hello

pleae help explaining this qustion in details

thanx

Deﬁnition: A simple graph G is called self-complementary if G and
G are isomorphic.
(a) Show that the following graph is self-complementary.

[img]http://img256.imageshack.us/img256/8049/52154988kv6.png[/img]

(b) Find a self-complementary simple graph on ﬁve vertices.

(c) Show that there is no self-complementary simple graph on six vertices. (Hint: Consider the number of edges in K6 .)

[quote="t0rajir0u"]Are you sure you don't mean $G, G'$?[/quote]

yes you are right thats what i meant but i did not know how to insert the simple like the way you did

thanx


inequalities
parameterization
combinatorics unsolved
combinatorics

$ \alpha$ is an irrational number. Prove that the set $ \{a\plus{}b\alpha|a,b \in\mathbb{Z},(a,b)\equal{}1\}$ is dense in $ \mathbb{R}$.

your post is nuts. It is wrong.
 :rotfl:

No, it is very much right.

[hide]
We claim that we can make it within $ \epsilon$ of any constant $ c$. After some rearrangment of the resulting inequalities, we see that it is enough to find integers $ a,b$ with $ \alpha$ within $ \epsilon/b$ (the $ /b$ doesn't actually matter since $ \epsilon$ is arbitrarily small) of $ \frac{a\minus{}c}{b}$, so this is just estimation of irrationals with rationals, which clearly follows by the density of rationals. The $ (a,b)\equal{}1$ restriction doesn't matter because arbitrarily large $ a,b$ satisfying this makes the $ c$ irrelevant, so we're just reducing to lowest terms in that case.

We can also use Kronecker's; set $ a$ so that it leaves only the fractional part of $ \alpha$, and since it's dense in $ [0,1]$ it's dense everywhere by varying $ b$. $ (a,b)\equal{}1$ is harder to deal with in this approach, or maybe I'm just stupid. [/hide]

[quote="fwolth"] so this is just estimation of irrationals with rationals, which clearly follows by the density of rationals. [/quote]

Good idea! Relating this problem to the 'irrationals estimation' was a good step. However, there is s bug in the proof: the estimation deviation (i.e. $ \epsilon/b$) of $ \alpha$ with the fraction $ \frac{a\minus{}c}{b}$ is not independent from the fraction! More formally, for an arbitrary small $ \delta>0$, let $ \frac{p_{\delta}}{q_{\delta}}$ be the estimation of $ \alpha$ within $ \delta$ (i.e. $ |\frac{p_{\delta}}{q_{\delta}}\minus{}\alpha|<\delta$). Suppose $ (p_{\delta},q_{\delta})\equal{}1$. Then if we use this estimation for $ \alpha$ and set $ \frac{a\minus{}c}{b}\equal{}\frac{p_{\delta}}{q_{\delta}}$, the consequence is (even by ignoring the condition $ (a,b)\equal{}1$), we should have: $ q_{\delta}<\epsilon/\delta$. Is it something easy to prove?

[hide]
Well, say we want it to be within $ \epsilon$ of $ \alpha$. Choose $ a,b$ so $ a/b$ is within tolerance $ \epsilon/2$ and so that $ \frac {c}{b} < \epsilon/2$ as well. Notice that we can do this since $ c$ is fixed and fractions with arbitrarily large denominators must be able to estimate irrationals to arbitrary precision. Then I don't think any of these issues come up.
[/hide]

The presence of $ c$ is not a problem at all and as you mentioned, since $ c$ is constant and $ b$ can be made large enough, it is ignorable. My argument is for the choice of the fraction $ \frac{a}{b}$. You said lets choose it within the tolerance $ \epsilon/2$ of $ \alpha$. But this is not the right the choice, the right choice is it should be within the tolerance $ \epsilon/2b$! And the challenging thing is, the parameter $ b$ is present in both the fraction and the tolerance. This means, as the fraction $ a/b$ approaches $ \alpha$ , the desired difference (i.e. $ \epsilon/2b$) also becomes small (as $ b$ probably grows). So we need a argument that there is a fraction $ a/b$ which lies within $ \epsilon/b$ of $ \alpha$. I found this argument a bit challenging. But most likely it is the key to the main problem.

Hmm I see what you are saying now. Unfortunately I can't find a way to fix this, and also no way to address $ (a,b)\equal{}1$ with Kronecker's (or any way to approach that condition with Weyl's). The $ (a,b)\equal{}1$ makes this problem tricky.


[quote="ProtestanT"][quote="Quevvy"]A fly starts at one vertex of a cube with edge length $1$ inch and randomly walks along edges of the cube. What is the number of inches in the longest distance that it could walk without retracing any edge?

I think I got the right answer but I just kept trying and failing so it took me a while. Is there a simple method?[/quote]
[hide]
it is simply asking for the highest number of edges. able to be walked without retracing.
i think it is when it walks the top 4 edges, 1 edge down, then the bottom 4 edges. that is $9$ edges
$9$ edges x $1$ inch per edge =
$9$ inches
[/hide][/quote]
That's what I got but I was wondering if there was any formula or something


One root of the equation $ 5x^2\plus{}kx\equal{}4$ is 2. What is the other?

By Vieta, the product of the roots of $ P(x)\equal{}5x^2\plus{}kx\minus{}4$ is $ \minus{}\frac45$.

Thus, if we let the unknown root be $ r$, then $ 2r\equal{}\minus{}\frac45\implies r\equal{}\boxed{\minus{}\frac25}$.


Are there consecutive four positive integers such that each of them has just four positive divisors?

Haven't done this kind of stuff in a while so I might be a bit rusty  :P .

[hide="Solution?"]

Well I just randomly called the four integers $ n,n\plus{}1,n\plus{}2$ and $ n\plus{}3$. If a number has $ 4$ divisors then it takes on one of two forms: $ p^{3}$ or $ pq$ where $ p,q$ are primes.

We either have $ (n, n\plus{}2)\equal{}2$ or $ (n\plus{}1, n\plus{}3)\equal{}2$. Notice however none of the numbers can possibly be $ 8$ (check manually). So we will have two numbers $ 2p$ and $ 2q$ such that $ 2q \equal{} 2p\plus{}2$ so $ q\equal{}p\plus{}1$ which means $ (p,q) \equal{} (2,3)$. However this gives two possible set of $ 4$ integers $ (3,4,5,6)$ or $ (4,5,6,7)$ clearly neither of which work.

So I'm guessing there's no such quadruples.[/hide]

That's right,BanishedTraitor :lol:
same as my solution.


Using a counting argument, prove that
\[{{n\choose 2}\choose 2}= 3{n+1\choose 4}.\]

May I please have a hint? Stuck...  :P

Hm, nice.  Is the following enough for you?

[hide="Little hint"]Both sides count the number of pairs of edges in a graph on $n$ vertices.  The way the RHS counts this is [hide="big hint"]via the addition of an extra vertex whose name is "Shareanendpoint."[/hide][/hide]

Ah, yes, the small hint did the trick. I wasn't sure how to interpret the double combination, so I tried starting on the RHS and got stuck. :P

For the counting argument, are you allowed to rewrite the given equation?

Like $\binom{n+1}{4}= \binom{n}{4}+\binom{n}{3}$?
(I mean, that's what I did right away because it made sense in the argument I thought up.)


The sum of three numbers is $ 98$. The ratio of the first to the second is $ \frac {2}{3}$, and the ratio of the second to the third is $ \frac {5}{8}$. The second number is:

$ \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33$

$ A\equal{}\frac23B,C\equal{}\frac85B$. So $ (\frac23\plus{}1\plus{}\frac85)B\equal{}98$ and $ B\equal{}30$. So it's C.


calculus
integration
calculus computations

Find the minimum value of $\int_{-1}^1 \sqrt{|t-x|}\ dt$

-1<=x<=1

break the integral into two
evaluate the integral dt, then do d/dx

min occur at x=0

minimun value is 4/3

Yes,your answer is correct, isaacchao. :)


function
calculus
integration
logarithms
number theory
Diophantine equation

A ball of mass $ m$ is projected vertically upwards with initial speed $ u$ in a resisting medium that produces a retardation force of magnitude $ kv^2$, where $ v$ is the ball's speed. What are the dimensions of $ k$?

Show by means of dimensional analysis that when the ball returns to its point of projection, its speed $ w$ can be written in the form
\[ w = u \, f(\lambda)\]where $ \lambda = \frac{ku^2}{mg}$

Integrate the equations of motion to show that $ f(\lambda) = (1 + \lambda)^{-\frac{1}{2}}$.

Discuss the cases (i) $ \lambda \ll 1$, (ii) $ \lambda \gg 1$.

So far, my work:
I have that the dimensions of $ k$ are $ ML^{-1}$ where $ M$ is units of mass, $ L$ is units of length. How do I do dimensional analysis to derive the $ w = u \, f(\lambda)$? I only know $ f(\lambda)$ has no units.

As for the equations of motion, I have $ v^2 = u^2 + 2as$, $ s = ut + \frac{1}{2} at^2$, but what do we integrate with respect to?

Any help/clues are appreciated! :)

Note than $ \lambda$ is dimensionless (check the units of each individual term).

All I can gather so far from reading about Dimensional Analysis is that if you have some dependent variable and you want to know the relation between it and independent variables, they are connected by a simple power law. We just have to identify which factors govern the dependent variable.

So in the context of this question I would say that the final speed when the ball returns to its point of projection is $ v_f \equal{} f(v_i, g,k)$ where $ v_i$ is initial velocity, $ g$ is acceleration due to gravity. Then $ v_f \equal{} [v_i]^{\alpha}[g]^{\beta}[k]^{\gamma} \equal{} [LT^{\minus{}1}]^{\alpha}[LT^{\minus{}2}]^{\beta}[ML^{\minus{}1}]^{\gamma}$. Comparing coefficients, $ \alpha \plus{} \beta \minus{} \gamma \equal{} 1, \alpha \plus{} 2\beta \equal{} 1, \gamma \equal{} 0$. Is this correct? From here how do we get a dimensionless quantity $ \lambda \equal{} \frac{ku^2}{mg}$?

However I still do not know how to integrate the equations of motion

[quote="aidan"]
So in the context of this question I would say that the final speed when the ball returns to its point of projection is $ v_f \equal{} f(v_i, g,k)$ where $ v_i$ is initial velocity, $ g$ is acceleration due to gravity. Then $ v_f \equal{} [v_i]^{\alpha}[g]^{\beta}[k]^{\gamma} \equal{} [LT^{ \minus{} 1}]^{\alpha}[LT^{ \minus{} 2}]^{\beta}[ML^{ \minus{} 1}]^{\gamma}$. Comparing coefficients, $ \alpha \plus{} \beta \minus{} \gamma \equal{} 1, \alpha \plus{} 2\beta \equal{} 1, \gamma \equal{} 0$. Is this correct? From here how do we get a dimensionless quantity $ \lambda \equal{} \frac {ku^2}{mg}$?
[/quote]
Well, your approach is correct.. But a slight error...$ v_f$ depends on mass also. So it is..
$ v_f \equal{} f(v_i, g,k,m)$ $ v_f \equal{} [v_i]^{\alpha}[g]^{\beta}[k]^{\gamma}[m]^{\delta} \equal{} [LT^{ \minus{} 1}]^{\alpha}[LT^{ \minus{} 2}]^{\beta}[ML^{ \minus{} 1}]^{\gamma} [M]^{\delta}$
Equating coefficients you get
\[ \alpha \plus{} \beta \minus{} \gamma \equal{} 1,\alpha \plus{} 2\beta \equal{} 1, \gamma \plus{} \delta \equal{} 0\]
[hide="The remaining is all math..."]
Observe that $ \alpha \plus{} 2\beta \equal{} 1$ is a Diophantine equation, general solution is $ \alpha \equal{} 1 \plus{} 2\eta , \beta \equal{} \minus{} \eta$ where $ \eta \in \mathbb{R}$
This gives $ \gamma \equal{} \eta$ and $ \delta \equal{} \minus{} \eta$ from the remaining 2 equations
Putting them back into the equation you have
\[ w \equal{} u\, \left(\frac {ku^2}{mg}\right)^{\eta}\]
It is in form
\[ w \equal{} u\ f(\lambda)\]
[/hide]

[quote="aidan"]
As for the equations of motion, I have $ v^2 \equal{} u^2 \plus{} 2as$, $ s \equal{} ut \plus{} \frac {1}{2} at^2$, but what do we integrate with respect to?
[/quote]
No, those equations canNOT be used here they've been derived assuming that acceleration of the particle is constant..
But this question is regarding non-uniform acceleration.

General Method to find equation of motion is using the differential forms of definitions of acceleration and velocity.
\[ a \equal{} \frac {dv}{dt}\]

\[ v \equal{} \frac {dx}{dt}\]
One more derived from the above 2 definitions is $ a \equal{} v.\frac {dv}{dx}$. I think you can figure out how it is derived...

