \equal{} 2$, then the hyperbola $ (a \plus{} 1)(b \plus{} 1)$ touches to the circle $ a^2 \plus{} b^2 \equal{} 1\ (a\geq 0,\ b\geq 0)$ at $ T$, we have $ k \equal{} \frac {3 \plus{} 2\sqrt {2}}{2}$. Since $ k$ varies continuously betwee the values (reffer to the attached file ''Graph 1''), yeilding $ 2\leq k\leq \frac {3 \plus{} 2\sqrt {2}}{2}\Longleftrightarrow 2\leq (1 \plus{} a)(1 \plus{} b)\leq \frac {3 \plus{} 2\sqrt {2}}{2}$.

[color=darkblue]Solution 2[/color]

Let $ a \plus{} b \equal{} s,\ ab \equal{} t$, $ a,\ b$ are the non negative roots of the quadratic equation:$ x^2 \minus{} sx \plus{} t \equal{} 0$.
$ \therefore a\geq 0$ and $ b\geq 0\Longleftrightarrow a \plus{} b\geq 0$ and $ ab\geq0$ and the discriminant $ D\geq 0$.
$ \Longleftrightarrow s\geq 0$ and $ t\geq 0$ and $ s^2 \minus{} 4t\geq 0\ \cdots [1]$.
$ a^2 \plus{} b^2 \equal{} 1\Longleftrightarrow s^2 \minus{} 2t \equal{} 1\Longleftrightarrow t \equal{} \frac {1}{2}s^2 \minus{} \frac {1}{2}\ \cdots [2]$ and
$ k \equal{} (1 \plus{} a)(1 \plus{} b) \equal{} 1 \plus{} s \plus{} t\Longleftrightarrow t \equal{} \minus{} s \plus{} k \minus{} 1\ \cdots [3]$. Thus our problem can be restated that : On $ s \minus{} t$ plane the region of the common part of $ [1]$ and $ [2]$, $ s\geq 0,\ t\geq 0$ have intersects with the family of the line $ [3]$ with the constant slope $ \minus{} 1$. As attached file ''Graph 2'' shows the red line, when the line passes through the point $ A(1,\ 0)$, we have $ k \equal{} 2$ and passes through the point $ B\left(\sqrt {2},\ \frac {1}{2}\right)$, we have $ k \equal{} \frac {3 \plus{} 2\sqrt {2}}{2}$. Since $ k$ varies continuously betwee the values (reffer to the attached file ''Graph 2''), yeilding $ 2\leq k\leq \frac {3 \plus{} 2\sqrt {2}}{2}\Longleftrightarrow 2\leq (1 \plus{} a)(1 \plus{} b)\leq \frac {3 \plus{} 2\sqrt {2}}{2}$.

[color=darkblue]Solution 3[/color]

Similar clue to the problem as Solution 1, 2, we can set the intersection points of the graphs of $ a^2 \plus{} b^2 \equal{} 1\ (a\geq 0,\ b\geq 0)$ and $ (a \plus{} 1)(b \plus{} 1) \equal{} k$ as $ a \equal{} \cos \theta ,\ b \equal{} \sin \theta\ \left(0\leq \theta \leq \frac {\pi}{2}\right)$. Then $ (1 \plus{} a)(1 \plus{} b) \equal{} (1 \plus{} \cos \theta)(1 \plus{} \sin \theta) \equal{} 1 \plus{} \sin \theta \plus{} \cos \theta \plus{} \sin \theta \cos \theta$

$ \equal{} 1 \plus{} t \plus{} \frac {t^2 \minus{} 1}{2} \equal{} \frac {1}{2}(t \plus{} 1)^2 \equal{} f(t)$, where $ t \equal{} \sin \theta \plus{} \cos \theta$. Since $ 1\leq t\leq \sqrt {2}$, yielding $ f(1)\leq f(t)\leq f(\sqrt {2})\Longleftrightarrow 2\leq (1 \plus{} a)(1 \plus{} b)\leq \frac {3 \plus{} 2\sqrt {2}}{2}$.

[color=darkblue]Solution 4[/color]

$ (a \plus{} b)^2 \plus{} (a \minus{} b)^2 \equal{} 2(a^2 \plus{} b^2) \equal{} 2\Longrightarrow |a \plus{} b|\leq \sqrt {2}$. Since $ a\geq 0,\ b\geq 0$ we have $ 0\leq a \plus{} b\leq \sqrt {2}$.

Thus use A.M.-G.M. we have $ (1 \plus{} a)(1 \plus{} b)\leq \left\{\frac {(1 \plus{} a) \plus{} (1 \plus{} b)}{2}\right\}^2 \equal{} \left(\frac {a \plus{} b \plus{} 2}{2}\right)^2\leq \left(\frac {\sqrt {2} \plus{} 2}{2}\right)^2$

$ \equal{} \frac {3 \plus{} 2\sqrt {2}}{2}$. The equality holds when $ 1 \plus{} a \equal{} 1 \plus{} b$ and $ a^2 \plus{} b^2 \equal{} 1\ (a\geq 0,\ b\geq 0)\Longleftrightarrow (a,\ b) \equal{} \left(\frac {1}{\sqrt {2}},\ \frac {1}{\sqrt {2}}\right)$.

[color=red]Solution 5[/color]

$ a^2 \plus{} b^2 \equal{} 1\ (a\geq 0,\ b\geq0)\Longrightarrow \minus{} 1\leq a \minus{} b\leq 1$ and $ 1\leq a \plus{} b\leq \sqrt {2}$,

Let $ f(a,\ b): \equal{} (1 \plus{} a)(1 \plus{} b) \equal{} \frac {1}{4}\{(a \plus{} b \plus{} 2)^2 \minus{} (a \minus{} b)^2\}$, $ f(a,\ b)$ is maximized when $ a \minus{} b \equal{} 0$ and $ a \plus{} b \equal{} \sqrt {2}\Longleftrightarrow a \equal{} b \equal{} \frac {1}{\sqrt {2}}$, yielding maximum value $ \frac {3 \plus{} 2\sqrt {2}}{2}$, then when $ |a \minus{} b| \equal{} 1$ and $ a \plus{} b \equal{} 1\Longleftrightarrow (a,\ b) \equal{} (1,\ 0),\ (0,1)$, yielding minimum value $ 2$. That is to say $ 2\leq (1 \plus{} a)(1 \plus{} b)\leq \frac {3 \plus{} 2\sqrt {2}}{2}$.

[hide="Algebraic solution for second half"]
By trivial inequality, $ a^2\plus{}b^2\ge 2ab$ or $ \frac{1}{2}\ge ab$. Note now that $ (1\plus{}a\plus{}b)^2\equal{}2(a\plus{}1)(b\plus{}1)$. By AM-GM, $ a\plus{}b\plus{}2\ge \sqrt{2(2)(a\plus{}1)(b\plus{}1)}\equal{}\sqrt{2}(1\plus{}a\plus{}b)$. Solving for $ a\plus{}b$, we get $ a\plus{}b\le\sqrt{2}$. Combining, $ ab\plus{}a\plus{}b\plus{}1\equal{}(a\plus{}1)(b\plus{}1)\le\frac{3\plus{}2\sqrt{2}}{2}$.[/hide]

Left inequalty: $ (1\plus{}a)(1\plus{}b)\equal{}1\plus{}a\plus{}b\plus{}ab\geq 1\plus{}a\plus{}b$. As $ 1\equal{}a^2\plus{}b^2\geq a^2$, $ a\in [0,1]$, similarly $ b\in [0,1]$ giving $ a\plus{}b\geq a^2\plus{}b^2\equal{}1\Rightarrow (1\plus{}a)(1\plus{}b)\geq 2$.

