logarithms
advanced fields
advanced fields unsolved

Hey,
I have a problem with proving a result of Yu Kunrui (1986) saying:
Let $ a$ be a positive integer not divisible by $ p$. Prove there exists  a constanst $ c >0$ depending on $ a$ and $ p$ such that for each positive integer $ b$ we have $ |a^b\minus{}1|_{p} \geq cb^{\minus{}1}$.

I have already proved the following:
Suppose that $ p^k \parallel{}(a\minus{}1)$ (i.e, $ p^k$ divides $ a\minus{}1$ but $ p^{k\plus{}1}$ does not) then $ |a^b\minus{}1|_{p} \geq |a\minus{}1|_{p} b^{\minus{}1}$
But I don't know how to use this to prove the result of Yu Kunrui.

Can anyone help?
Thanks

I'll sketch the proof (if you need more details,feel free to ask): suppose that our statement is not true, then we can find a sequence of natural numbers $ n_k\to\infty$ such that $ a^{n_k} \minus{}1 \equal{}0(mod p^k)$ and $ \frac{n_k}{p^k}\to 0$. We can suppose that $ n_k\equal{} ord_{p^k} a$. We want to prove now that for $ \frac{ord_{p^l} a}{p^l} >c$ for some $ c>0$. We can write $ ord_{p^l} a\equal{} p^{t_l}x_l$ where $ t_l \le l\minus{}1$ and $ x_l\le p\minus{}1$ (it follows from the fact that $ a^{p^{l\minus{}1}(p\minus{}1)}\equal{}1(mod p^l)$). We now have $ ord_{p^l} a|ord_{p^{l\plus{}1}} a$, hence $ t_{l\plus{}1}\ge t_l$ and $ x_l |x_{l\plus{}1}$. Suppose, that $ t_{l\plus{}1}\equal{} t_l$ then we can write $ x_{l\plus{}1}\equal{}kx_l$.
Now $ a^{ord_{p^{l\plus{}1} a}}\minus{}1\equal{} (a^{ord_{p^{l} a}})^k\minus{}1\equal{}(a^{ord_{p^l} a}\minus{}1)(a^{ord_{p^{l}} a})^{k\minus{}1}\plus{}....\plus{}1)\equal{}0(mod p^{l\plus{}1})$,hence $ k\equal{}0 (mod p)$,contradiction. Hence $ t_{n\plus{}1}>t_n$ which immediately gives desired result.


John surveyed a group of people about their knowledge of rats. To the nearest tenth of a percent, he found that $ 86.8\%$ of the people surveyed thought rats carried diseases. Of the people who thought rats carried diseases, $ 45.7\%$ said that rats frequently carried rabies. Since rats do not frequently carry rabies, these 21 people were mistaken. How many total people did John survey?

This is mostly about calculations.  First, we multiply $ 0.868$ and $ 0.457$.  We get $ 0.396676$.  We can then write a ratio:

$ \frac {21}{X} \equal{} \frac {0.396676}{1}$

where $ X$ is the total number of people John surveyed.  We cross-multiply and divide both sides by $ 0.396676$, getting

$ X \equal{} \frac {21}{0.396676}$

$ X \equal{} 52.94 \approx \boxed{53}$

Checking our work, we find that our answer indeed rounds down to $ 21$.

here is another that i need help on. The ratio of the number of games won to the number of games lost(no ties) by the Middle School Middies is $11/4$. To the nearest whole percent, what percent of its games did the team lose?

Please begin a new topic if you want help.


i hav found a crank homepage http://kryptochef.net/index2e.htm

[quote]Why are 256 bits the technically highest coding depth at all on computers possible are ?

A computer knows only 256 different indications.
1 indication = 1 byte has 8 bits in binary the number system exactly.
1 bit knows only the switching status:
on or out or 0 or 1 by the combination of these 8 bits results 256 bits.
The computation in addition:
2 switching status highly 8 bits = 256 bits these 256 bits
is addressed in decimally the number system from 0 to 255 = 256 bits.
Computers work however in in hexadecimals the number system.
There these 256 bits designated above are addressed from 00 to FF = 256 bits.
A byte cannot be thus under bits 0 or over bits 255.
Therefore 256 bits are the technically highest coding depth at all on computers
possible are.[/quote]

see also http://www.schneier.com/blog/archives/2006/06/the_doghouse_kr.html and http://krueptoscheff.dyndns.org/index-en.html

Welcome to AoPS. Are you hairuo (ignore this if you aren't)?

yes, I'm hairuo and my A T O M I C F I S T is waiting (charged in NRDkrorororea)

F E A R

[quote="dragon96"]Nice to meet you hairou. Guess who I am.  :D[/quote]
You know hairuo?