Find $ a$ from the $ a \equal{} \frac {F_{net}}{m}$ which is the Newton's Second Law.. It turns out to be a function of $ v$ in this case. So, take the equation $ a \equal{} v.\frac {dv}{dx}$ and integrate to find relation between $ v$ and $ x$. Set x=0 find $ v$.

Well that is the idea.. If you have any problem anywhere feel free to ask.. I'll post the detailed working if you want :)

[quote="savy2020"][quote="aidan"]
So in the context of this question I would say that the final speed when the ball returns to its point of projection is $ v_f \equal{} f(v_i, g,k)$ where $ v_i$ is initial velocity, $ g$ is acceleration due to gravity. Then $ v_f \equal{} [v_i]^{\alpha}[g]^{\beta}[k]^{\gamma} \equal{} [LT^{ \minus{} 1}]^{\alpha}[LT^{ \minus{} 2}]^{\beta}[ML^{ \minus{} 1}]^{\gamma}$. Comparing coefficients, $ \alpha \plus{} \beta \minus{} \gamma \equal{} 1, \alpha \plus{} 2\beta \equal{} 1, \gamma \equal{} 0$. Is this correct? From here how do we get a dimensionless quantity $ \lambda \equal{} \frac {ku^2}{mg}$?
[/quote]
Well, your approach is correct.. But a slight error...$ v_f$ depends on mass also. So it is..
$ v_f \equal{} f(v_i, g,k,m)$ $ v_f \equal{} [v_i]^{\alpha}[g]^{\beta}[k]^{\gamma}[m]^{\delta} \equal{} [LT^{ \minus{} 1}]^{\alpha}[LT^{ \minus{} 2}]^{\beta}[ML^{ \minus{} 1}]^{\gamma} [M]^{\delta}$
Equating coefficients you get
\[ \alpha \plus{} \beta \minus{} \gamma \equal{} 1,\alpha \plus{} 2\beta \equal{} 1, \gamma \plus{} \delta \equal{} 0\]
[hide="The remaining is all math..."]
Observe that $ \alpha \plus{} 2\beta \equal{} 1$ is a Diophantine equation, general solution is $ \alpha \equal{} 1 \plus{} 2\eta , \beta \equal{} \minus{} \eta$ where $ \eta \in \mathbb{R}$
This gives $ \gamma \equal{} \eta$ and $ \delta \equal{} \minus{} \eta$ from the remaining 2 equations
Putting them back into the equation you have
\[ w \equal{} u\, \left(\frac {ku^2}{mg}\right)^{\eta}\]
It is in form
\[ w \equal{} u\ f(\lambda)\]
[/hide][/quote]

Well I totally forgot about the mass of the particle - could we instead have said that $ v_f$ is dependent on the net force experienced by the object? Though of course it seems wiser to separate into $ k, g, m$

[quote="savy2020"][quote="aidan[quote="aidan"]
As for the equations of motion, I have $ v^2 \equal{} u^2 \plus{} 2as$, $ s \equal{} ut \plus{} \frac {1}{2} at^2$, but what do we integrate with respect to?
[/quote]
No, those equations canNOT be used here they've been derived assuming that acceleration of the particle is constant..
But this question is regarding non-uniform acceleration.

General Method to find equation of motion is using the differential forms of definitions of acceleration and velocity.
\[ a \equal{} \frac {dv}{dt}\]

\[ v \equal{} \frac {dx}{dt}\]
One more derived from the above 2 definitions is $ a \equal{} v.\frac {dv}{dx}$. I think you can figure out how it is derived...[/quote]

I see, thanks.. Yes you just divide $ a$ by $ v$ and then $ a \equal{} v \frac{dv}{dx}$.

[quote="savy2020"]Find $ a$ from the $ a \equal{} \frac {F_{net}}{m}$ which is the Newton's Second Law.. It turns out to be a function of $ v$ in this case. So, take the equation $ a \equal{} v.\frac {dv}{dx}$ and integrate to find relation between $ v$ and $ x$. Set x=0 find $ v$.

Well that is the idea.. If you have any problem anywhere fell free to ask.. I'll post the detailed working if you want :)[/quote]

However, $ kv^2$ is in a different direction when the ball is going downwards - do we need 2 integrals then?

[quote="aidan"]
Well I totally forgot about the mass of the particle - could we instead have said that $ v_f$ is dependent on the net force experienced by the object? Though of course it seems wiser to separate into $ k, g, m$
[/quote]
I think it would be better to take them as seperate.
[quote="aidan"]
However, $ kv^2$ is in a different direction when the ball is going downwards - do we need 2 integrals then?
[/quote]
Oh yeah! we need to take 2 integrals..hmm...that'd complicate things I think...

Also, why would we want $ v\frac{dv}{dx}$? We can't get this when the ball reaches the highest point in its trajectory - $ \frac{dx}{dt}$ would be zero and we can't divide the equation $ a \equal{} \frac{dv}{dt}$ by $ v \equal{} \frac{dx}{dt}$.

I still can't do the integration though!

[hide=" For the upward motion, "]
\[ F_{net} \equal{} \minus{} (mg \plus{} kv^2)\]

\[ a \equal{} v\frac {dv}{dx} \equal{} \minus{} \frac {mg \plus{} kv^2}{m}\]

\[ \int^{0}_{u}{\frac {vdv}{mg \plus{} kv^2}} \equal{} \minus{} \int^{x_{max}}_{0} {\frac {dx}{m}}\]
Evaluating the integrals (I'll leave out the math part-- hint is-- substitute $ y \equal{} mg \plus{} kv^2$..That'll make the integration easy.)
\[ \frac1{2k} \ln \left(\frac {mg}{mg \plus{} ku^2} \right) \equal{} \minus{} \frac {x_{max}}{m} \implies x_{max} \equal{} \frac {m}{2k} \ln \left(\frac {mg \plus{} ku^2}{mg} \right)\]
[/hide]

[hide="For the downward motion"]
\[ F_{net} \equal{} mg \minus{} kv^{2}\]

\[ a \equal{} v\frac {dv}{dx} \equal{} \frac {mg \minus{} kv^2}{m} \implies \int^{v_f}_{0}\frac {vdv}{mg \minus{} kv^2} \equal{} \int^{x_{max}}_{0} \frac {dx}{m}\]
Again evaluating the integral (I'm leaving the math part - use the substitution $ y \equal{} mg \minus{} kv^2$)
\[ \frac { \minus{} 1}{2k} \ln \left( \frac {mg \minus{} k{v_f}^2}{mg} \right) \equal{} \frac {x_{max}}{m} \implies x_{max} \equal{} \frac {m}{2k} \ln \left( \frac {mg}{mg \minus{} k{v_f}^2} \right)\]
[/hide]

[hide="final solution"]
Using the 2 equations from the 2 cases, dividing them you'll have
\[ \left(\frac {mg \plus{} ku^2}{mg} \right) \equal{} \left( \frac {mg}{mg \minus{} k{v_f}^2} \right)\]
Solving for $ v_f$ you have the required..

If you cannot comprehend any of the above steps you may ask.. I'll try to clear that :)
[/hide]

I think I get the whole thing, thanks! Nice of you to think of this  :)


Define a sequence $\{a_i\}$ by $a_1=3$ and $a_{i+1}=3^{a_i}$ for $i\geq 1$. Which integers between $00$ and $99$ inclusive occur as the last two digits in the decimal expansion of infinitely many $a_i$?


inequalities
function
inequalities unsolved

This problem is posted but i don't have any solution:
find the least value m satisfy the inequality:
[tex]( \frac{x^m+y^m+z^m}{3})^2 \ge ( \frac{xy+yz+zx}{3})^m [/tex]

Thank you!

Oh,your problem is:
If $x+y+z=1$ then $(xy)^m+(yz)^m+(zx)^m \le max(3,(\frac{3}{2})^{2m})$(*)
And we need tp find m:
$3=(\frac{3}{2})^{2m}$
$\leftrightarrow ln3=2m(ln3-ln2)$
$\leftrightarrow m=\frac{ln3}{2(ln3-ln2)}$
(*) is very nice!

[quote="hungkhtn"]Oh,your problem is:
If $x+y+z=1$ then $(xy)^m+(yz)^m+(zx)^m \le max(3,(\frac{3}{2})^{2m})$(*)
[/quote]
But how to prove (*)! Please tell me more.

$x \ge y\ge z$
You try to consider the function:
$f(u)=z^m((t-u)^m+(t+u)^m)+(t^2-u^2)^m$
It's quiet complex,to prove $f(u)\le f(0)$
 :)  :)  :)  :)  :)


Prove that if $ x,y,z$ are positive numbers with $ x\plus{}y\plus{}z\equal{}1$ then the inequality \[ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}\sqrt{12xyz}\leq 1\] holds true.

An easy one:
$ \sum x^{2}\plus{}\sqrt{12xyz}\leq 1\Longleftrightarrow\sum x^{2}\plus{}\sqrt{12xyz(x\plus{}y\plus{}z)}\leq x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2 (xy\plus{}yz\plus{}zx)\Longleftrightarrow 12xyz(x\plus{}y\plus{}z)\leq 4(xy\plus{}yz\plus{}zx)^{2}\Longleftrightarrow$
$ x^{2}yz\plus{}xy^{2}z\plus{}xyz^{2}\leq x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}x^{2}z^{2}$.Which is true by Muirherd. :lol:

Detailed proof, without using any "extra inequalities" : LaTex4.pdf.

Yes, It is easy but as an exercize you can try it using Blundon's inequality: http://www.mathlinks.ro/Forum/viewtopic.php?t=164095

We only need to prove
$ ab\plus{}bc\plus{}ac\geq\sqrt{3abc}$,which follows since
$ \left(x\plus{}y\plus{}z\right)^{2}\geq3\left(xy\plus{}yz\plus{}zx\right).$ :)


inequalities
calculus
derivative
inequalities unsolved

[quote="cezar lupu"]For what $a, b, c>0$ does the following inequality holds:

                  \[ \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq\ 6\cdot\frac{ a^{2}+b^{2}+c^{2}}{ab+bc+ca}.  \][/quote]

Hmm, it doesn't even hold for a, b, c being sides of a triangle (a = 1, b = 1, c = 1.9 being a counterexample)...

  Darij

I got a conclusion:
Without loss of generation,we assume that a+b+c=1,$a\geq b\geq c$.
Then if and only if
\[ \frac{b}{c}\geq\frac{1}{2}(x+\sqrt{x^2-4}) \]
and where
\[ x=\frac{1}{2a}[9a^2-11a+4+(3a-1)\sqrt{9a^2-12a+12}] \]
The given inequality holds.

I got it after a very very long compute process.I use derivative in my solution,I've no patient to post it here.
But it really has no skill in it.


For any two subsets $ A,B \subset \mathbb{R}$ of the real numbers let $ A \plus{} B$ be the set of all sums $ a \plus{} b$ where $ a \in A$ and $ b \in B$.

(i) Show that if $ A$ and $ B$ are open subsets of $ \mathbb{R}$ then so is $ A \plus{} B$
(ii) Show that if $ A$ and $ B$ are compact subsets of $ \mathbb{R}$ then so is $ A \plus{} B$
(iii) If $ A$ and $ B$ are closed subsets of $ \mathbb{R}$ is $ A \plus{} B$ also a closed subset of $ \mathbb{R}$?

[hide](i) it suffices to assume that $ A$ is open. Write $ A\plus{}B$ as the union of translated copies of $ A$.
(ii) consider $ A\plus{}B$ as the image of $ A\times B$ under a continuous map.
(iii) in the light of (ii), a counterexample must involve unbounded sets. Try to make $ A$ unbounded in the positive direction, and $ B$ unbounded in the negative direction. Further hint: consider partial sums of the harmonic series.[/hide]
Bonus: (iv) prove that if $ A$ and $ B$ are closed, and one of them is bounded, then the sum $ A\plus{}B$ is also closed.

I am confused. what is a continuous map? We haven't learned that in class yet.

Then you can argue like this: [hide]let $ (x_n)$ be a sequence of elements of $ A \plus{} B$. For each $ n$ there exist $ a_n\in A$ and $ b_n\in B$ such that $ a_n \plus{} b_n \equal{} x_n$. Choose a convergent subsequence $ (a_{n_k})$. From $ (b_{n_k})$ choose a convergent subsequence $ (b_{n_{k_j}})$. Then $ (x_{n_{k_j}})$ is a convergent subsequence of $ (x_n)$, and its limit lies in $ A \plus{} B$.[/hide]


Suppose that: $\ a^{log_bc}+b^{log_ca}=m$
Find the value of: $\ c^{log_ba}+a^{log_cb}$

 ;)  :huh:

[hide]one of the properties of logs is that $a^{log_bc}=c^{log_ba}$, so

$c^{log_ba}+a^{log_cb}=m$[/hide]

[quote="hydro"]$a^{log_bc}=c^{log_ba}$[/quote]

[hide="maybe out of topic"]Have you got a proof?[/hide]

[quote="paladin8"]$\log_b{c} \cdot \log_c{a} = \log_b{a}$ suffices.[/quote]

Still don't get it

[quote="joml88"]That comes from the change of base theorem which is a standard theorem that a quick google search could give you more information on.[/quote]

Instead of putting in that tone and way "a standar theorem...a quick google search could give you more info", why don't you post a link????