Right inequality: $ (1\plus{}a)(1\plus{}b)\leq \frac{(1\plus{}a)^2\plus{}(1\plus{}b)^2}{2}$. By Cauchy, $ (1\plus{}a)^2\leq (\frac{\sqrt{2}}{2}\plus{}a^2)(\sqrt{2}\plus{}1)$ and a similar inequality for $ b$, giving
\[ \frac{(1\plus{}a)^2\plus{}(1\plus{}b)^2}{2}\leq \frac{\sqrt{2}\plus{}1}{2}(\sqrt{2}\plus{}a^2\plus{}b^2)\equal{}\frac{(\sqrt{2}\plus{}1)^2}{2}\equal{}\frac{3\plus{}2\sqrt{2}}{2}\]

[quote="Virgil Nicula"]$ \{\begin{array}{c} a\ge 0\ ,\ b\ge 0 \\
 \\
a^2 + b^2 = 1\end{array}\ \ \implies\ \ 2\le (1 + a)(1 + b)\le \frac {3 + 2\sqrt 2}{2}$ .[/quote]
[color=darkblue][b][u]An algebraical proof.[/u][/b] $ \{\begin{array}{c} 0\le 2ab\implies 1\le (a + b)^2\implies 1\le a + b \\
 \\
(a + b)^2\le 2(a^2 + b^2) = 2\implies a + b\le\sqrt 2\end{array}\|$ $ \implies$ $ (a + b)\in [1,\sqrt 2]$ .

Observe that $ 2(1 + a)(1 + b) =$ $ 1 + 2(a + b) + (a + b)^2 = [1 + (a + b)]^2$ . In conclusion,

$ 2(1 + a)(1 + b)\in [\ (1 + 1)^2\ ,\ (1 + \sqrt 2)^2\ ]$ , i.e. $ 2\le(1 + a)(1 + b)\le\frac {3 + \sqrt 2}{2}$ .

Similarly prove easily the following bilateral conditioned inequality with three variables :[/color]

[quote="Virgil Nicula"]$ \{\begin{array}{c} a\ge 0\ ,\ b\ge 0 \ ,\ c\ge 0\\
 \\
a^2 + b^2 +c^2= 1\end{array}\ \ \implies\ \ 2\le (1 + a)(1 + b)(1+c)\le \frac {18 + 10\sqrt 3}{9}$ .[/quote]

[quote="Virgil Nicula"]$ \{\begin{array}{c} a\ge 0\ ,\ b\ge 0 \ ,\ c\ge 0 \\
 \\
a^2 + b^2 + c^2 = 1\end{array}\ \ \implies\ \ 2\le (1 + a)(1 + b)(1 + c)\le \frac {18 + 10\sqrt 3}{9}$ .[/quote]

Let $ a + b + c = k$, we have $ ab + bc + ca = \frac {1}{2}\{(a + b + c)^2 - (a^2 + b^2 + c^2)\} = \frac {k^2 - 1}{2}$.

Thus $ a,\ b,\ c$ are the roots of the cubic equation : $ (x - a)(x - b)(x - c) = \boxed{x^3 - kx^2 + \frac {k^2 - 1}{2}x - abc = 0}$.

Plugging $ x = - 1$ into the equation gives $ (1 + a)(1 + b)(1 + c) = \frac {1}{2}(k + 1)^2 + abc$.

Since $ a\geq 0,\ b\geq 0,\ c\geq 0\Longrightarrow ab + bc + ca\geq 0\Longleftrightarrow\frac {k^2 - 1}{2}\geq 0\ (k\geq 0)$ and

$ 3(a^2 + b^2 + c^2) \geq (a + b + c)^2\Longleftrightarrow (a - b)^2 + (b - c)^2 + (c - a)^2\geq 0\Longleftrightarrow 0\leq k\leq \sqrt {3}$.

Thus the range of $ k$ is $ 1\leq k\leq \sqrt {3}$. Furthermore we have $ 0\leq abc\leq (\frac {a + b + c}{3})^3\Longleftrightarrow 0\leq abc\leq \frac {1}{27}k^3$.

$ \therefore \frac {1}{2}(1 + 1)^2 + 0\leq (1 + a)(1 + b)(1 + c)\leq \frac {1}{2}(\sqrt {3} + 1)^2 + \frac {1}{27}\cdot (\sqrt {3})^3$, yielding $ 2\leq (1 + a)(1 + b)(1 + c)\leq \frac {18 + 10\sqrt {3}}{9}$.


function algebra domain Asymptote logarithms

Determine the domain, range and vertical asymptote of each log function.

(1) g(x) = -3logx

(2) G(x) = log(5x)

Note that in $ \log x$,
x must be greater than zero (there is no log of negative numbers or zero)

If this is that hard for you that it must be posted on here, please explain what part of the problem you need help with. Otherwise, stop trying to get other people to do your homework.


factorial
function
combinatorics proposed
combinatorics

We have a set 1, 2, 3, ..., n. Consider all the r-element subsets of
this set. Prove that if we take the smallest element of each of these
subsets, the average of these smallest elements will be (n+1)/(r+1).

Maybe we could make this a combinatorics problem:

$C^{n-1}_{r-1} \times 1 + C^{n-2}_{r-1} \times 2 +...$

Not sure if this'll help.

I've managed to get an expression that seems to be the
same as (n+1)/(r+1), but I don't know how to prove that it's the same,
since my expression uses factorials (from the Choice function) and
sigma. This expression that I've obtained is the following sum:

Sum of k*Choose(n-k, r-1) for k = 1 to n

divided by Choose(n, r). I don't know how to show that my expression
is equal to (n+1)/(r+1) which, through my many test cases, it seems to
be.

Hm, this isn't that easy, this was a problem at IMO 1981...

I'll move it to Olympiad Section.

There's a very elegant proof on the Kalva website along the following lines:
Let $F(n,r)$ be the desired average.  Then ${n\choose r}F(n,r)=\sum_{i=1}^{n-r+1}i{n-i\choose r-1}$ as mann already had.

But now consider the set of all $r+1$ element subsets of $1,2,\cdots,n+1$.  There are $i{n-i\choose r-1}$ such subsets with $i+1$ as the second smallest element, so $\sum_{i=1}^{n-r+1}i{n-i\choose r-1}$ must be the total number of such subsets, which is simply ${n+1\choose r+1}$.

So ${n\choose r}F(n,r)={n+1\choose r+1}$, and the result $F(n,r)=\frac{n+1}{r+1}$ follows.