Um, wow. Either you're really lazy or you just want an excuse to add another post to your post count. Either way, it doesn't take that long to type "logs and change of base formula" on google.

Jose, your posts in this topic are obviously [b]spam.[/b] (Would you like that term looked up on Google as well?) Not only are you wasting other peoples' time, but you're ruining the quality of the forum.

Also, If you don't know the change of base theorem, I suggest you stay in a forum that's more appropriate for you.

Regarding this post,
[quote="Jose"]"Instead of putting in that tone and way 'a standar theorem...a quick google search could give you more info', why don't you post a link????"[/quote]you should stop posting in that tone. It's irritating. If you're going to [b]spam,[/b] at least make your [b]spam[/b] more courteous. Next time I see this, I'm deleting it.

I have to agree with everyone about this Jose. Why should Joe post a link for you? His tone was not inappropriate at all. He has a right to be upset b/c you have done this in various forums. Please do not complain this time about how everyone in your country does not speak English because this is a matter of common sense.

I'm sorry for offending you Jose.  I'll show you how hydro and paladin's comments work:

[hide]
We want to show that $a^{\log_b{c}}=c^{\log_b{a}}$.  We can see this as follows:

\begin{eqnarray*} a^{\log_b{c}} &=& a^{\frac{\log_a{c}}{\log_a{b}}\\ &=& \left(a^{\log_ac}\right)^{\frac 1{\log_ab}}\\ &=& c^{\frac 1{\log_ab}}\\ &=& c^{\log_ba} }\end{eqnarray*}[/hide]


Prove that
\[ \sum\frac{a^3\plus{}27bcd}{(b\plus{}c\plus{}d)^3} \ge \frac{11}{8}
\]
for any $ a,b,c,d \ge 0.$

[quote="can_hang2007"]Prove that
\[ \sum\frac {a^3 \plus{} 27bcd}{(b \plus{} c \plus{} d)^3} \ge \frac {11}{8}\]
for any $ a,b,c,d \ge 0.$[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-51-n-v-thach.html


That only works for 1/4, 1/2, and 3/4 I think.

[quote="Rep123max"]does anybody know how to make a fraction on microsoft word.[/quote]

The general solution is to use the Microsoft Equation Editor, accessed by Insert|Object from the pull-down menus, which then gives a list of choices of objects to insert, of which one is Microsoft Equation. That opens the equation editor that Microsoft licensed from another company.

Hope this helps!

In WordPerfect (which, by the way, is a far superior program to Word),  you should be able to find "Equation" on the "Insert" menu.  It is much the same as the Word Equation Editor.

rep123max microsoft installs the minimal stuff available if you don't select full install. insert your office disk and it should autorun or (use control panel add remove) and change your installation.

jbl microsoft office offers more linking with other software i have maple export into excell, and other programs use the word spell checker.

[quote="Osiris"]Sorry if I veered a bit off course, but that's why I don't use Word (or any other word processor, in fact) to type up math stuff. I always TeX it. Even though word processors offer (limited) capabilities for math typesetting, for complex expressions (sums, products, integrals, nested fractions, etc) it can get pretty confusing with built-in editors. ;)

TeX exports to PDF which is another big bonus.[/quote]

I like to use TeX too...except there's just a horrible amount of typing that we have to do...

You could also use MathType, which is in Word. I think it costs money, but it's got more options than equation editor.

I tried MathType on my calc teacher's computer a few years ago, and it seemed to be only very slightly more advanced than the free equation editor that comes with Word. I also recommend LaTeX, and I think that once you get used to it, it's much faster to write something with equations in LaTeX than it is to use the equation editor, especially since the equation editor messes up so much stuff in terms of formatting since Word isn't designed for math or even graphics.

Math Type is mostly nicer due to convenience then equation editor -- you can insert symbols as you're typing, you don't have to go into this seperate window.  Also, I think it's somewhat easier to use.  I do think it costs money, though.  And everything is easier to use than Tex if you (like me) don't know anything.


Let $ a,b,c$ be positive real numbers such that $ abc\equal{}1$. Determine the maximum value of
\[ P\equal{}\frac{1}{(a^2\minus{}a\plus{}1)^2}\plus{}\frac{1}{(b^2\minus{}b\plus{}1)^2}\plus{}\frac{1}{(c^2\minus{}c\plus{}1)^2}\]

[hide="@@Zaizai"]Em nên hạn chế spam lại Hoàng ạ![/hide]

yes, I know. Can you show me where is my mistake ?

[quote="zaizai-hoang"]My idea is considering the following function:
\[ f(x) = 1 - 2\ln{x} - {\frac {1}{(a^2 - a + 1)^2},x\ge 0
}\]
We have:
\[ f'(x) = - \frac {2(x^4 - x^3 + 3x^2 + 1)(x - 1)^2}{x(x^2 - x + 1)^3}\le 0, \forall x\ge 0
\]
So we have $ f(x)\le f(0) = 0$  :) Similar to $ b,c$ and adding up them we have $ P\ge 3$
Maybe I have made mistake in somewhere   :( I am trying to find out the maximum value[/quote]
The inequality $ P \ge 3$ is not true, Hoang. For example $ a=b=2,c=\frac{1}{4}$.
By the way, the problem asked to find the maximum value, indeed. :)

[quote="dduclam"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Determine the maximum value of
\[ P \equal{} \frac {1}{(a^2 \minus{} a \plus{} 1)^2} \plus{} \frac {1}{(b^2 \minus{} b \plus{} 1)^2} \plus{} \frac {1}{(c^2 \minus{} c \plus{} 1)^2}
\]
[hide="@@Zaizai"]Em nên hạn chế spam lại Hoàng ạ![/hide][/quote]

Nice and not easy,Can  :) We have a similar one,but not hard:

Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$.Prove that
$ \frac {1}{a^2 \minus{} a \plus{} 1} \plus{} \frac {1}{b^2 \minus{} b \plus{} 1} \plus{} \frac {1}{c^2 \minus{} c \plus{} 1}\le 3$[/quote]
[url=http://www.mathlinks.ro/viewtopic.php?t=199301]See here[/url] :)

[quote="zaizai-hoang"]It make me remember an old problem of you  :wink:
Let $ a,b,c,d > 0$ and $ a,b,c\ge 1,abcd \equal{} 1$
\[ \sum \frac {1}{(a^2 \minus{} a \plus{} 1)^2}\le 4
\]
Can you give a hint? For example: the maximum value (I think it is {MaxValue=3.561772058842,b=c=0.5058794241135})  :P[/quote]
My solution is the following :)

[quote="phanquockhanh"]These problems are too easy !!![/quote]
May I see your solution?

Ohh, it is really hard solution with a lot of cases  :wink: In the morning, I have tried it for one lesson at my class but I could not complete it. I have used some idea of four-varialbes and used to jensen's inequality. I devided them into some cases but not full as your solution. Anyway, If I have enought time to think, I will try find out another solution although I know it is not easy to find out another one. Thank you for a very nice solution  :)


Find all function $\large f: R to R$ satisfying:
$\large f(x^2+x+2)+f(x^2-x+2) = x^4+2x^3+4x^2+3x+4$ for all $\large x \in R$

its easy to see that this fuction dont have solution

just put x=-1 and u have f(2) + f(4)= 4
and put x=1 and u have f(2) + f(4)= 14
 :P


:lol: Yes it is another warm-up.

Solve the equation;

  3x - 5 = x + 4789

please don't solve it you are a master at this. It is just practice for younger ones.  :)

No it's not, it's called subtracting the same thing from an equation.  The subtraction property of equality.

Combining like terms is like $ 3x\plus{}x\equal{}4x$.

Combining like terms would be on only one side of the equation or in an expression.

The step inbetween where you have 3x - x , it needs combining like terms to get it to 2x.

[quote="Wickedestjr"]Yes, that would work also. When you go from 3x - 4794 = x to 2x - 4794 = 0, that is called [b]combining like terms[/b]. For those that don't know.[/quote]

But this ISN'T combining like terms.

Whatever, it does not matter. The question has been answered anyway.

I hope this doesn't count as reviving a thread, but...

Technically if you are subtracting x from 3x, isn't that just combining (3) times x and (-1) times x?

Yes, because $ \minus{}x\equal{}(\minus{}1)x$.

@waffle - If you have 3x + -x, basically what you are doing is factoring it into x(3 - 1), and then finding the difference, x(2), and then multiplying to  get 2x. And no, it is not reviving, because it was only about a week after. It would be reviving if it were about a month.

I believe you are misunderstanding what AIME15 said. Wickedestjr mistakenly called the Subtraction Property of Equality combining like terms. AIME15 corrected him.

[quote="mathgeniusxx"][quote="AIME15"]No it's not, it's called subtracting the same thing from an equation.  The subtraction property of equality.

Combining like terms is like $ 3x \plus{} x \equal{} 4x$.[/quote]

Its called combining like terms, regaurdless of subtracting or adding... UR WRONG![/quote]

Please don't post on this thread anymore. We have solved the question.


superior algebra
superior algebra solved

Let $A \in M_n(R)$ such that: $Trace(A)=Trace(A^2)=\cdots =Trace(A^n)=0$.
Prove that $A^n=0$

You can take $A$ in a base $B$: $A'=\mat(A)_B$ with $a_{ij}=0$ for $i>j$ so $Tr(A)=Tr(A')$
Now see the diagonals elements of $A'^k$, you will obtain $n$ equations with $n$ unknowns (the $a_{ii}$)

You cannot deduce that A is zero, the only thing you can deduce is that $A^n=0$. THis can be done using the Newton formulae and the fact that $Tr(A^k)=x_1^k+...+x_n^k$ where $x_i$ are the eigenvalues of A. You will find that $x_i=0$. In fact, the matrices that satisfy this condition are the nilpotent matrices.


algebra
polynomial
function
algebra unsolved

Let r be a nozero number. Find polynomail P of degree n or less so P(j)=r^j

Find a unique polynomial P of degree n so P(j)=1/j_1

p(j) = a^j is the definition of an exponential function base a.  I thought exponential functions were not equal to any polynomial for the reason that the power series for exponential functions is infinite i.e. there is no N such that all n > N, the coefficient of j^n is 0.  Any truncation of the series to just N terms results in an error O(j^(n+1)).

e.g. let a = e.  p(j) = e^j is certainly not a finite series of powers of j.

I am even more perplexed about the case a < 0, j not an integer e.g. let a = -1.  Now evaluate p at j = 1/2.  This results in multivaluedness.

Is j supposed to be a non-negative integer?  I still think you'd have problems constructing a polynomial which coincided with any given exponential function at infinitely many points.  Can you explain how I am either misreading this, or what is meant in more detail.  Many thanks!


geometry rectangle quadratics MATHCOUNTS

[hide]Um, well, if it's a 9x4 rectangle, surely you can get a 4x4 square.

The shapes are 4x4 square and 4x5 rectangle[/hide]

[hide]You can always make a square out of the smaller of the measures.

2343x323 rectangle has a 323x323 square.[/hide]

[hide="a solution"]Well,
$(x+3)(x+3)=99$
and we solve.
$x^{2}+6x+9=99$
$x^{2}+6x-90$
and we solve by quadratic equation.
$\frac{-6\pm\sqrt{36+360}}{2}=\frac{-6\pm\sqrt{396}}{2}=\frac{-6\pm4\sqrt{99}}{2}=-3\pm6\sqrt{11}=x$
Therefore,
$-3+6\sqrt{11}=x$
since x has to be greater than -3. We square to get the area, but I'm too lazy to do it right now. *yawn*[/hide]

[quote="most-wanted"]thanks once again
i think i am too lazy for this.
i have the same kind of question and struggling with it.
i hope if you can help once again.
A sqaure is enlarged by 3cm each sides, which cuase the area to be        $99cm^{2}$
what was the Original square's area and sides.[/quote]
[hide]
Well,
$(x+3)^{2}=99$
$x+3= \pm \sqrt{99}$
$x= \pm \sqrt{99}-3$

$x$ has to be positive, so we take the positive square root, and simplifing we have $\boxed{x=3\sqrt{11}-3}$, the original side length.

Squaring $3\sqrt{11}-3$, we have $99-18\sqrt{11}+9$, which simplifies to $\boxed{108-18\sqrt{11}}$ as the area.[/hide]

[quote="most-wanted"]A rectangle(a=9,b=4) should be divided into 2 shapes.
one of its shapes a Square,
i applied the following eqaution.
i don't what is wrong with that.

$(9-x )*(4-x ) = x^{2}$
$36-9x-4x+x^{2}= x^{2}$
$36-13x = 0$
$x = 36/13$
which is 2.769

i am sure that is really stupid answer.

can you please solve it ?[/quote]

What your equations are saying:
[code]
 _____________________________
|          2        |         |
|___________________|_________|
|                   |    1    |
|___________________|_________|
[/code]

Area 1 is a square with side length 2.769. And Area 1 = Area 2.

This topic is way too hard for Classroom Math. The second question was a typical MathCounts question.

[quote="most-wanted"]Thanks, but what if the number are bigger like
2343 X 323

how would i have got that ?[/quote]

[hide="you dont need algebra for this..."]
make a 323x323 square, because 323 is the smaller number. the next rectangle would be (2343-323)X323 which is 2020X323[/hide]


[quote="drst123"]a,b,c,d are positive real,
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d} \equal{} 4.
\]
Prove that:
\[ \sqrt [3]{\frac {a^{3} \plus{} b^{3}}{2}} \plus{} \sqrt [3]{\frac {b^{3} \plus{} c^{3}}{2}} \plus{} \sqrt [3]{\frac {c^{3} \plus{} d^{3}}{2}} \plus{} \sqrt [3]{\frac {d^{3} \plus{} a^{3}}{2}}\leq 2(a \plus{} b \plus{} c \plus{} d) \minus{} 4.
\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135374


What is the remainder when 359,975 is divided by 12?

Since $ 359976$ is divisible by $ 12$, $ 359975$ divided by $ 12$ will yield a remainder of $ \fbox{11}$.

But how do you know 359976 is divisible by 12?

because 359976 is divisible by three(add the digits) and 4(last two digits)  :D

But how would you solve this problem fast enough for Countdown round? I'm sure I wouldn't have the time to find that 359976 is divisible by 3 and 4.

Well, you could see that the last two digits are 76, which is divisible by 4.  Also, $3 + 5 + 9 + 9 + 7 + 6 = 39$ which is divisible by 3.  So $\boxed{11}$.

Also note 360000 is divisible by 12, and since 359975 is 25 less than 360000 the remainder when 359975 is divided by 12 is 11


Show that $ y \equal{} 3\sin x \plus{} 2\cos x$ is a sinusoid in the form $ y \equal{} a\sin (x \plus{} b)$ by finding a and b

y=a sin(x+b)
y=a cos(b)sin(x)+a sin(b)cos(x) =3sin(x)+ 2cos(x)
a cos b=3
a sin b=2
so
tan b=2/3
b= arctan 2/3
now we have a right triangle with sides 2, 3, and $ \sqrt {13}$, so sin b=2/$ \sqrt {13}$ and a=$ \sqrt {13}$
y=$ \sqrt{13}\sin{(x\plus{}\arctan{\dfrac{2}{3}})}$

[quote="funtwo"]y=a sin(x+b)
y=a cos(b)sin(x)+a sin(b)cos(x) =3sin(x)+ 2cos(x)
a cos b=3
a sin b=2
so
tan b=2/3
b= arctan 2/3
now we have a right triangle with sides 2, 3, and $ \sqrt {13}$, so sin b=2/$ \sqrt {13}$ and a=$ \sqrt {13}$
y=$ \sqrt {13}\sin{(x \plus{} \arctan{\dfrac{2}{3}})}$[/quote]

How did you end up with $ a$ as $ \sqrt{13}$? how can you know that the hypoteneuse is $ a$?

Notice that cos b = adjacent / hypotenuse, so hypotenuse * cos b = adjacent (in a right triangle). Similarly, hypotenuse * sin b = opposite. If you make a right triangle with sides $ a \cos b \equal{} x$, $ a \sin b \equal{} y$, then the hypotenuse will be equal to a, since $ a^2 \cos^2 b \plus{} a^2 \sin^2 b \equal{} a^2 (\frac {x^2}{a^2} \plus{} \frac {y^2}{a^2}) \equal{} a^2$ by Pythagoras. We know that $ a^2 \equal{} 13$ because $ x \equal{} 3, y \equal{} 2 \Longrightarrow x^2 \plus{} y^2 \equal{} 13$


geometry
trapezoid
circumcircle
parallelogram
analytic geometry
power of a point
radical axis

Let $ ABCD$ be a trapezium where $ AB // CD$ and $ F$ be the point on $ AB$ such that $ FC \equal{} FD$. $ AC$ intersects $ BD$ at $ E$. Prove that $ EF$ is the radical axis of $ (FAD)$ and $ (FBC)$, where $ (XYZ)$ stands for the circumcircle of $ \triangle XYZ$.

$ (P), (Q)$ are the circumcircles $ \odot(FAD), \odot(FBC).$ They cut $ AC$ at $ A, C$ and again at $ U, V,$ respectively. Power of $ A$ to $ (Q)$ is $ \overline{AV} \cdot \overline{AC} \equal{} \overline{AF} \cdot \overline{AB}$ and $ \frac{\overline{EA}}{\overline{EC}} \equal{} \frac{\overline{AB}}{\overline{CD}}$ $ \Longrightarrow$

$ \frac{\overline{EV}}{\overline{EA}} \equal{} 1 \minus{} \frac{\overline{VA}}{\overline{EA}} \equal{} 1 \minus{} \frac{\overline{AF} \cdot \overline{AB}}{\overline{CA}} \cdot \frac{\overline{AB} \plus{} \overline{DC}}{\overline{CA} \cdot \overline{AB}} \equal{}  1 \minus{} \frac{\overline{AF} \cdot (\overline{AB} \plus{} \overline{DC})}{\overline{CA}^2}.$

$ (P)$ cuts $ CD$ at $ D$ and again at $ X.$ $ AFDX$ is cyclic trapezoid, isosceles $ \Longrightarrow$ $ AX \equal{} FD \equal{} FC$ $ \Longrightarrow$ $ AFCX$ is a parallelogram $ \Longrightarrow$ power of $ C$ to $ (P)$ is $ \overline{CU} \cdot \overline{CA} \equal{} \overline{CD} \cdot \overline{CX} \equal{} \overline{CD} \cdot \overline{FA}$ $ \Longrightarrow$

$ \frac{\overline{EU}}{\overline{EC}} \equal{} 1 \minus{} \frac{\overline{UC}}{\overline{EC}} \equal{} 1 \minus{} \frac{\overline{CD} \cdot \overline{FA}}{\overline{AC}} \cdot \frac{\overline{BA} \plus{} \overline{CD}}{\overline{AC} \cdot \overline{CD}} \equal{}  1 \minus{} \frac{\overline{FA} \cdot (\overline{BA} \plus{} \overline{CD})}{\overline{AC}^2}.$

As a result, $ \frac{\overline{EV}}{\overline{EA}} \equal{} \frac{\overline{EU}}{\overline{EC}}$ or $ \overline{EV} \cdot \overline{EC} \equal{} \overline{EU} \cdot \overline{EA},$ powers of $ E$ to $ (P), (Q)$ are equal, $ E$ is on their radical axis.

[b][size=100][color=DarkBlue]LEMMA. – Through the vertex $ A$ of a given isosceles triangle $ \bigtriangleup ABC,$ we draw a line parallel to its base $ BC$ and let $ D,\ E$ two arbitrary points on it, from both sides of $ A.$ We denote as $ K,$ the intersection point of the circucircles $ (O_{1}),\ (O_{2}),$ of the triangles $ \bigtriangleup ABD,\ \bigtriangleup ACE$ respectively, the other than $ A$ and let be the points $ M\equiv DE\cap CK$ ( between $ A,\ D$ ) and $ N\equiv DE\cap BK.$ Prove that $ \frac {AM}{MD} \equal{} \frac {AN}{NE}.$[/color][/size][/b]

[b]PROOF OF THE PROPOSED PROBLEM.[/b] - Let $ (O_{1}),\ (O_{2})$ be, the circucircles of the triangles $ \bigtriangleup FAD,\ \bigtriangleup FBC$ respectively and let $ K$ be, their intersection point, the other than $ F.$

We denote as $ M,\ N,$ the intersection points of the side-segment $ AD$ of the given trapezium $ ABCD,$ from the line segments $ CK,\ DK$ respectively and it is enough to prove that the points $ F,\ K,\ E,$ are collinear.

$ \bullet$ We consider the pencil $ C.FMAD$ which is intersected from the line segment $ FA\parallel CD$ and so, we have $ (C.FMAD) \equal{} (F,M,A)$ $ ,(1)$ as well.

Similarly, because of the pencil $ D.FNBC$ is intersected from the line segment $ FB\parallel DC,$ $ \Longrightarrow$  $ (D.FNBC) \equal{} (F,N,B)$ $ ,(2)$

Based on the above [b]Lemma[/b], we have that $ (F,M,A) \equal{} (F,N,B)$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ (C.FMAD) \equal{} (D.FNBC)$ $ ,(4)$

From $ (4)$ and because of the pencils $ C.FMAD,\ D.FNBC,$ have the line segment $ CD$ as their common ray, we conclude that the points $ F\equiv CF\cap DF,\ K\equiv CM\cap DN$ and $ E\equiv CA\cap DB,$ are collinear and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Mihalis Lambrou.[/size][/b]

kostas Vittas.

PS. I will post here later the proof of the above [b]Lemma[/b] I have in mind.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. – Through the vertex $ A$ of a given isosceles triangle $ \bigtriangleup ABC,$ we draw a line parallel to its base $ BC$ and let $ D,\ E$ two arbitrary points on it, from both sides of $ A.$ We denote as $ K,$ the intersection point of the circucircles $ (O_{1}),\ (O_{2}),$ of the triangles $ \bigtriangleup ABD,\ \bigtriangleup ACE$ respectively, the other than $ A$ and let be the points $ M\equiv DE\cap CK$ ( between $ A,\ D$ ) and $ N\equiv DE\cap BK.$ Prove that $ \frac {AM}{MD} \equal{} \frac {AN}{NE}.$[/color][/size][/b][/quote]
[b][size=100]PROOF OF THE LEMMA.[/size][/b] - Let be the points $ B'\equiv (O_{1})\cap BC$ and $ C'\equiv (O_{2})\cap BC$ and $ P\equiv DB'\cap EC'.$

It is easy to show that the triangle $ \bigtriangleup PED$ is isosceles, because of $ \angle DEP \equal{} \angle ACB \equal{} \angle ABC \equal{} \angle EDP$ $ ,(1)$

So, we have that the trapezium $ DB'C'E$ is isosceles and hence, it is cyclic with circumcircle so be it $ (O).$

We conclude now, that the line segment $ AK,$ passes through the point $ P,$ as the radical center of the circles $ (O),\ (O_{1}),\ (O_{2})$ and let be the point $ X\equiv BC\cap AKP.$

$ \bullet$ From $ (1)$ and because of $ \angle ABC \equal{} \angle BAD,$ we have that $ PE\parallel AB$ and similarly, $ PD\parallel AC.$

From $ BC\parallel DE$ $ \Longrightarrow$ $ \frac {AM}{AN} \equal{} \frac {XC}{XB}$ $ ,(2)$

From the similar triangles $ \bigtriangleup ABC,\ \bigtriangleup PED$ and because of $ \angle BAX \equal{} \angle EPA$ and $ \angle XAC \equal{} \angle APD,$ we conclude that $ \frac {XC}{XB} \equal{} \frac {AD}{AE}$ $ ,(3)$

From $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {AM}{AN} \equal{} \frac {AD}{AE}$ $ \Longrightarrow$ $ \frac {AM}{MD} \equal{} \frac {AN}{NE}$ $ ,(4)$ and the proof is completed.

Kostas Vittas.

We define rectangular coordinates as follow

$ A:(b,h) \ , \  F:(0,h) \ , \ B:(c,h) \ , \ C:(a,0) \ , \ D:(\minus{}a,0)$

Coordinates of the circumcenters $ O_1$ and $ O_2$ of $\triangle AFB$ and $\triangle BFC$ are given by

$ O_1 \left (\frac{b}{2} \ , \ \minus{}\frac{a(a\plus{}b)}{2h} \plus{}\frac{h}{2} \right) \ , \ O_2 \left (\frac{c}{2} \ , \ \frac{a(c\minus{}a)}{2h} \plus{}\frac{h}{2} \right)$

$\Longrightarrow \delta  \overline{O_1O_2}\equal{}\frac{a(b\plus{}c)}{h(c\minus{}b)}$

Lines $ AC$ and $ BD$ intersect at $ P \left ( \frac{a(b\plus{}c)}{c\minus{}b\plus{}2a} \ , \ \frac{2ah}{c\minus{}b \plus{}2a} \right)$

$\Longrightarrow \delta \overline{FP}\equal{}\frac{h(b\minus{}c)}{a(a\plus{}b)} \ \Longrightarrow \  \delta \overline{O_1O_2} \cdot \delta \overline{FP}\equal{}\minus{}1,$

which means that $ O_1O_2 \perp FP$ $\Longrightarrow$ $FP$ is radical axis of  $\odot (AMB),\odot( BMC)$

Let (ADF) cuts AC at A and M while (BCF) cuts it at C and N. From CF = DF we get $ \angle BFC \equal{} \angle AFD, hence \angle BNC \equal{} \angle AMD$ ( 1 ), i.e. BN || DM, consequently $ \frac {NE} {ME} \equal{} \frac {BE} {DE}$  ( 2 ), but  $ \frac {BE} {DE} \equal{} \frac {AE}{CE}$  ( 3 ), that is, $ \frac {NE}{ME} \equal{} \frac {AE}{CE}$  ( 4 ). The relation (4) shows that E belongs to the radical axis of the subject circles.