This approach also makes is (relatively) simple to tackle a more general problem:
Let $F(n,r,m)$ be the average value of the $m^{th}$ smallest element, taken over all $r$ element subsets of $1,2,\cdots,n$. Prove that $F(n,r,m)=m\frac{n+1}{r+1}$.


find all functions from integers to integers such that
  f(x+f(y))=f(x)- f(y) for all integer values of x,y

 [hide] i think anwer is f(x)=f(0)-x or a null function, but it it's rather hard to prove that   i've found that if it's not f(0)-x then there is an integer d such that for all x f(x) is divisible by d (d is not 1), otherwise if there is at least one pair of integes x,y such  that gcd(f(x),f(y))=1 then it's solved[/hide]

Let R be the range of f(x). Also Let x be in R.
then f(x) = f(0 + f(y)) = f(0) - f(y)= a -x where a=f(0)
Note that this implies that f(a)=0.
Also if x belongs to R: f(a+x)= -x. so that if x in R so is -x.
If y belongs to R and x be arbitrary then f(x+y)=f(x)-y.
we also know that if x1 and x2 belong to the set, so does x1 -x2 and x2 - x1.
Hence if the set R be {...,y2,y1,0,x1,x2,...} where 0<x1<x2<... and yi<0.
we have that x2-x1=x1 and x3-x1=x2, x4-x1=x3,... so that R= {yi,...,0,x1,2x1,3x1,...,nx1,...}
in fact one can show that yi is the set -x1,-2x1,...
so that R= {....,-mx1,...,-x1,0,x1,2x1,...}
Now for any x, we have x=tx1 + r for some r,t, where r<x1
So f(x)=f(r) - tx1.
so all we have to do is the following:
Define g(x): A -> B arbitrarily, where A={1,2,3,...,x1-1} and B={...,-2x1,-x1,0,x1,2x1,...}
Then f(x)=g(x(mod x1)) - (x-x1) for all x.
So f(x) is not nesssacarily a-x for all x. The above construction gives all the solutions.
-Ali

thanks for your solution   that was a really good function
  there's another one here, i 've made it yesterday
it's simpler than this one(i've made in 10 mins)  but i liked it
  f(x^2-y^2)=xf(x)-yf(y)     f is from reals to reals

Well, the solution took much less than  10 minutes :lol:

Note that f(x+y)=f(x)+f(y) and f(x^2)=xf(x)
so f((x+y)^2) = (x+y)f(x+y) = xf(x)+yf(y)+ yf(x)+xf(y)
so xf(x)+y(fy)+2f(xy) = xf(x)+yf(y)+yf(x)+xf(y)
2f(xy)=yf(x)+xf(y)
now let y be rational:
so 2yf(x)=yf(x)+xf(y)
so yf(x)=xf(y)=xyf(1)
so f(x)=xf(1) so all x.
-Ali

Probably, I'm asking something stupid  :oops:
But how did you got f(x+y)=f(x)+f(y) ?

put y=0, to get f(x^2)=xf(x)
also from here it is easy to see that f(-x)=-f(x)
so f(x^2-y^2)=f(x^2) - f(y^2)
so f(a)+f(b)=f(a+b) for all b>=0. and a>=-b
but even if b<0 f(a) +f(b) = f(a) - f(-b)=f(a-(-b))=f(a+b)
so no matter what , f(a+b)=f(a)+f(b)...
-Ali

Ah ! I just didn't relate the fact that f(x^2)=x*f(x) to f(x^2-y^2)=x*f(x)-y*f(y)=f(x^2)-f(y^2)
Thanks !

i've did it in the same way :) by hte way 10 minutes was a symbolic time unit :) it's too easy for thinking about it 10 mins :)

try  this one if u want, but this one is really hard
it;s from the IMO 2004 shortlist
  f is from reals to reals
f(x^2+y^2+2f(xy))=f(x+y)^2

This was discussed in much detail here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=33734

and this one(quite easy
f is from reals to reals
f(x)^f(y)=f(x^y)

[hide] show that f is multiplicative, additive and increasing[/hide]


[quote="Thorn"]I would like to compile a list of problems (with solutions) using the number 2009.  For example:

1. How many integer solutions, (x,y), does the equation $ x^2 \minus{} y^2 \equal{} 2009$ have?

2. The product of two numbers is 2009 and one number is 8 less than the other.  Compute the sum of these two numbers.

3. How many digits does the 2009th Fibonacci number have?  (Hint: Use log and Binet's formula)[/quote]

Number 2:

$ x(x \minus{} 8) \equal{} 2009\equal{}x^2 \minus{} 8x \minus{} 2009$

We solve it and find out that $ x \equal{} 49$ or $ x \equal{} \minus{} 41$.

If $ x \equal{} 49$, then the other number is $ 41$, and the sum is $ \boxed{90}$

If $ x \equal{} \minus{} 41$, then the other number is $ \minus{} 49$, and then the sum is $ \boxed{ \minus{} 90}$

The answer is either $ 90$ or $ \minus{} 90$.

[hide="1"]
$ x^2 \minus{} y^2 \equal{} (x \plus{} y)(x \minus{} y) \equal{} 41.(7)^2$
$ (x,y)$ has three intiger solutions[/hide]

[hide="2"]
$ xy \equal{} 2009, x \minus{} y \equal{} 8$
$ x^2 \minus{} 2xy \plus{} y^2 \equal{} 64$
$ x^2 \plus{} 2xy \plus{} y^2 \equal{} 64 \plus{} 4(2009)$
$ x \plus{} y \equal{} \sqrt {64 \plus{} 4(2009)} \equal{} \pm 90$[/hide]

[hide="3"]
For such a large number, and approximation will work
$ F_n \approx \dfrac{\Phi^n}{\sqrt {5}}$
Number of digits: $ 1 \plus{} \log_{10} (F_n)$

$ 1 \plus{} \lfloor 2009\log_{10}(\Phi) \minus{} \frac {1}{2}\log_{10}(5) \rfloor \equal{} 420$
[/hide]

That's a good start!
But problem 1 requires more explanation, and did we say x,y are positive?

Is there anything special or interesting about 2009?  Can we turn that interesting thing into a problem?

There must be, since all the positive integers are interesting:

Clearly 1 is interesting for many reasons (neither prime nor composite,  the multiplicative inverse of itself,...)
2 is the only even prime, which is very interesting.
Now suppose that some integer n>2, n is uninteresting.
That would make n the least uninteresting number, which is [b]very interesting![/b]
Therefore no such n exists, and natural numbers are interesting.  Especially 2009.

[quote="Illos"][hide=" 1."]Since $ 2009 \equal{} 41 \cdot 7^2$, it has 6 divisors and $ x^2 \minus{} y^2 \equal{} 2009$ has 3 solutions in positive integers. Each of these solutions has 4 permutations when one considers all integers so there are a total of 24 ordered pairs $ (x, y)$ which satisfy the equation.[/hide][/quote]

You mean 12, right? :)

[quote="Thorn"]

Clearly 1 is interesting for many reasons (neither prime nor composite,  the multiplicative inverse of itself,...)
2 is the only even prime, which is very interesting.
Now suppose that some integer n>2, n is uninteresting.
That would make n the least uninteresting number, which is [b]very interesting![/b]
Therefore no such n exists, and natural numbers are interesting.  Especially 2009.[/quote]

Linguistic induction. That's new.. haven't heard of that one  :wink:

I like the joke. I think it's been done, but that's a fresh rendition of it :)


Hi guys,
I'd like to better format the following code for a post here.  Any way I can make the \sum appear in \displaystyle inside the \begin{align}\begin{cases} construct?  \displaystyle does not seem to work.

Also, any way I can line up the "for n odd" and "for n even" in both sets?  Too much to ask?

\[ \begin{align*}
a_{n,k}&\equal{}\begin{cases}\displaystyle 2\sum_{h\equal{}1}^{k} (\minus{}1)^h (h\pi)^{n\minus{}1} & \text{for } n \text{ odd}\vspace{10pt} \\
\phantom{llll}0 & \text{for } n \text{ even}
\end{cases}
\\
\\
b_{n,k}&\equal{}\begin{cases}\cfrac{(\minus{}1)^{\frac{3\plus{}n}{2}} 2 (2^{n\minus{}1}\minus{}1) B_{2n}}{(1\plus{}n)!}\plus{}\displaystyle 2\sum_{h\equal{}1}^{k} \cfrac{(\minus{}1)^h}{(h\pi)^{n\plus{}1}} & \text{for } n \text{ odd}\vspace{10pt} \\
\phantom{\minus{}\minus{}\minus{}\minus{}}0 & \text{for } n \text{ even}
\end{cases}
\end{align*}\]

Thanks,
Shaw

Ok, mega-use of \phantom for the alignment of "for n even" and "for n odd", but that's a trial and error thing and if I reformat the sums correctly then I'll have to go back and alter the \phantom code.  No way to do it automatically?

\begin{align*}
    a_{n,k}&=\begin{cases}\displaystyle 2\sum_{h=1}^{k} (-1)^h (h\pi)^{n-1} &\phantom{xxxxkkjjkkxxxxxxx} \text{for } n \text{ odd}\vspace{10pt} \\
    \phantom{llll}0 & \phantom{xxxxkkjjkkxxxxxxx} \text{for } n \text{ even}
    \end{cases}
    \\
    \\
    b_{n,k}&=\begin{cases}\cfrac{(-1)^{\frac{3+n}{2}} 2 (2^{n-1}-1) B_{2n}}{(1+n)!}+\displaystyle 2\sum_{h=1}^{k} \cfrac{(-1)^h}{(h\pi)^{n+1}} & \text{for } n \text{ odd}\vspace{10pt} \\
    \phantom{----}0 & \text{for } n \text{ even}
    \end{cases}
    \end{align*}

Ok, I think I have it -- I'm sure you guys would have figured it out though: I used \mathop\sum\limits for the sums.

\begin{align*} a_{n,k} & = \begin{cases} 2\mathop\sum\limits_{h = 1}^{k} ( - 1)^h (h\pi)^{n - 1} & \phantom{xxxxkkjjkkxxxxxxx} \text{for } n \text{ odd}\vspace{10pt} \\
\phantom{llll}0 & \phantom{xxxxkkjjkkxxxxxxx} \text{for } n \text{ even} \end{cases} \\
 \\
b_{n,k} & = \begin{cases}\cfrac{( - 1)^{\frac {3 + n}{2}} 2 (2^{n - 1} - 1) B_{2n}}{(1 + n)!} +2 \mathop\sum\limits_{h = 1}^{k} \cfrac{( - 1)^h}{(h\pi)^{n + 1}} & \text{for } n \text{ odd}\vspace{10pt} \\
\phantom{ - - - - }0 & \text{for } n \text{ even} \end{cases} \end{align*}

Do not use \mathop and \phantom. Just use \limits

\[ a_{n,k} \& = \begin{cases} 2\sum\limits_{h = 1}^{k} ( - 1)^h (h\pi)^{n - 1} \&  \text{for } n \text{ odd}\vspace{10pt} \\
0 \&  \text{for } n \text{ even} \end{cases}\]

Ok.  But I still seem to need \phantom to line up the first set of "for n odd" and "for n even" with the second set (just a personal preference, I think it makes it a little easier to read and less cluttered):

\begin{align*}
a_{n,k}&=\begin{cases}2\sum\limits_{h=1}^{k} (-1)^h (h\pi)^{n-1} & \phantom{aaaaaaaaaaaaaaaaaa}\text{for } n \text{ odd}\vspace{10pt} \\
0 & \phantom{aaaaaaaaaaaaaaaaaa}\text{for } n \text{ even}
\end{cases}
\\
\\
b_{n,k}&=\begin{cases}\cfrac{(-1)^{\frac{3+n}{2}} 2 (2^{n-1}-1) B_{2n}}{(1+n)!}+2\sum\limits_{h=1}^{k} \cfrac{(-1)^h}{(h\pi)^{n+1}} & \text{for } n \text{ odd}\vspace{10pt} \\
0 & \text{for } n \text{ even}
\end{cases}
\end{align*}

[quote="Valentin Vornicu"]I think think it's easier to read.[/quote]

"think think" as in not easier to read?  I don't know for sure.  Still learning.

\hspace is better than \phantom here:
$ \begin{cases}
a,&\hspace{100pt}b\\
cxx,&\hspace{100pt}d
\end{cases}$
I suspect that stevem can come up with even better solution though :)

Thanks for the vote of confidence fedja :)
The problem here is that the 2 cases environments are not connected so LaTeX can't tell in the first one how wide the second one is going to be.
It's possible to get round this by abandoning the align and cases environments and putting everything in a table to ensure it all lines up. But it would be complicated and would need the multirow and bigdelim packages so wouldn't work on the forum.
It's also possible to set up boxes containing some of the formulae, measure the width of the boxes and use that to find the amount of horizontal space required.
But neither method would be as simple as using \hspace with a bit of trial and error, so I suggest sticking with that.

In a document, instead of \text{for } n \text{ odd}, you can write \text{for $ \$$n$ \$$ odd} but that won't work on the forum. However, you can replace $ \$$...$ \$$ by \(...\) (plus an extra space that has been swallowed up) to get \text{for \(\ n\) odd}.
\begin{align*}
a_{n,k}&=\begin{cases}2\sum\limits_{h=1}^{k} (-1)^h (h\pi)^{n-1} & \hspace{9.5em}\text{for \(\ n\) odd}\\[10pt]
0 & \hspace{9.5em}\text{for \(\ n\) even}
\end{cases}
\\
\\
b_{n,k}&=\begin{cases}\cfrac{(-1)^{\frac{3+n}{2}} 2 (2^{n-1}-1) B_{2n}}{(1+n)!}+2\sum\limits_{h=1}^{k} \cfrac{(-1)^h}{(h\pi)^{n+1}} & \text{for \(\ n\) odd}\\[10pt]
0 & \text{for \(\ n\) even}
\end{cases}
\end{align*}

I bet a package could be written, say "multicases" to do just this.  Here's the pseudo-code:

[code]\begin{multicases}
a_1=\begin{cases} long text here & n odd \\
            0& n even
       \end{cases}
a_2=\begin{cases} 2x& n odd \\
            more long text & n even
       \end{cases}
\end{multicases}[/code]


A long wire is known to have a radius greater than 4.0 mm and to carry a current uniformly distributed over its cross section. If the magnitude of the magnetic field is 0.285 mT at a point 4.0 mm from the axis of the wire and 0.200 mT at a point 10 mm from the axis, what is the radius of the wire?

and

A long solenoid (n = 1200 turns/m, radius = 2.0 cm) has a current of a 0.30 A in its winding. A long wire carrying a current of 20 A is parallel to and 1.0 cm from the axis of the solenoid. What is the magnitude of the resulting magnetic field at a point on the axis of the solenoid?

Thanks for the help

okay its really embarrassing I forgot how to do this?  can anybody help, I think you using the line $ \oint \vec{B} \cdot \vec{dl} \equal{} \mu_0 I_{\text{enclosed}}$

Therefore, your problem is almost done except you have to make calculations. Don’t confuse a(distance from the axis) with radius of the wire r. You are given two values of B dependent on the distance from axis so you will have 2 equations of 2 unknowns. Solve the equation and obtain the result.