Best regards,
sunken rock


If $a,b,c$ are positive real numbers and $abc=2$ prove that

$a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b}$.

Could this one solved by bunching ?Thanks

By AM-QM and by Shur:
${\sum a\sqrt{b+c}=\sum \sqrt{a^2b+ca^2} \leq \sqrt{3(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2)}} \leq \sqrt{3(a^3+b^3+c^3+3abc)}$

So it is enough to prove that:
$a^3 + b^3 + c^3 \geq \sqrt{3(a^3+b^3+c^3+3abc)}$

But after squaring and homogenizing we get:
$2 \sum a^6 + 4 \sum a^3b^3 \geq 3 \sum a^4bc + 9(abc)^2$

And this is true because following inequalities holds:
$3\sum a^3b^3 \geq 9(abc)^2$
and
$2 \sum a^6 + \sum a^3b^3 \geq 3\sum a^4bc$

First one is pure AM-GM and for the second one uses inequalities like $2a^6 + b^3c^3 \geq a^4bc$ which are also true by AM-GM.


I competed in bronze division being so confident, and after the contest found myself so broken and so empty within!
I solved two probs and solved another but somehow it didnt fit in time limit.... I wish you guys can show some mercy on this poor soul by helping me... here is the problemset of usaco open bronze division -->

[quote]
*********************************************************************
                           BRONZE PROBLEMS
**********************************************************************
                  Five problems numbered 11 through 15
**********************************************************************

Problem 11: Bedtime Reading, I [Traditional, 2006]

Farmer John was performing his nightly bedtime reading duties for
Bessie. "Oh, this gives me a headache," moaned Bessie.

"But it's just simple number theory," replied FJ. "Let's go over
it again. The sigma function of a number is just the sum of the
divisors of the number. So, for a number like 12, the divisors are
1, 2, 3, 4, 6, and 12. Summing them we get 28."

"That's all there is to it?" asked Bessie.

"Yep." replied FJ. "Perhaps someone will write a program to calculate
the sigma function of an integer I (1 <= I <= 1,000,000)."

PROBLEM NAME: br1

INPUT FORMAT:

* Line 1: A single integer, I

SAMPLE INPUT (file br1.in):

12

OUTPUT FORMAT:

* Line 1: The sum of all of I's divisors.

SAMPLE OUTPUT (file br1.out):

28

**********************************************************************

Problem 12: Eenie Meenie Miney Moe [Rob Kolstad, 2006]

"Eenie, meenie, miney, moe... catch a heifer by the toe," an
exasperated Bessie muttered. She could never get the syllables right
when choosing calves for game playing. Today she is trying to see
which of N (1 <= N <= 2,300) heifers conveniently numbered 1..N
will get a special hunk of hay made from tender spring grass.

Thus, Bessie is leaving it up to you. You will be provided a list
of integers. This list is of length L (1 <= L <= 10) and describes
a sort of "elimination sequence" containing integers in the range
(1..N). If the N heifers are lined up in order in a circle, then
the elimination count starts at heifer number 1. The L1th
heifer is eliminated. Start at the next heifer, the count proceeds
for L2 more heifers to find the baby cow to be eliminated.
Both the line of heifers and the elimination sequence are circular,
wrapping around as necessary.

By way of example, consider a set of 8 heifers and an elimination
sequence of {5, 3}. Initially, the 8 heifers form a circle as
shown in Figure A.

Starting at heifer 1, count 5 cows and eliminate the 5th one (Figure
B). Now count 3 more and eliminate the 3rd cow, which is number 8
(Figure C).

The elimination list is exhausted so it is recycled. Count five
more heifers and eliminate the fifth one, which is number 6 (Figure
D) since number 5 is no longer present. Now count three more to
eliminate the third cow, number 2 (Figure E). Recycle the list and
count five more to eliminate number 3 (Figure F). Count three more
to eliminate number 1 (Figure G), and finally count five more to
eliminate number 4 (Figure H).

Heifer 7 is left standing.

PROBLEM NAME: eenie

INPUT FORMAT:

* Line 1: Two space-separated integers: N and L

* Line 2: L space-separated integers (range 1..N) that respectively
        compose the elimination sequence.

SAMPLE INPUT (file eenie.in):

8 2
5 3

OUTPUT FORMAT:

* Line 1: A single integer that is the number of the heifer left
        standing (who will receive the special hay treat)

SAMPLE OUTPUT (file eenie.out):

7

**********************************************************************

Problem 13: Cow Pizza [Traditional, 2006]

Oh how the cows love their pizza! Even though they are picky, they
do love variety, too. They order from the local pizza parlor that
features T (1 <= T <= 20) different toppings in addition to a
complete set of soft drinks and healthy salads.

The toppings are conveniently numbered 1..T so the cows can order
by number. Your job is to calculate how many possible pizzas can
be created given that some cows will not tolerate various combinations
of toppings (e.g., some cows simply will not eat Anchovies or the
combination of Mushrooms and Asparagus).

Given a set of N (1 <= N <= 52) constraints, figure out how many
pizzas can be made using all possible combinations of the ingredients
(which, of course, includes no ingredients at all). Each constraint is
a set of numbers of size 1..T that lists the ingredients that disqualify
a pizza from being considered.

A constraint like "5 3" means that no pizza can contain ingredient
#5 and also ingredient #3. This means a pizza with ingredients 3,
5, and 6 will not be counted as acceptable.

PROBLEM NAME: pizza

INPUT FORMAT:

* Line 1: Two space-separated integers: T and N

* Lines 2..N+1: Each line describes a constraint using space-separated
        integers. The first integer is the number of ingredients in
        the constraint, Z (1 <= Z <= T). The subsequent Z integers
        (which are unique) list the ingredient(s) whose combination
        disqualifies a pizza from consideration for the cows.

SAMPLE INPUT (file pizza.in):

6 5
1 1
2 4 2
3 3 2 6
1 5
3 3 4 6

INPUT DETAILS:

Six ingredients numbered 1..6. Five constraints: No pizza can contain
ingredient 1, no pizza can contain ingredients numbered 2 and 4, etc.

OUTPUT FORMAT:

* Line 1: A single integer that is the total number of pizzas that can
        be created using the number of ingredients and constraints.

SAMPLE OUTPUT (file pizza.out):

10

OUTPUT DETAILS:

This includes pizzas with: no ingredients; ingr 2; ingr's 2 & 3;
ingr's 2 & 6; ingr 3; ingr's 3 & 4; ingr's 3 & 6; ingr 4; ingr's 4
& 6; and ingr 6.

**********************************************************************

Problem 14: Bedtime Reading, II [Traditional, 2006]

Bessie rubbed her eyes again. "The headache's gone. Can you go
over that part about number chains again, Farmer John?"

"Sure!" FJ exclaimed. "Let's continue with that sigma function, the
one that is the sum of a number's divisors. First, though, let's
exclude the number itself from the sum and call our new function
bess(), in your honor. Now bess(12) = 1+2+3+4+6 = 16. If we continue
and compute bess(16), we find it is 1 + 2 + 4 + 8 = 15. Then bess(15)
= 1 + 3 + 5 = 9 and bess(9) = 1 + 3 = 4. Then bess(4) = 1 + 2 = 3
and bess(3) = 1. For better or worse, bess(1) is 0. Not much of a
chain, really."

"But look," continued FJ. "We see that bess(6) = 1 + 2 + 3 = 6 and
bess(28) = 28. Those form little tiny chains called 'perfect numbers'.
Moving to bigger numbers, bess(220) = 284 and -- get this -- bess(284)
= 220 -- a chain of length 2! Some chains are even longer."

"It's so late; I'm not sure I can do the math," mooed Bessie.

"No problem," said FJ. "We'll ask the programmers to write a program
that accepts a start and end integer (1 <= start <= end <= 1,000,000)
and finds all the possible number chains whose starting value is
in the interval (start..end). No intermediate value is allowed to
exceed 2,000,000, though some might be larger than the 'end' integer.
No chain is longer than 50 elements. Print only chains whose smallest
element comes first (i.e., no intermediate value is less than the
start value of the chain). We must ask them, though, to compute the
answers with their program and not just print them. We'll supply
test data that is reasonable to finish in 1.0 CPU seconds."

PROBLEM NAME: br2

INPUT FORMAT:

* Line 1: Two space-separated integers: start and end

SAMPLE INPUT (file br2.in):

200 1200

OUTPUT FORMAT:

* Lines 1..?: Each line contains a space-separated list that
        represents one chain of numbers calculated using the described
        bess() function.

SAMPLE OUTPUT (file br2.out):

220 284
496
1184 1210

**********************************************************************

Problem 15: Cows on a Leash [Brian Dean and Rob Kolstad, 2006]

Farmer John has forced indignity onto the N (1 <= N <= 32,000) cows:
he has purchased a number of leashes and has leashed the cows to
stakes (conveniently located at integer locations) next to a long,
linear fence that is no longer than 5,300,000 meters and runs
east-west. Every single cow is straining her leash as much as she
can to the east (though never beyond the end of the fence).

Happily, the cows have acquired an ally: FJ's wife thinks that
leashes are barbaric and has come to the fence with her dull butcher
knife. She can perform only a few cuts with it and wants to do the
best job she can. She can locate herself halfway between two integer
points along the fence and cut all the leashes that cross in front
of her at that place.

Given the length of each leash and the location of its stake in the
ground, what is the minimal number of cuts she must perform to free
the cows?

PROBLEM NAME: leash

INPUT FORMAT:

* Line 1: A single integer, N

* Lines 2..N+1: Each line contains two space-separated positive
        integers that describe a leash. The first is the location of
        the leash's stake; the second is the length of the leash.

SAMPLE INPUT (file leash.in):

7
2 4
4 7
3 3
5 3
9 4
1 5
7 3

INPUT DETAILS:

Here is a simple graphical representation of the locations of the leashes:
                  1 1 1 1
1 2 3 4 5 6 7 8 9 0 1 2 3
-------------------------
. 111111111 . . . . . . .
. . . 222222222222222 . .
. . 3333333 . . . . . . .
. . . . 4444444 . . . . .
. . . . . . . . 555555555
66666666666 . . . . . . .
. . . . . . 7777777 . . .

OUTPUT FORMAT:

* Line 1: A single integer that is the minimum number of cuts so that
        each leash is cut at least once.

SAMPLE OUTPUT (file leash.out):

2

OUTPUT DETAILS:

Cutting at 5.5 cuts leashes 1, 2, 3, 4, and 6. A second cut at 9.5 cuts
leashes 5 and 7. Other locations are surely possible, but no single cut can
penetrate leashes 5 and 6, so at least two cuts are necessary.

**********************************************************************
[/quote]

The first problem is silly.
Second problem, i used linked list and got the job done very fast.
Third problem made me wondered, i thought i will think about it after solving the four. But the thinking didnt happen.
Fourth one wasted my three hours! I tried and tried but as i checked till $\frac{n}{2}$, nothing could help me! Later on i checked till $\sqrt{n}$ and got only 4 testcase done. One was WA, and all others were TLE.
Fifth one, i could smell some brute force, i can put flags, then find the point of maximum intersecting. max++, then remove all the leashes from there. Then again find the maximum ... max++ ... until there is no leash left to intersect. But look at the range, this algo wont fit for the real good testcases!   :mad:

I am not happy with my work... what can i do ?

Check here: http://ace.delos.com/OPEN06.htm

The analysis are available now... :)

http://train.usaco.org

Best way to practice, took me from Bronze to USAICO in about a year.


group theory
abstract algebra
advanced fields
advanced fields unsolved

Are there any additive subgroups of $ \mathbb R$ completely ordered (except the trivial ones) s.t. every point is an accumulation point?

What are the trivial ones¿
And you may take a Hamel basis $ B$ of $ \mathbb R$ over $ \mathbb Q$ and take those $ r \equal{} \sum a_i b_i$ with $ a_i \in \mathbb Q$ and $ b_i \in B$ such that $ a_j\equal{}0$ for one fixed $ j$.


A segment of length $ 2$ is divided into $ n$, $ n\geq 2$, subintervals.  A square is then constructed on each subinterval.  Assume that the sum of the areas of all such squares is greater than $ 1$.  Show that under this assumption one can always choose two subintervals with total length greater than $ 1$.

The solution becomes apparent from drawing a diagram. Put the largest square on the far left, the second largest right next to it, largest having length x, second having length y.

Then, if $ x \plus{} y \le 1$ then consider the maximal possible area of the squares. Clearly, it's just $ x^2 \plus{} (2 \minus{} x)y$, since no square past the y square can have height greater than y, so the maximal area looks like x^2 on the left, and a long rectangle with height y extending the rest of the way.

So, we just have to show that $ x^2 \plus{} (2 \minus{} x)y \le 1$ where $ x\ge y$ and $ x \plus{} y\le1$. Let $ x \plus{} y \equal{} k\le1$. Then, $ x\equal{}k/2\plus{}m$ $ y\equal{}k/2\minus{}m$ for some $ 0 \le m \le k/2$

$ k^2/4 \plus{} km \plus{} m^2 \plus{} k \minus{} 2m \minus{} k^2/4 \plus{} m^2 \equal{} 2m^2 \plus{} km \plus{} k \minus{} 2m$

So we now have to show that

$ 2m^2 \plus{} km \plus{} k \minus{} 2m \le 1$

But $ 2m^2 \plus{} km \plus{} k \minus{} 2m \le 2m^2 \plus{} m \plus{} 1 \minus{} 2m \equal{} 2m^2 \minus{} m \plus{} 1 \equal{} m(2m \minus{} 1) \plus{} 1 \le 1$ since $ m \le k/2 \le 1/2$

So $ 2m \minus{} 1 \le 2*1/2 \minus{} 1 \equal{} 0$ so $ m(2m \minus{} 1)$ is non-positive, so we are done.


limit
function
logarithms
algebra unsolved
algebra

solve the equation
$x^{x+1}=(x+1)^{x}$
x >0 and in R

your equation is equivalent to $\frac{\ln(x+1)}{\ln(x)}=\frac{x+1}{x}$
see that $\lim_{x\to\infty}\frac{\ln(x+1)}{\ln(x)}=\lim_{x\to\infty}1+\frac{\ln(1+\frac{1}{x})}{\ln(x)}=\lim_{x\to\infty}\frac{x+1}{x}=1$ kepping that and using the graph of the two functions $\log_{x}(x+1)$
and $\frac{x+1}{x}$ wich inform that there is an intersection in $\infty$
so there is a solution but in $\infty$.

$x\sim 2.293$. What else do you want?
$f(x)=x^{1\over x}=e^{\ln x\over x}$ is strictly increasing on $(0,e]$ and strictly decreasing on $[e,\infty)$
with $\lim_{x\to 0}f(x)=0$ and $\lim_{x\to \infty}f(x)=1$
and $x^{x+1}=(x+1)^{x}\Leftrightarrow f(x)=f(x+1)$.
So $x\sim 2.293$ is only solution.


function
inequalities
induction
number theory theorems
number theory

I am having problem with one step in the famous paper "The product of consecutive integers is never a power". The paper can be found here: http://www.math-inst.hu/~p_erdos/1975-46.pdf, but I will try to describe the step I got stucked with as best as I can.

The authors are trying to prove the following estimate (labelled $ (13)$): for each $ k \geqslant 30000$ it holds\[ \prod_{i \equal{} 1}^{k \minus{} \pi(k)} a_i > k!,\]where $ \pi(k)$ is prime-counting function, and $ a_i$ sequence defined earlier in the paper. Then they manage to prove that $ a_i \geqslant 4.3402(i \minus{} 1492)$ (and two other estimates for smaller indices, but this is the key one), and say: "With these inequalities, a routine calculation using Stirling's formula suffices to verify $ (13)$ when $ k \equal{} 30000$, and $ (16)$ ensures that $ (13)$ holds when $ k > 30000$" (where $ (16)$ is the mentioned estimate for $ a_i$).

The second part of the quoted sentence is the step I am having problem with. As I understand, they are trying to use induction, saying that the estimate for $ a_i$ ensures that stepping from $ k$ to $ k \plus{} 1$ adds larger factor to the left side than to the right side. However, if $ k \plus{} 1$ is a prime number, then the left side remains exactly the same, since $ k \minus{} \pi(k) \equal{} (k\plus{}1) \minus{} \pi(k\plus{}1)$.

Am I getting something wrong? Thank you in advance for your help.

So step from $ k$ to $ k \plus{} 2$.  (I haven't checked to see if that works.)

Thank you for your response, but I also thought about this and immediately saw that it didn't work. Indeed, stepping from $ k$ to $ k\plus{}2$ multiplies the right side with $ (k\plus{}1)(k\plus{}2)\equal{}k^2\plus{}3k\plus{}2$, and we obviously can not guarantee that only [b]one[/b] new $ a_i$ will beat this (since the estimate we have is only of first degree in $ k$).


Consider $ P(X) \equal{} \det\left [\begin {array} {ccccc} 1890 \plus{} X & 19 \plus{} X & 19 \plus{} X & ... & 19 \plus{} X \\
5 \plus{} X & 1891 \plus{} X & 19 \plus{} X & ... & 19 \plus{} X \\
5 \plus{} X & 5 \plus{} X & 1892 \plus{} X & ... & 19 \plus{} X \\
... & ... & ... & ... & ... \\
5 \plus{} X & 5 \plus{} X & 5 \plus{} X & ... & 2008 \plus{} X \end {array} \right ]$.

This is a linear polynomial in $ X$, $ P(X) \equal{} aX \plus{} b$ so to evaluate it one needs two values: say $ X \equal{} \minus{} 5, X \equal{} \minus{} 19$.
$ S \equal{} P( \minus{} 5) \equal{} 1885\cdots 2003$. $ T \equal{} P( \minus{} 19) \equal{} 1871\cdots 1989$. Now use the formula $ P(0) \equal{} b \equal{} {\frac {x_{{1}}T \minus{} Sx_{{2}}}{ \minus{} x_ {{2}} \plus{} x_{{1}}}} \equal{} {\frac {1}{14}}\,(19S \minus{} 5T).$


"Bella's Lullaby" from Twilight

[youtube]4PZb0Ecl1Sw[/youtube]

Nice.. I like this one

http://www.youtube.com/watch?v=dZn_VBgkPNY

This is such a enchanting piece!

I love the Bernstein conducting Tchaikovsky Sym 5

I also like "It ends tonight" by All-American Rejects.

Pirates of the Caribbean : The Black Pearl Played on Piano

[youtube]itK6FvWde1c[/youtube]

You must enable BBCode.

Either that, or your browser just doesn't support youtube embed/your filter blocked youtube.

[color=orange][b]Toccata & Fugue[/b][/color] is one of my favourites, too

If it includes non-classical music, then probably Crazier by Taylor Swift...
http://www.youtube.com/watch?v=KfdxeymNwjU&feature=PlayList&p=27886EC594CA51FD&index=5&playnext=6&playnext_from=PL
(sorry it's so long; it's in my playlist) :)

Jonathan Rhys Meyers-Something Inside

http://www.youtube.com/watch?v=LOO2N1mwD8o

I like :

Animusic

1. [url]http://www.youtube.com/watch?v=3ahoqR6OGdM[/url] -Future Retro
2. [url]http://www.youtube.com/watch?v=Cgt4DEBQy50&feature=related[/url] -Aqua Harp
3. [url]http://www.youtube.com/watch?v=OROAGNbsikQ&feature=related[/url] -Acoustic Curves

RuneScape

I like songs from RuneScape, mainly:

Harmony, YesterYear, Scape Main, Scape Summon, Start, Doorways, and The Mentor.

You can search on Youtube.
Some good channels are Bigstingman, RuneScapeSoundtrack, and RunescapeOST.

I like the Fantasie Impromptu played by Yundi Li. I also like the La Campella (spelling error) also played by Yundi Li.


Let A, B, C are angels in the triangle. Prove
$cos{\frac{A}{3}}+cos{\frac{B}{3}}+cos{\frac{C}{3}}>\frac{5}{2}$


a lamp post stands by the side of a straight road, the nearest point A on the road from the lamp post being at a distance $a$.A man of height $b$ is walking along the road . When he is at B , the length of his shadow is 3/2 times the length of the shadow when he is at C, the middle point of AB. Find the height of the lamp post if the length of the shadow when he is at C is $a$

My book doesnt say where is point B , but i guessits a point somewhere before A , right??

any way my answers's not matching with the book's  :(  :(  :(   :?  :?  :mad:  :mad:  :mad:

Something is wrong with that problem...the guy's shadow and the distance the lamp is from the road is independent of the height of the lamp post.

gee calc,I dont know, maybe my book is wrong , but that's a remote possibilty  :? . Lets see what others have to say..........

Oops thought it meant the shadow by the sun, not by the lamp....

Anyway I got $ab(1+2\sqrt{\frac{3}{7}})$. I called AB $\phi$ and the lamp's height h and set up triangles. The triangles gave me two proportions: \[ \frac{h}{a+\sqrt{{\phi}^2+a^2}} = \frac{b}{a} \] and \[ \frac{h}{\frac{3a}{2}+\sqrt{4{\phi}^2+a^2}} = \frac{2b}{3a} \]. I then solved and found that ${\phi}^2 = \frac{5a^2}{7}$ and plugged in to get my answer.

calc , the answer is $b(1+2\sqrt{3}/\sqrt{7})$

when you plug 5a^2/7in the first equation, $a$ would get cancelled with the $a$on the RHS :P , i think you overlooked that  ;)  :)

nice soln :)


In one step it is possible to change a number $ a$ on the blackboard to $ \frac{a}{2}$ or $ 1\minus{}a$. Prove that for any given $ a$, it is possible to obtain on a blackboard some number $ b$ which differs from $ \frac{1}{2007}$ by no more than $ \frac{1}{2^{2007}}$

I actually think that we can get arbitrarily close to any number $ \alpha\in [0,1]$.  Given some finite subset of natural numbers $ \{ n_1,n_2,\dots , n_k\}$, denote the sum $ S_k\equal{}\sum_{i\equal{}1}^k (\minus{}1)^{i\plus{}1}2^{\minus{}n_i}$.  I claim that for any $ \epsilon>0$, there is some appropriately chosen subset of natural numbers $ \{n_1,\dots, n_k\}$ where $ |S_k\minus{}\alpha|<\epsilon$.

We proceed by induction.  First, there is some non-negative integer $ n_1$ such that $ \frac{1}{2^{\minus{}(n_1\plus{}1)}}<\alpha\le\frac{1}{2^{\minus{}n_1}}$.  Thus if $ S_1\equal{}\alpha$ or $ \epsilon\ge 2^{\minus{}(n_1\plus{}1)}$, we are finished.  Otherwise, let $ n_2\equal{}min \{N\in\mathbb{N}: |S_1\minus{}\alpha|\le 2^{\minus{}N}\}$.  Then we can be sure that $ n_2>n_1$, so that $ |S_2\minus{}\alpha|<2^{\minus{}(n_2\plus{}1)}.$  Continuing inductively, we will eventually produce a subset such that $ |S_k\minus{}\alpha|<2^{\minus{}(n_k\plus{}1)}<2^{\minus{}k}<\epsilon$.  This completes the induction.

Let $ \{n_1,\dots, n_k\}$ be given so that $ |S_k\minus{}\alpha |<\frac{\epsilon}{2}$.  Hence for any $ \epsilon>0$, we know that there is some integer $ N$ such that $ \left |\frac{a}{2^N}\right |<\frac{\epsilon}{2}$, and then if we let $ n_{k\plus{}1}\equal{}min \{N, n_k\plus{}1\}$, we have that

$ |S_k\plus{}(\minus{}1)^k\frac{a}{2^{n_{k\plus{}1}}}\minus{}\alpha|\le |S_k\minus{}\alpha|\plus{}\frac{|a|}{2^{n_{k\plus{}1}}}<\frac{\epsilon}{2}\plus{}\frac{\epsilon}{2}\equal{}\epsilon$.

Because any sequence of moves on the blackboard is a one-to-one correspondence with finite subsets of natural numbers and the appropriate sum $ S_k$, this proves the desired result.


In an acute angled triangle ABC, measure of angle A is 30 degree.  H is the ortho centre and M is the mid point of BC. On the line HM, take a point T such that HM=MT. Show that AT= 2BC [/code]

[hide]
Put $ABC$ on the complex plane such that the circumcenter of $ABC$ is the origin $O$. Then the points $A,B,C$ are on a circle centered at $O$.

We know $H = A+B+C$ and $M = \frac{B+C}{2}$. The we calculate $T = M+(M-H) = -A$, so $T$ is opposite $A$ on the circle. Then the length of $AT = |A-T| = |2A| = 2|A|$, the length of the diameter of the circle.

But since $\angle A = 30$, we have $\angle BOC = 60 \Rightarrow BOC$ is equilateral so the length of $BC$ is one radius. Then $AT = 2BC$ as desired.