Επανειλθα και γω παιδεια υστερα απο διακοπες που ειχα παει σε ενα ομορφο χωριο εξω απο το Γυθειο....Ομολογω παντος πως μου ελειψε ιδιαιτερος το βιβλιο Θ.Α του Πουλου(ετσι αρκεστη μονο στο χαρτι και τον στυλο...)Εκει βεβαια σκεφτικα καποιες ασκησεις.Ας προτεινω λοιπον την πρωτη ασκηση:Nα βρειτε ολους τους ακεραιους που μπορουν να παρασταθουν με την μορφη (α^2+1)/(β^2-1) οπου οι α,β ειναι θετικοι ακεραιοι.Περιμενω.... :lol:

Καμια ιδεα?When (a^2+1)/(b^2-1) is an integer?(a,b are possitive integers) :huh:

Λοιπόν μου θυμίζειτα εξής προβλήματα....
ΙΜΟ 1988  Αν  $\frac{a^{2}+b^{2}}{ab+1}=q$  όπου $q$ ακεραιος τότε ο $q$ είναι τέλειο τετράγωνο .

(Και ένα παραπλήσιο)
Αν $\frac{a^{2}+b^{2}}{ab-1}=m$ όπου $m$ ακέραιος τότε $m=5$

Kai τα δύο λύνονται με τον ίδιο τρόπο (χρησιμοποιώντας υπερβολές) .....
Βλέπω ότι και το δικό σου ανήκει στην ίδια κατηγορία οπότε πιστεύω  ότι λύνεται παρόμοια......
Αλλά αυτό είναι απλά σκέψη με το πρόβλημα δεν έχω αχοληθεί ακόμα.... :roll:

Πολυ ενδιαφερον αυτο που λες αν και η λυση μου μιας και εγω μονος μου το εφτιαξα(μπορει σε καποιον να ειναι γνωστο) ειναι διαφορετικη.Θα ηθελα παντος να γινεις πιο αναλυτικος. :)

όχι οτι το έψαξα αλλά αποκλείεται να έχει λύση με καθαρή θεωρία αριθμών διοτι δεν ξέρω υπερβολές :D ?
ερωτω

:)  Οποιος εχει καμια ιδεα παιδια ας την πει.

λοιπον εχω μια ιδεα  . να  την πω?

Φυσικα Κωστα!Περιμενω.... :)

ο α^2+1 εχει διαιρετες τις μορφης 4κ+1.αρα και ο β^2-1 πρεπει να εχει ολους τους πρωτους διαιρ. στη μορφη 4κ+1. σωστα?

Οχι! Υπαρχει αντιπαραδειγμα σε αυτο που ειπες.π.χ για β=6 τοτε ο β^2-1=6^-1=0mod7 ο πως βλεπεις και συ 0 7 δεν ειναι τις μορφης 4κ+1. :)

ρε δημητρη αν d ενας πρωτος διαιρετης του β^2-1 τοτε d διαιρει και το α^2+1 αρα d = 4κ+1. υπαρχει δηλ η επιπλεον συνθηκη β^2-1 διαιρει α^2+1.

Σωστος.Για συνεχισε....

αν ν=α^2+1/β^2-1 θεωρω το ν+1 . εχω ν+1=α^2+β^2/β^2-1 τοτε α^2+β^2=α^2+2=ν+1mod4 .μηπως κανω λαθος?

Δεν σε καταλαβαινω...

προσπαθησε να δεις που το παω. βγαινει στο αποτελεσμα.εξαλου ισως καποιος να την προσπαθει ισως να μην πω τιποτα  :)

Κωστα ασε τα υπονοουμενα! :diablo:

:harhar: παντως ειτετην εχω λυσει ή οχι αυτη η ασκηση ειναι υπεροχη.θα ηταν περιζητητη στο αdvanced.οσο για η λυση... αυτο κι αν ειναι καλο(υπονοουμενο Νο2)

Ωστε δεν εχεις να πεις τιποτε αλλο Κωστη? :gleam:

εχω διαβασμα δεν προλαβαβαινω να πω πολλα.να δωσω τη λυση εγω (αν την δωσω) δεν εχει σημασια.το σιγουρο ειναι οτι βγαινει με αυτα που ειπα.αν τελοςπαντων καποιος απο τα παιδια που την προσπαθησαν οπως ο σιλουαν θελει να πω την δικη μου προσεγγιση θα την πω.οκ?  :blush:

Ο.Κ!Παιδεια αν ειναι καποιος που την εχει προσπαθησει ας πει κατι.... :wink:

Εγώ την έχω ψάξει με καθαρή θεωρια αριθμων κι διοφαντικές κι δεν μου εβγαιναν λύσεις μάλλον θα πρεπει να το ξεχάσω έτσι παιδιά?

ομως αμβροσιε πρεπει να βρεις τη σχεση των α,β.ισως αν δοκιμαζες διαιρετοτητα.

Φίλε κώστα προσπαθώ ειτε να βρώ σχέση ειτε να φράξω το ζητούμενο κλάσμα μεταξύ ακεραίων κι δεν βγαίνει με τίποτα θα το συνεχίσω πιστεύω κάπως να βγαίνει

Παιδια αν κανεις εχει βρει καμια λυση θα ηθελα πολυ να την πει. :lol:

[size=150][color=darkblue][b]Πρόβλημα:[/b][/color][/size] Να βρεθούν όλα τα $a,b \in \mathbb Z$ τέτοια ώστε ο αριθμός $N=\frac{a^{2}+1}{b^{2}-1}$ να είναι ακέραιος.

[color=green][b]Λήμμα:[/b][/color] Αν $d|(a^{2}+1)$ ,όπου $a,d \in \mathbb Z$ τότε ο $d$ είναι της μορφής $4k+1$ ή $8k+2$ με $k \in \mathbb Z$.

[u][color=blue][b]Λύση:[/b][/color][/u]
Έστω ότι $(b^{2}-1)|(a^{2}+1)$ τότε:

- Αν ο αριθμός $a$ είναι άρτιος τότε εύκολα προκύπτει ότι και ο $b$ είναι άρτιος, αφού αν ήταν περιττός θα ίσχυε ότι άρτιος διαιρεί περιττό (άτοπο). Έστω λοιπόν $a=2r$ και $b=2m$  τότε θα ισχύει $(4m^{2}-1)|(4r^{2}+1)$ λόγω όμως του παραπάνω λήμματος ο αριθμός $4r^{2}+1$ έχει μόνο διαιρέτες της μορφής $4k+1$ ή $8k+2$, άρα ο $4m^{2}-1$ έχει μια από αυτές τις μορφές (άτοπο).

- Αν ο αριθμός $a$ είναι περιττός διακρίνουμε δύο περιπτώσεις:

   1)Αν ο $b$ είναι περιττός, έστω  $a=2r+1$ και $b=2m+1$  τότε θα ισχύει $(4m^{2}+4m)|(4r^{2}+4r+2)$ δηλαδή $4|(4r^{2}+4r+2)$ (άτοπο).