The same method can be applied with vectors, see my [url=http://wangsblog.com/jeffrey/]blog[/url] for more information.[/hide]


algebra
function
domain
vector
real analysis
real analysis unsolved

Let $ u$ be a $ C^1$ solution of:

$ a(x,y)u_x \plus{} b(x,y) u_y \equal{} \minus{}u$

defined in a neighborhood of the unit disk $ D_1$ in the xy-plane. Assume that $ xa(x,y) \plus{} yb(x,y) > 0$ on the boundary of $ D_1$. Show that $ u$ must be identically $ 0$ in $ D_1$.

maybe something along these lines:
write it as $ A(x,y).\nabla u \equal{} \minus{}u$ where $ A(x,y).\nabla(x^2\plus{}y^2) > 0$. Notice that if $ u$ has a local  extremum, then this extremum is null. Moreover, if $ (x,y) \in \partial D_1$ is such that $ u(x,y)>0$, then one can see from the condition on $ A$ that $ u$ grows when one goes inside $ D_1$ (there exists $ (x_n,y_n) \in D_1$, arbitrarily close to $ x,y$ st. $ u(x_n,y_n)>u(x,y)$). Similarly, if $ (x,y) \in \partial D_1$ satisfies $ u(x,y)<0$, then $ u$ decreases when one goes inside $ D_1$. So if there exists a positive value on the boundary of $ D_1$, then the maximum of $ u$ is achieved inside $ D_1$, hence the maximum of $ u$ on $ D_1$ equals $ 0$, a contradiction since one assumed that there was a point on the boundary of $ D_1$ where $ u$ was positive. One shows similarly that it cannot exist any negative value on the boundary. Hence $ u$ is null on the boundary, and the same argument shows that $ u$ is null inside $ D_1$

[quote="alekk"]maybe something along these lines ...[/quote]
Here's a streamlined version of this argument:

First set $ w = u^2$. Then $ 2w +A\cdot \nabla w=0$ and $ w \ge 0$

Now consider $ \max_{D_1} w$. If this maximum is attained in the interior of $ D_1$, then $ w = 0$ there and hence everywhere. So assume it's only attained on the boundary, say at $ (x_0,y_0)$. The Kuhn-Tucker conditions imply that at this point $ \nabla w(x_0,y_0) =  \lambda (x_0,y_0)$ and  $ \lambda \ge 0$. Therefore at this point
\[ 2w = -A \cdot \nabla w = - \lambda A \cdot (x_0,y_0) \le 0 \, .\]
Therefore $ w = 0$ everywhere in $ D_1$ and also $ u = 0$ in $ D_1$.

The argument carries over to equations $ A(x)\cdot \nabla u(x) + a(x)u(x) = 0$ in a general bounded domain $ \Omega \subset R^n$ with smooth boundary, assuming now that $ a > 0$ in $ \Omega$ and $ A\cdot \nu > 0$ on the boundary of $ \Omega$, where $ \nu$ is the exterior normal vector field.


let $f(x)= \sum_{P}\frac{1}{P^{x}}$ where the sum is over all primes.

Let $x=s$ be the smallest positive value that $P(x)$ converges. Can $s$ be rational?

There is no smallest value for that it converges: it converges for all $x>1$ and diverges otherwise.


Working out of an old handbook, did this question, but don't like how I got the answer.  Please show how you did it...

A lamppost in a park can illuminate an area of radius 40 feet.  What is the minimum number of lampposts that must be installed in the park so that the entire region is illuminated?

Park is a rectangle side lengths 200 sq.rt. 2 by 120 sr.rt. 2.

Draw the largest square that can be inscribed in the 40 ft radius, and its side length is 20:sqrt:2. Therefore, it takes 10 x 6, or [b]60 lampposts[/b].

We have to have one on evry corner of a 40' side equilateral triangle


algebra
polynomial
calculus
integration
group theory
number theory proposed
number theory

Let $ P(x)\in Z[x]$ $ deg P>0$
Prove that there exist infinite $ p\in P$ such that for each k there exist $ a_k\in N$ such that
$ P(a_k)\equiv 0(\mod p^k)$
2.Prove that for each m
$ x^n\equiv 1(\mod p^m)$ has most n solution.

[quote="TTsphn"]Let $ P(x)\in Z[x]$ $ deg P > 0$
Prove that there exist infinite $ p\in P$ such that for each k there exist $ a_k\in N$ such that
$ P(a_k)\equiv 0(\mod p^k)$
2.Prove that for each m
$ x^n\equiv 1(\mod p^m)$ has most n solution.[/quote]
1. If equation $ P(x)\equiv 0\mod p$ had solution only for finite number of primes $ p_1,p_2,...,p_m$, then $ P(x)\equal{}\prod_{i\equal{}1}^m p_i^{k_i} \ \forall x\in Z$. But numbers $ P(x)<y$ most then $ Cy^{1/n}$, were n is deg(P(x)) and numbers $ \prod_{i\equal{}1}^m p_i^{k_i} <y$ less than $ C_2(lny)^m$ contradition.
2. I don't undestand your question. If x is solution $ x\plus{}p^m$ is solution too, therefore infinetely many solutions.
If you find only solutions $ 0<x<p^m$ it may be if only $ p|n$.

when we call solution it mean that class reside solution ($ x\in[1,p^m]$)
The 2 problem is true.Can you find out a example if it is not true.?

[quote="ZetaX"]@1:
Take any prime $ p$ not dividing the discriminant of $ P$ and dividing $ P(m)$ for some $ m$ (such $ p$ exists as any nonconstant polynomial has $ \infty$ different prime divisors of it's values). Then by Hensel's lemma, there is a solution $ \mod p^k$ for all $ k$.

@2:
The prime residue classes $ \mod p^k$ build a cyclic group, the rest follows.[/quote]
Yes solution is quite easy.
Other problem
Consider the equation
$ P(x)\equiv 0(\mod p)$
Suppose equation has exactly solution $ x_1,..,x_r$ and $ \gcd(f'(x_i),p) = 1,\forall i = 1..,r$
Prove that the equation
$ P(x)\equiv 0(\mod p^m)$ has exactly r solution $ x_1^{(k)},...,x_r^{(k)}$ and ${ \gcd(f'(x_i^{(k)},p)} = 1$
This result prove (2)
2.Prove that the equation $ x^{p - 1}\equiv 1(\mod p^m)$ has exactly $ p - 1$ solution.
3)(Interest)
Find all $ (m,n)$ such that
$ m|a^{n} - 1,\forall a = 1,..,n$

4)$ P(x)\in Z[x]$ and Irreducible
Prove that there exist infinite $ m\in N$ such that
$ \prod_{i \equal{} 1}^{m} P(i)$ not divide by $ P(m \plus{} 1)$

1) Comes from $ p$-adical standard facts (to be more exact, Hensel agein).
2) Follows from 1) or is well known.
4) We do not need $ P(x)$ to be irreducible, to be nonconstant suffices. Then the well knwon fact that $ P(k)$ has $ \infty$ many different prime divisors strikes again: just take $ m$ such that $ P(m\plus{}1)$ has a prime divisor not occuring before.

are  you sure ?
If we chose
$ P(x)\equal{}(x\minus{}a)h(x)$ then it is not true
we can edit
$ P(x)$ doesn't has natural solution :wink:

Yes, nonconstant and no (nonnegative integral) roots is necessary and sufficient. But then your statement must be corrected as $ x$ is irreducible ;)

Prove that for each m there exist $ (a,b,k)\in Z$ such that
$ 2m\equal{}a^{19}\plus{}b^{99}\plus{}k2^{1999}$
This use some result above.

Using some result above is not sufficient ;)
Otherwise I could start a thread containing $ 0 \equal{} 0$ and everything could be posted there ;)

We can find solution (b=1) $ 2m\minus{}1\equal{}a^{19}\mod 2^N$ (it is easy by Gensel's lemm).


limit
trigonometry
logarithms
calculus
calculus computations

a) Find $ \lim \frac{e^x\minus{}e^{tanx}}{x\minus{}tanx}$ as $ x\rightarrow0$

b) If $ f(0) \equal{}0$, $ f'(x)$ is continuous in the neighborhood of $ 0$, and $ f'(0)\neq0$, prove that $ \lim x^{f(x)}\equal{}1$ as $ x\rightarrow0$

$ \lim_{x \to 0} \frac{e^x\minus{}e^{tanx}}{x\minus{}tanx}\equal{}\lim_{x \to 0} \frac{e^x\minus{}e^{tanx}\sec^2 x}{1\minus{}\sec^2 x}\equal{}$

$ \lim_{x \to 0} \frac{e^x\cos^2 x\minus{}e^{tanx}}{\cos^2 x\minus{}1}\equal{}\lim_{x \to 0} \frac{e^x(\cos^2x\minus{}2\cos x\sin x)\minus{}e^{tanx}\sec^2 x}{\minus{}2\sin x\cos x}\equal{}$

$ \lim_{x \to 0} \frac{e^x(\sin 2x\minus{}\cos^2 x)\plus{}e^{tanx}\sec^2 x}{\sin 2x}\equal{}\lim_{x \to 0} e^x \plus{} \frac{\minus{}e^x\cos^2 x\plus{}e^{tanx}\sec^2 x}{\sin 2x}\equal{}$

$ 1\plus{} \lim_{x \to 0} \frac{\minus{}e^x\cos^2 x\plus{}e^{tanx}\sec^2 x}{\sin 2x}\equal{} 1\plus{} \lim_{x \to 0} \frac{\minus{}e^x\cos^4 x\plus{}e^{tanx}}{\sin 2x \cos^2 x}\equal{}$

$ 1\plus{} \lim_{x \to 0} \frac{e^x(4\cos^3 x \sin x \minus{}\cos^4 x)\plus{}e^{tanx}\sec^2 x}{2\cos 2x \cos^2 x\minus{}\sin^2 2x}\equal{}$

$ 1\plus{} \lim_{x \to 0} \frac{e^x\cos^2 x(2\sin 2x  \minus{}\cos^2 x)\plus{}e^{tanx}\sec^2 x}{\cos 4x\plus{}\cos 2x}\equal{}1$

$ \lim_{x \to 0} \frac{e^x \minus{} e^{\tan x}}{x \minus{} \tan x} \equal{} \lim_{x \to 0} \frac{1 \plus{} x \minus{} e^{x \plus{} x^3/3}}{x \minus{} (x \plus{} x^3/3)} \equal{} \lim_{x \to 0} \frac{1 \plus{} x \minus{} (1 \plus{} x \plus{} x^3/x)}{\minus{}x^3/3}$

$ \equal{} \lim_{x \to 0} \frac{\minus{}x^3/3}{\minus{}x^3/3} \equal{} 1$

$ \lim_{x \to 0} \frac{e^x\minus{}e^{tanx}}{x\minus{}tanx}\equal{}\lim_{x \to 0} {e^{tanx}}\lim_{x \to 0} \frac{e^{x\minus{}tanx}\minus{}1}{x\minus{}tanx}\equal{}lim_{x \to 0} {e^{tanx}}\lim_{u \to 0} \frac{e^{u}\minus{}1}{u}\equal{}(e^0)\lim_{u \to 0} \frac{e^u}{1}\equal{}1$
Where $ u\equal{}x\minus{}tanx$

[quote="shape_shift2005"]$ \lim_{x \to 0} \frac {e^x \minus{} e^{tanx}}{x \minus{} tanx} \equal{} \lim_{x \to 0} {e^{tanx}}\lim_{x \to 0} \frac {e^{x \minus{} tanx} \minus{} 1}{x \minus{} tanx} \equal{} lim_{x \to 0} {e^{tanx}}\lim_{u \to 0} \frac {e^{u} \minus{} 1}{u} \equal{} (e^0)\lim_{u \to 0} \frac {e^u}{1} \equal{} 1$
Where $ u \equal{} x \minus{} tanx$[/quote]

nicely done.  :)   :lol:

For [b](b)[/b], we can equivalently show that the limit of the logarithm approaches 0.

$ \lim \ln x^{f(x)}$
$ \equal{} \lim f(x) \ln x$
$ \equal{} \lim \frac {f(x)}{1/ \ln x}$
$ \equal{} \lim \frac {f'(x)}{ \minus{} 1/(x ( \ln x)^2)} \qquad \qquad \text{\color{blue}{L'Hopital's Rule: }}$
$ \equal{} 0 \qquad \qquad \qquad \qquad \qquad \text{\color{blue}{ Numerator given to be bounded, denominator unbounded}}$


geometry
geometric transformation
combinatorics unsolved
combinatorics

30 people sit around a round table, each of them is either wise or fool. Ask everyone" is your right neighbor wise or fool?" The answer of a wiseman is correct and that of a fool is either correct or wrong. The number of fools is no more than F. Find the maximum value of F such that it is always possible to identify wisemen.

The number of fools is no more than F.  Call the number of fools D (D is less than or equal to F). For every F, D = 0 is possible.  If D = 0 then 30 wisemen will correctly identify 30 wisemen but this gives no information since 30 fools could incorrectly identify 30 wisemen.  For any given value of F greater than zero one might fail to find any wisemen.  The maximum value of F is 0 !

I am SURE this is not what the problem means and it is perhaps a translation problem.  Should it be 'the number of fools is NO LESS than F'?

If someone could clarify that would be great.  Also,
is it required only that at least one wiseman can be identified or all of them?

At least one wiseman can be identified is enough!

[quote="hhaller"]We can probably assume that $F$ is known. That should rule out the case suggested. For example if it is known that $F = 1$, then it is trivially possible to identify the wisemen.[/quote]

Hmm... the wiseman will say that his neighbour is fool, but what if all the fools also say the same? Then you can't identify the wiseman. I think there is something wrong with the question's phrasing. If however the question is something like "Find the minimum $F$ such that a wiseman can always be identified if there are $F$ wisemen and wisemen always give correct answers" then it will be an interesting problem.

[quote="JBL"]If there are 15 or more fools, there are many situations in which each wiseman has a fool to his right, in which case it is possible for everyone to say, "there is a fool to my right," in which case we can't identify any wisemen.  So, the upper limit on $F$ is 14, and the question is if, in every arrangement of 16 wisemen and 14 fools, it is possible to identify for certain a wiseman, and if not, how many additional wisemen we need before it becomes necessarily possible.[/quote]

Oh sorry, this is what I meant. I mistook $F$ for the number of wisemen :P.


This one is well known, and it was probably posted before, but still I can't find it:

Prove that from 2n-1 natural numbers, one can choose n numbers, which sum is divisible by n.

Maybe some hints? I got stuck  :wink:

[quote="ZetaX"]Erdös
Ginzburg
Zif[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=57546
@ZetaX:  Myszon11 is a new member therefore I must help!  :wink:


So I am thoroughly frustrated...sigh...

57) The set of real numbers $ x$ for which the series $ \sum_{1}^{\infty}{\frac{n!x^{2n}}{n^{n}(1\plus{}x^{2n})}}$ converges is:

A) $ x\equal{}0$

B) $ \minus{}1 < x < 1$

C) $ \minus{}1 \le x \le 1$

D) $ \minus{}\sqrt{e} < x < \sqrt{e}$

E) All reals

I know how to do this problem without the $ (1\plus{}x^{2n})$ in the denominator, but I don't know how to deal with it, it's 2am, and there is not a single example like it in my calculus text or my analysis text.  Yay...

Hint:
Did you notice that
$ |\frac{x^{2n}}{1\plus{}x^{2n}}| \equal{} |1\minus{}\frac{1}{1\plus{}x^{2n}}|$ for all $ x$?

I'm quite sure this has been posted here very recently but..

Does the series with nth term n!/n^n converge? What is the GRE, by the way?

OK,

$ |\frac {n! x^{2n}}{n^{n}(1 \plus{} x^{2n})}|$
$ \equal{} \frac {1}{n}\cdot\frac {2}{n}\cdots\frac {n}{n} |1 \minus{} \frac {1}{1 \plus{} x^{2n}}|$
$ < \frac {1}{n}\cdot\frac {2}{n}\cdot |1|$ for all $ x$

$ \sum \frac {2}{n^2}$ converges,
so the original series converges for all $ x\in\mathbb{R}$.


Let $ ABC$ be a triangle whose circumcircle is $ (O)$ and such that $ AC < AB$.THe tagent lines to $ (O)$ at $ B;C$ intersect at $ T$.The line through $ A$ and perpendicular to $ AT$ meet $ BC$ at $ S$ .Choose the points $ B_1$ and $ C_1$ on $ ST$ such that $ TB_1\equal{}TC_1\equal{}TB$ and $ C_1$lies between $ S$ and $ T$.Prove that $ ABC$ and $ A_1B_1C_1$ are similar

To mod: Please lock this topic because this problem is on [b]Mathematics and Youth Magazine[/b] (Maths magazine in Viet Nam).
To nhanmot: Are you Vietnamese??  :mad:

Please moderator of mathlinks lock this topic immediately.

@nhanmot: You should check your behavior again if you are a Vietnamese. I feel sad for you  :mad:


Let a_1,a_2,..... be a sequence of real numbers satisfying a_i+j \leq a_i+a_j for all i,j=1,2,....
Prove that

a_1+(a_2)/2 + (a_3)/3+...+(a_n)/n   \geq a_n

for each integer positive n.


What is the sum of the integer solutions to $ |x\plus{}2| < 5$?


23. A box contains some green marbles and exactly four red marbles.  The probability of selecting a red marble is x%.  If the number of green marbles is doubled, the probability of selecting one of the four red marbles from the box is (x-15) %.  How many green marbles are in the box before the number of green marbles is doubled?

29. The digits 2,3,4,7 and 8 will be put in random order to make a positive five-digit integer. What is the probability that the resulting integer will be divisible by 11?  Express your answer as a common fraction.

15. The integers from 1 through 15 are written in numerical order in pencil going clockwise around a circle. A student begins moving clockwise around the circle erasing every third integer that has not yet been erased until only the integer 11 remains. Which integer did the student erase first?

29. $ \frac{1}{10}$.

Using the divisibility rule for $ 11$, the $ 5$ digit number will be divisible if the sum of the tens and thousands digits is equal to the sum of the other three digits. In other words, the number is divisible if the sum of the tens and thousands digits is $ 12$. This can only happen one way, if the tens and thousands digits are $ 4$ and $ 8$. Thus we have $ 2$ possibilities for the tens and thousands places and $ 3$ for the other three. Using this format, we can make $ 12$ different numbers $ \frac{7!}{2!3!}$. A total of $ 120$ different numbers can be created $ 5!$. Thus the probability is $ \frac{12}{120}\Longrightarrow\frac{1}{10}$.

15. If the first erased is $ 3$, the "survivor" is $ 7$.  Thus, we shift everything over by $ 4$ so the first erased is $ \boxed{7}$ and the survivor is $ 11$.

[quote="jogaman323"]23. A box contains some green marbles and exactly four red marbles.  The probability of selecting a red marble is x%.  If the number of green marbles is doubled, the probability of selecting one of the four red marbles from the box is (x-15) %.  How many green marbles are in the box before the number of green marbles is doubled?

Let there be $ k$ green marbles.

Thus, $ \frac{4}{4\plus{}k}\equal{}\frac{x}{100}$
$ 400\equal{}4x\plus{}xk$

After doubling the green marbles,
$ \frac{4}{4\plus{}2k}\equal{}\frac{x\minus{}15}{100}$
$ 400\equal{}4x\minus{}60\plus{}2kx\minus{}30k$
$ xk\equal{}60\plus{}30k$
$ k(x\minus{}30)\equal{}60$.
We can quickly check the solutions, to get $ k\equal{}6$, $ x\equal{}40$ is the only working solution. Checking,

$ \frac{4}{10}\equal{}40 \%$
$ \frac{4}{16}\equal{}25 \%\equal{}40\minus{}15$.

Thus, there are $ 6$ marbles before doubling.

29. The digits 2,3,4,7 and 8 will be put in random order to make a positive five-digit integer. What is the probability that the resulting integer will be divisible by 11?  Express your answer as a common fraction.

15. The integers from 1 through 15 are written in numerical order in pencil going clockwise around a circle. A student begins moving clockwise around the circle erasing every third integer that has not yet been erased until only the integer 11 remains. Which integer did the student erase first?[/quote]

15. First, use letters to represent the numbers. a b c d e f g h i j k l m n o. Then cross out every second letter starting from the letter a. You should end up with letter c. If 11 is c, then the number you crossed out first, a, is 9. Therefore, 7, as someone put, is incorrect.


Find $\displaystyle a,b,c$ such that $\displaystyle \lim_{n \to \infty} n^a \left( n+\sqrt[3]{bn^3+cn+1} \right) = 1$.

There's more than one family of solutions here. As TheBB suggests, $a=-1,b=0$ works (with arbitrary $c.)$

Now: what can you come up with for $b=-1?$


The concept of Temperature that is explained by Zemansky's classic "Heat and Thermodynamics" and other books is based on thermal equillibrium.

So,

[quote][i]Temperature is that property of a body which determines whether it will be in thermal equilibrium with another body. By thermal equlibrium we mean that, if 2 bodies are kept in contact for a very long time, no transfer of heat from one body to another takes place.[/i][/quote]

[quote="chesspro"]Temperature is defined as the average kinetic energy (on a molecular scale) of an object. Entropy is the amount of disorder in a system.[/quote]Can you give some exact formula? Does this work for ideal gas only or can you use it on liquids and solid objects as well? Because molecules often also have potential energy which transforms to kinetic energy and back all the time so it doesn't seem to make too much sense to only take the kinetic energy. And also, this potential energy is quite impossible to find and really depends on where you set the zero potential...
Is there any way to measure entropy? Does it have some kind of units?

Rushil - is there any way to quantify the temperature (based on your definition)? I.e. find the numerical value of temperature?

[quote="bus"]Recently when I was thinking about the Maxwell-Boltzmann distribution and about the flask question in the previous threads, I also started thinking about how we define tempreture of some object. Once I heard a definition from kubus which involved terms such as entropy but I didn't understand it because I don't know what entropy really is. Could somebody explain entropy and/or define temperature? How is according to this definition the temperature of ideal gas related to mean square velocity of its particles? What about other substances?[/quote]bus,

Let see if I can be a bit more analytic about things. First, entropy is a difficult concept to grasp the first time around. According to the second law of thermodynamics, it is impossible to get as much work out of a thermodynamic system as you put into it. That is because some of the work you put into the system only serves to increase the randomness of the motions of the molecules in the system. This randomness is called entropy. Finally if the system is closed, meaning that it does not permit new molecules to enter nor permit old ones to leave then the entropy cannot decrease. Ultimately a closed thermodynamic system will cease functioning altogether because the entropy that must be overcome will exceed the work that can be done.

That said, entropy is usually denoted by $S$. And the first law of thermodynamics in differential form is: $\mathrm{d}E = T\mathrm{d}S -P\mathrm{d}V$, $E$ is internal energy of the system, $T$ is temperature, $S$ is entropy, $P$ is pressure (exerted by the system on its surroundings), and $V$ is the volume occupied by the system.

From calculus we know that $\mathrm{d}E =\dfrac{\partial E}{\partial S}\mathrm{d}S + \dfrac{\partial E}{\partial V}\mathrm{d}V$.

Equating like terms in the two expressions for $\mathrm{d}E$, we find $T = \dfrac{\partial E}{\partial S}$.

To answer your question then, "Temperature is the rate of change of internal energy with respect to entropy."

PS Boltzmann's constant for statistical reasons has units of entropy $\text{ Joules/Kelvin}$.

[quote="Rushil"]The quantification of temperature based on my ( Zemansky's ) definiton is that you consider isotherms obtained by chaniging some property of some system. Then, you establish a standard and find all tempertures relative to that.[/quote]Yes, but the question is can one just go about changing the values of system variables ad hoc and claim that doing so constitutes a temperature change? My answer is no, one cannot.

Incidentally, [i]A Dictionary of Physics[/i] (ISBN 0-19-280030-2) defines temperature as:

[i]The property of a body or region of space that determines whether or not there will be a net flow of heat into it or out of it from a neighboring body or region and in which direction (if any) the heat will flow.[/i]

I certainly would not dispute Professor Zemansky, he wrote a book on the subject, a copy of which I happen to own, [i]Temperatures (Very High and Very Low)[/i] [ISBN 0-486-24072-X]. In this book he spends the entire first chapter telling the reader what temperature is qualitatively and how to measure it, but he doesn't define what one measures. Finally, in chapter 2 he gives a formula equivalent to the one I derived [Zemansky, Eq. 2-6]. Temperature is the rate of change of internal energy with respect to entropy.

[quote="perfect_radio"]Can anyone explain to me why is there a minimal temperature ($0 ~ K$)? Why can't we go any lower? (I'm guessing that this should have connection with the definition, but I didn't really understand it)

On the other hand, can we have temperatures as high as we want? (of course, i'm interested in some explanation for this also)[/quote]Okay, here goes ...

Considering an ideal gas, $S(V,T) = \int_{0}^{T} C_V \dfrac{\mathrm{d}T}{T}\; \& \; S(P,T) = \int_{0}^{T} C_P \dfrac{\mathrm{d}T}{T}$

Therefore, in general $S(X,T) = \int_{0}^{T} C_X \dfrac{\mathrm{d}T}{T}$

If we now consider two adiabatic processes $S(X_a,T_a) = \int_{0}^{T_a} C_{X_a} \dfrac{\mathrm{d}T}{T}\; \& \; S(X_b,T_b) = \int_{0}^{T_b} C_{X_b} \dfrac{\mathrm{d}T}{T}$

Since the processes are adiabatic $S(X_a,T_a) = S(X_b,T_b)$

$\Rightarrow \int_{0}^{T_a} C_{X_a} \dfrac{\mathrm{d}T}{T} = \int_{0}^{T_b} C_{X_b} \dfrac{\mathrm{d}T}{T}$

The integrals necessarily converge, and if we let $T_b \longrightarrow 0$ then

$\int_{0}^{T_a} C_{X_a} \dfrac{\mathrm{d}T}{T} = 0$, but since $C_{X_a} \neq 0$ and $T_a \neq 0$ this is a contradiction.

Therefore, $T_b \neq 0$.

In English, an absolute temperature of zero is absolutely unachieveable -- this is known as the third law of thermodynamics.