   2)Αν ο $b$ είναι άρτιος, έστω $a=2r+1$ και $b=2m$ τότε θα ισχύει $(4m^{2}-1)|(4r^{2}+4r+2) \Leftrightarrow (4m^{2}-1)|(2r^{2}+2r+1)$. Σύμφωνα όμως       με το παραπάνω λήμμα ο αριθμός  $4r^{2}+4r+2=(2r+1)^{2}+1$ έχει μόνο διαιρέτες της μορφής $4k+1$ ή $8k+2$ άρα ο αριθμός             $2r^{2}+2r+1$ θα έχει μόνο διαιρέτες της μορφής $4k+1$, οπότε ο $4m^{2}-1$ έχει αυτή τη μορφή (άτοπο).

Άρα δεν υπάρχουν $a,b \in \mathbb Z$ τέτοια ώστε ο αριθμός $N=\frac{a^{2}+1}{b^{2}-1}$ να είναι ακέραιος.

Stelios...

στελιο και αν παρεις  μονο d=4k+1 παλι δε βγανει?

Δεν μπορώ να πάρω ότι $d=4k+1$ αφού π.χ. ο αριθμός $105^{2}+1=11026$ διαιρείται από τον $74$ που δεν είναι της μορφής $4k+1$. Γι' αυτό έγραψα ότι ο $d$ είναι είτε της μορφής $4k+1$ είτε της μορφής $8k+2$.  :wink:

ναι αλλα ο 74 δεν πρεπει να ειναι της μορφης ν^2-1  ? εφοσον αυτο το κλασμα εχω

Ναι, όμως δεν μπορείς να πεις κατευθείαν ότι αν ο d διαιρεί το α^2+1 τότε ο d είναι της μορφής 4κ+1. Το λήμμα δηλαδή είναι σωστό έτσι όπως το έγραψα.

εχεις δικιο.εγω εννοουσα τους πρωτους διαρετες.ωραια λυση :)

[quote="Ι"]
[color=green][b]Λήμμα:[/b][/color] Αν $d|(a^{2}+1)$ ,όπου $a,d \in \mathbb Z$ τότε ο $d$ είναι της μορφής $4k+1$ ή $8k+2$ με $k \in \mathbb Z$.
[/quote]

Μπορεί μήπως κάποιος να γράψει μια απόδειξη σ'αυτό το λήμμα, γιατί αυτή που έχω βρει κάπου κολλάει...  :maybe: .

Να η απόδειξή μου.
Ο  $a^{2}+1$ δεν μπορεί να έχει διαιρέτες της μορφής $4k+3$. Πράγματι, αν είχε θα είχε και ένα πρώτο διαιρέτη της μορφής 4t+3 αφού κάθε αριθμός της μορφής 4k+3 έχει έναν τυλάχιστον πρώτο διαιρέτη της μορφής $4t+3=p$ .Άρα έχουμε ότι
 $a^{2}\equiv-1 \mod{p}$. Άρα από εκεί θα έχουμε $(a^{2})^{\frac{p-1}{2}}=a^{p-1}\equiv 1 \equiv (-1)^{\frac{p-1}{2}}\mod{p}$.
Άρα ο $\frac{p-1}{2}$ πρέπει να είναι άρτιος που είναι αδύνατο για  $p=4t+3$ .

Ελπίζω να είμαι σωστός....

η περιπτωση για 8κ+2 ισχυει μονο για το 2? και κατι αλλο  Σιλουανε προφανες χρησιμοποιεις το μικρο θεωρημα του Φερμα.please μην ξεχνας να το αναφερεις γιατι μπορει να μην το γνωριζει καποιος .

Μπορούμε νομίζω να συντομέψουμε λίγο την λύση.

Αν $b =2k+1$ τότε $b^{2}-1 \equiv0\mod 4$ ενώ $a^{2}+1 \not \equiv 0 \mod 4$ .

Αν $b$ είναι άρτιος  τότε θα είναι $b=4k$ ή  $b=4k+2$  άρα το $b^{2}-1=(b-1)(b+1)$

θα έχει και στις $2$ περιπτώσεις παράγοντα της μορφής  $4m+3$

αδύνατο σύμφωνα με το λήμμα .

-------------------------------------------------------------------

Ναι Κώστα η μορφή $8k+2$ χρειάζεται στην περίπτωση που ο $d$ είναι άρτιος .

μια ενδιαφερουσα ασκηση:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5781[/url]

[quote="Αρχιμήδης 6"]να βρεθουν ολοι οι ακεραοι θετικοι οι οποιοι ειναι τις μορφης: $\frac{a^{2}+1}{b^{2}-1}$, με $a,b$ φυσικοι[/quote][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=784285#p784285]εδώ[/url]


I don't know either, and I am in geometry. sad.

I hope someone answers this.

I just know that it has something to do with origins or something but have no idea.

[quote="Ihatepie"][quote="puuhikki"]Let $m,n\in \mathbb{Z},n\ne 0$ and $a\geq 0$. Then $a^{m/n}=\sqrt[n]{a^{m}}$. For irrational number $x$ we have defined \[a^{x}=\sup \,\{a^{q}| q\in\mathbb{Q}\text{ and }q> x\}.\] [/quote]

If what you are saying is right then this is way beyond my level[/quote]

The first two sentences are definitely within grasp - it's saying in math language that when m and n are real and a is greater than (or equal to) 0, a to the $\frac{m}{n}$ power is the nth root of a to the mth power.

[quote="Ihatepie"]In my class as a bonus my teacher gave 3to the 1.5 power. I didn't know so I guessed. Now that I think of it the answer might be 4.5 because 3 to the 1.5 power + 3 to the 1.5 power = 3 to the 3rd power which is 9. Am I right? and how do you do harder questions like 9 to the 7.8 power?[/quote]
A couple of points I wanted to make about your reasoning:
$a^{b}+a^{c}\neq a^{b+c}$ You can check this with $2^{2}+2^{3}\neq 2^{5}$
So $3^{1.5}+3^{1.5}\neq 3^{3}$.
Also, $3^{3}\neq 9$.


Let $A\subset [0,1]$ be the set of all numbers with a decimal expansion that has only digits $0$, $5$, and $9$. (An infinite tail of $9$s is allowed.) Let $B\subset [0,1]$ be the set of all numbers with a decimal expansion that has only digits $0$, $8$, and $9$.

#1. Find a Lipschitz map from $A$ onto $B$.
#2. Is there a Lipschitz map from $B$ onto $A$?

Warning: #2 is a version of an open problem in the book "Fractured fractals and broken dreams" by David and Semmes. A strange title, but it fits the book.

EDIT: Three Chinese mathematicians (Rao, Ruan, and Xi) recently solved #2: there is even a bi-Lipschitz map between $A$ and $B$.   :o

I think the most obvious approach is to read [url=http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6X1B-4HYV07W-6&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1032703163&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=2eaf0af5170e97dccd5e3df3a99569b2]the original article[/url] written by Rao, Ruan, and Xi.


function
Support
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real analysis unsolved

Hi,

I came across an interesting problem in Rudin's PMA, for which I haven't been able to find a solution. Help is needed.

Here goes: Is it possible for each closed subset E of R1, to find a real valued f defined on R1 such that

                  (1)   f is infinitely differentiable on R1
                  (2)   the zero set of f is exactly E         ?

I think I have been able to find an f which is n times continuously differentiable, as follows:

suppose E such that E is closed and an arbitrary integer n.

=> R1\E is open.

R1\E is an open subset of R1 => it is a union of disjoint open segments (ak,bk) ; k=1,2,3...

Define f(x)=
                   0                               ;when x belongs to E
                  ((x-ak)*(bk-x))^(n+1) ;when x belongs to (ak,bk).

Now f(x) is n times continuously differentiable on R1 and the zero set of f is exactly E.

But when it's time to find a function that is infinitely differentiable, I can't find one:
I came to the conclusion that if the Taylor's serie of f converges to f, we have for every x belongs to (ak,bk)

f(x)= Sum[n goes from zero to infinity] ( f(n) (ak)/n!* (x-ak)^n)=0 ,
          since f(n)(ak)=0 for every ak,
          (because ak belongs to E and f(n) is continuous).

=> f does not fulfill the conditions wanted because f is zero also on some point of R1\E.

So then this would mean that f must not be representable by its Taylor's serie. But this makes everything pretty difficult...

The only infinitely continuosly differentiable function I know which is not representable by it's taylor serie is the famous exp(-1/x²),
but is it the only function with such properties?

Can anyone come up with a function? Or perhaps a mistake in my reasoning?

Thanks a lot in advance for any help
-Laura

Here's how to build it:
Step 1: Construct a $C^\infty$ function $g$ with compact support (zero outside the interval $[0,1]$).
Let $g(x)=0$ if $x\not\in(0,1)$. Let $g(x)=e^{-\frac1x}\cdot e^{-\frac1{1-x}}$ if $x\in(0,1)$. This is basically the example you already know.

Now, we have $\mathbb{R}\setminus E=\cup_n (a_n,b_n)$, where $(a_n,b_n)$ are disjoint open intervals. Choose nonzero constants $c_n$, and let $f(x)=\sum_n c_ng\left(\frac{x-a_n}{b_n-a_n}\right)$. That's almost the answer, but it doesn't deal with infinite intervals properly. If $b_n=\infty$, we use $e^{-\frac1{x-a_n}}$ instead and if $a_n=-\infty$, we use $e^{-\frac1{b_n-x}}$ instead. If the only interval is $(-\infty,\infty)$, use $e^x$.

Thank you very much jmerry for such a swift response.

Kisses
-Laura :)

uhm...
there's a simpler solution, that can be easierly generalized to functions from other sets to $\mathbb{R}$:
since $E$ is closed, the distance function: $d_E(x) = \inf_{y \in E} |x-y|$ is well defined, and is $0$ exactly on the set $E$, and positive everywhere else.
now, we can define $f_E|_E = 0$ (meaning that the function $f_E$ restricted to $E$ is $0$), and $f_E|_{E^c}(x) = e^{-\frac1{d_E(x)^2}}$.
it's quite easy to check that $f_E \in C^{\infty}$.

another solution (which i don't know much about, though), would be considering the convolution product of two functions..
if anyone knows more about it, please tell us something :)

Convolution "smears out" functions and spreads the support around. It's not good for controlling the exact zero set of a function.

Also, your example has a problem. Suppose we're working on $\mathbb{R}$ and the set $E=\{-1,1\}.$ Then your example gives a function which is $e^{-\frac1{(x+1)^2}}$ for $x\in(-1,0]$ and $e^{-\frac1{(1-x)^2}}$ for $x\in[0,1).$ That function fails to be differentiable at $x=0.$

In fact, it ahs been shown on this forum (I believe I showed it, in a thread started by sam-n :)) that given a closed subset $E$ of $\mathbb R^n$, the function $d_E$ defined by ma_go is differentiable at $x\not\in E$ iff there is only one point $y\in E$ with $d_E(x)=\|x-y\|$.

yet, there is a way to fix it...
take a closed set $E'$ such that, chosen any two points $x<y \in E$ either $\left[x,y\right] \subset E$ or there's a point  $x' \in E' \cap [x,y]$, and such that no point of $E'$ is an isolated point ($\forall x' \in E' \exists a',b' \in E' \mid x' \in \left[a',b'\right] \subset E'$).
it's not difficult to show such a set exists.
given a set $A \subset \mathbb{R}$ (or in any metric space, after all), define the function $\displaystyle e_A(x) = e^{-\frac1{d(A,x)^2}}$, and consider the further function: $\displaystyle f(x) : = \frac{e_E(x)}{e_{E'}(x) + e_E(x)}$.
this should be a $C^\infty$ limited functions, which values are $0$ on $E$, and $1$ on $E'$.
(i hope this time i'm not wrong again!)

That's just not worth it in $\mathbb{R}^1$- and it's not at all clear how your construction generalizes.


Hi all,

I am graduating from college having discovered my liking for mathematics too late. However, I am determined to pursue this newfound passion and see where it leads me.

In particular, what draws me to college level mathematics is the level of abstraction and in particular, proofs. At this point in time, I happen to find proofs the most enjoyable(albeit the most frustrating as well) part of mathematics.

I have purchased most of the AoPS books in an attempt to solidify my mathematical foundation so as to learn elementary mathematical thinking and to get a good grasp of the basic tools and ideas needed to tackle more abstract problems. However, I notice a dearth of what I would call "direct proof problems", by which I mean problems where the question asks for an explicit proof of a result.

Does anyone know any books in particular that are in the style of AoPS for teaching and learning how to do proofs? Also, if there are books which simultaneously teach higher mathematics (e.g. abstract algebra, number theory etc.) while teaching problem solving tactics for proofs, that would be a double bonus.

(In fact, are there textbooks where, instead of the author giving the proof immediately after presenting a theorem, he chooses to guide the reader step by step through the proof, letting the reader derive it himself with suitable hints?)

Looking forward to your recommendations
Thanks!

I don't have any specific suggestions, but a good way to get better might be to find a source of proof problems, whether it be a problem book or a textbook. If you want to improve your proof-writing skill, you should try to work on problems that are easy for you but not trivial. You want it easy so that you won't be struggling just to solve it, but you want it hard enough so that the proof actually requires some delicate explanation and the practice is not too tedious.

Olympiad problems might be too hard to start with right away, but I know of a few books containing relatively easy proof-based problems: Math Olympiad Treasures, Math Olympiad Challenges, and some title along the lines of 500 Math Challenges. If you work on enough number theory problems, it will probably make learning introductory number theory and abstract algebra fairly easy, since you will already be comfortable with many of the objects and concepts they use.

Yeah, that might be helpful for writing proofs, but I find that the best way to improve problem solving skills in general is to do problems that you have to struggle to get. Not to mention, you will probably find these more fun.


Sally's salary increased from $ \$180$ to $ \$200$ per week, and then decreased to $ \$190$ per week. Express the net percent of change as a mixed number.


linear algebra
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linear algebra unsolved

Hi,

It just crossed my mind while I was doing another thing. I believe it is true.

Supose $\{v_1,v_2,\dots,v_n\}$ is an orthonormal basis of $\mathbb{R}^n$, and suppose that $\alpha_i\in\mathbb{R},\ \alpha_i\neq0\forall i$ and $\alpha_1^2+\alpha_2^2+\cdots+\alpha_n^2=1$. use Gram-Schmidth orthonormalization process on $\{u_1,v_2,\dots,v_n\}$, starting with $u_1$, where $u_1=\alpha_1v_1+\cdots+\alpha_nv_n$, obtaining the orthonormal basis $\{u_1,u_2,\dots,u_n\}$. Prove that if $B$ is the matrix of change of coordinates from the first to the second set of coordinates, then, none of the entries of $B$ is zero.

Best,

I still don't have a solution for this problem. Does anyone have one?

although it was not rigorously verified, it I think the following loosely explains it:
the (k,j) entry of B is: <v_k,u_j> (where v_k is the k-th vector in the V basis and u_j is the j-th vector in the U basis).
Is can be shown that for all k,j v_k and u_j are not orthogonal:
when you take any v_k (a vector of the orthonormal V basis) and subtruct a linear combination the same basis vectors: alfa_1*v_1+alfa_2*v_2+...+alfa_n*v_n, where 0<alfa_k<1 you get a different linear combination of {V_j} of unit magnitude with all coefficients beta_j s.t. 0<beta_j<1 => the inner product of this result with any v_k is larger than zero.
The Gram-shmidt procedure can be viewed as repeating the above procedure multiple n times.


Given 2 lines intersecting at $O$, a length $z$, and a point $P$. Construct a line passing through the point $P$ that intersect the two lines at $A$ and $B$ such that $OAB$ has a perimeter of $z$.

Here is the crux of the construction; the determination is boring and will be omitted.

Let X and Y be two points on the two given lines such that $OX=OY=\frac{z}{2}$. (Actually, there are two different possible positions for the point X and two different possible positions for the point Y - try out all of them: sometimes the one, sometimes another will yield a triangle OAB with perimeter z.)

Since OX = OY, there exists a circle k which touches the line OX at the point X and the line OY at the point Y. Let t be one of the two tangents from the point P to this circle k, and let A and B be the points of intersection of this tangent t with the lines OX and OY. Then, depending of the arrangement of the points, this circle k is either the incircle or one of the excircles of triangle OAB. If the circle k happens to be the excircle of triangle OAB opposite to the vertex O, then triangle OAB has the perimeter z, and thus, the line t is one of the required lines. [The proof of this is given in the spolier.]

[hide="Proof"][i]Proof:[/i] If the circle k is the excircle of triangle OAB opposite to the vertex O, the point X is the point of tangency of this excircle with the sideline OA of the triangle OAB, and thus, by a well-known theorem about excircles, the length OX equals the semiperimeter of triangle OAB. In other words, the perimeter of triangle OAB equals $2\cdot OX$. But since $OX=\frac{z}{2}$, the perimeter of triangle OAB thus equals $2\cdot\frac{z}{2}=z$. Hence, the line t is one of the lines we have to construct.[/hide]

Conversely, we can show by redoing our argumentation backwards that each required line can be found as a tangent from the point P to the circle k (don't forget that there are four different circles k, depending on the different possible positions of the points X and Y, and we have to check all of them).

  Darij


[quote=GameBot]What is the remainder when $ 13^{51}$ is divided by 5?[/quote]

$\phi(5)=4$

This reduces to $3^3 \equiv \boxed{2} \pmod 5$.

@advanture
I made a mistake. 13 = 3 mod 5, not -3 mod 5 lol.


a 5.00 g sample of aluminum pellets and a 10.00 g sample of iron pellets are placed together in a dry test tube, and the test tube is heated in a boiling water bath to 100 degree celcius,.  The mixture of hot iron and aluminum is then poured into 97.3 g of water at 22.5 C.  Towhat final temperature is the water heated by the metals?

thanks

Hello,
  You could post it in the Physics section as it has more to do with Thermal Physics.Infact,there is no chemistry required to solve the problem.You have to apply the Principle of mixtures.And ofcourse,the specific heats of the metals need to be given.Just apply,heat lost by the metals equals the heat gained by the water and you are through.
Sathej

In case you need more detail, all three relevant items (alum, iron and water) start at their initial temperature.

For each degree of temperature change for each material a certain amount of heat is needed or generated.  (This can be calculated by multiplying the heat capacity by the weight.)  One then knows, that since no reaction occurs, that there is no net heat change.  The amount heat released by the alum and iron equals the increase in heat by the water at some specific temperature (between 100 and 22.5).  This is the final temperature of the mixture.  The critical aspect is that the alum and iron start from 100 and move down to the final temperature, while the water starts at 22.5 and moves up to the final temperture.  The tricky part is geting both changes covered by the same one temperature variable.

If you have trouble, try to have the one variable defined that works for a specified temperature you select between 100 and 22.5.  It may help to select random temperatures and construct a data table then constuct the temperature variable from that data table.


Given 5 circumferences, every four of them have a common point.
Prove that there exist a point that belongs to all five circumferences.

Perform an inversion having as pole the common point of four of the circles. It turns four circles into lines and the fifth circle into a circle (if it's also a line then all five of them pass through the pole). If there are three lines which are not concurrent, then the system formed by them and the circle doesn't satisfy the conditions. This means that all the lines are concurrent, and the circle passes through the concurrence point, and that's it.


inequalities
logarithms
inequalities proposed

I actually spent all night on this problem and couldn't find any solution :(
Can,I was wondering if you have anything in your mind different than Arqady's solution?(which is nice but it has too many computations).
Could you give us a hint?Could anyone think a solution of this inequality just using Cauchy-Schwarz etc.Thanks

~Kostas

[quote="can_hang2007"]My proof is not very nice. I used the inequality:
$ \frac {a \plus{} b}{2} \plus{} \frac {3(a \minus{} b)^2}{a \plus{} b} \ge \sqrt [13]{\frac {a^{13} \plus{} b^{13}}{2}},$ (*)
to reduce the problem into proving that
$ \frac {5a^2}{b} \plus{} \frac {3b^3}{a^2} \ge 4(a \plus{} b) \plus{} \frac {24(a \minus{} b)^2}{a \plus{} b},$
which is true and it is not hard to prove.

By the way, I think (*) has a proof by AM-GM. I hope someone will find it out.[/quote]
$ \frac {5a^2}{b} \plus{} \frac {3b^3}{a^2} \ge 4(a \plus{} b) \plus{} \frac {24(a \minus{} b)^2}{a \plus{} b}$

[b]Proof?[/b]

Thanks.

[quote=can_hang2007]Let $ a$ and $ b$ be positive real numbers such that $ a^{13}\plus{}b^{13}\equal{}2.$ Prove that
$ \frac{5a^2}{b}\plus{}\frac{3b^3}{a^2} \ge 8.$[/quote]
[quote=arqady]The following inequality is also true.
Let $ a$ and $ b$ are positive numbers such that $ a^{14}\plus{}b^{14}\equal{}2.$ Prove that:
\[ \frac {5a^2}{b} \plus{} \frac {3b^3}{a^2} \ge 8\][/quote]
Solution of kuing:

Let $ a,b $ are positive real numbers such that $a^{12}+b^{12}=2. $ [url=https://artofproblemsolving.com/community/c6h2869319p25486803]Prove that[/url]$$\frac{3a^2}{b}+\frac{b^5}{a^4} \geq 4$$

[quote=can_hang2007]Let $ a$ and $ b$ be positive real numbers such that $ a^{13}\plus{}b^{13}\equal{}2.$ Prove that
$ \frac{5a^2}{b}\plus{}\frac{3b^3}{a^2} \ge 8.$[/quote]

Splendid


Let [tex] a,b,c [/tex] be three positive real numbers. Prove that:
[tex] a^2+b^2+c^2+2abc+1 \geq 2(ab+ac+bc) [/tex]  :)

This problem is posted on ML many times . It's mentioned in my recent topic .
 By AM-GM : $ 2abc + 1 \geq 3 {^3}\sqrt{a^2b^2c^2} $
 After that , use Schur 's ineq  ;)
  the equality holds for $ a=b=c=1 $