[quote="dragon96"]Yea. Wait, how do you know that that is hairou for sure?[/quote]
Actually the reason I asked in the first place was because I was chatting with him on google and he used the term roflcopter. I was all like, what in the name of omgswhatbbq/omgswhatrice/omgswhatnoodles is roflcopter supposed to be?

roflcopter is an online game if i remember correctly...
it is made entirely in ascii


try this one(really easy :) )
prove that  the product of four consecutive numbers plus one is a perfect square ;)

this can be geberalized to four consecutive terms of an arithmetic progression. the solution is trivial. just multiply the first and the fourth factor and the second with the third. we can see that our new 2 factors differ by and hence their product is a square minus one. and were done.


Let $ \{a_n\},\ \{b_n\},\ \{c_n\}$ be sequences such that

(i) $ a_n\plus{}b_n\plus{}c_n\equal{}2n\plus{}1$

(ii) $ a_nb_n\plus{}b_nc_n\plus{}c_na_n\equal{}2n\minus{}1$

(iii)$ a_nb_nc_n\equal{}\minus{}1$

(iv) $ a_n<b_n<c_n$

Find $ \lim_{n\to\infty} na_n$.

[hide="Solution"]$ a_n, b_n, c_n$ are the roots of the cubic $ t^3 \minus{} (2n \plus{} 1)t^2 \plus{} (2n \minus{} 1)t \plus{} 1 \equal{} 0$.  Note that the LHS factors as $ (t \minus{} 1)(t^2 \minus{} 2n t \minus{} 1)$, so we have roots $ 1, \frac {2n \pm \sqrt {4n^2 \plus{} 4}}{2} \equal{} n \pm \sqrt {n^2 \plus{} 1}$.  It follows that $ a_n \equal{} n \minus{} \sqrt {n^2 \plus{} 1}$.  Then the limit we seek is $ \lim_{n \to \infty} n(n \minus{} \sqrt {n^2 \plus{} 1}) \equal{} \lim_{n \to \infty} n\left(\frac { \minus{} 1}{n \plus{} \sqrt {n^2 \plus{} 1}}\right) \equal{} \minus{} \lim_{n \to \infty}\frac {1}{1 \plus{} \sqrt {1 \plus{} 1/n^2}} \equal{} \minus{} \frac {1}{2}$[/hide]


For any positive integer $a$, denote $f(a)=|\{b\in\mathbb{N}| b|a$ $\text{and}$ $b\mod{10}\in\{1,9\}\}|$ and $g(a)=|\{b\in\mathbb{N}| b|a$ $\text{and}$ $b\mod{10}\in\{3,7\}\}|$. Prove that $f(a)\geq g(a)\forall a\in\mathbb{N}$.

This problem is true. Let $a=2^{k}a_{0}$, were $a_{0}$ is odd. If $a=\pm 3(mod \ 10)$, and $d$ divisor, suth that $d=\pm 3(mod \ 10)$,
then $s(d)=\frac{a_{0}}{d}$  is divisor, suth that $s(d)=\pm 1(mod \ 10)$. Therefore $f(a)=g(a)$.
 If $a_{0}=\pm 1(mod \ 10)$, then $a_{0}=a_{1}a_{3}, \ a_{1}=\prod_{j}p_{j}^{k_{j}}, \ a_{3}= \prod_{i}q_{i}^{m_{i}}$,
were $p_{j}=\pm 1(mod \ 10), \ q_{i}=\pm 3(mod \ 10)$ and $\sum_{i}m_{i}$ is even. We have $f(a)=\tau(a_{1})f(a_{3}),g(a)=\tau(a_{1})g(a_{3})$.
If $a=bc, (b,c)=1)$, then $f(a)=f(b)f(c)+g(a)g(c),g(a)=g(b)f(c)+g(c)f(b)$.
We have $f(a)-g(a)=f(c)(f(b)-g(b))-g(c)(f(b)-g(b))=(f(c)-g(c))(f(b)-g(b)).$
Because $f(q^{k})-g(q^{k})=1+k(mod \2)\ge 0$, it is true for all numbers. It proved, that $f(a)>g(a)$ if and only if $ord_{p}(a)$ are even for all prime divisor $p=\pm 3(mod \ 10)$.


1>,Evaluate the following limits, or say why they do not exist.

When x approaches 2 , find the limit of:[square root(6-x)]-2/[square root(3-x)]-1

2>Let f:[-1,1] to R be given by f(x)=-x (0<=x<=1)
                                                     1-x (-1<=x<0)
  Use the Squeeze Theorm to prove that when x approaches 2 the limit of x*f(x) is 0

you mean this: $\sqrt{6-x}-\frac{2}{\sqrt{3-x}}-1$? when $x\to 2$ then limit equal $-1$

[quote="mlok"]1) $\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}= \frac{(\sqrt{6-x}-2)(\sqrt{6-x}+2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)(\sqrt{3-x}+1)(\sqrt{6-x}+2)}$
$=\frac{(6-x-4)(\sqrt{3-x}+1)}{(3-x-1)(\sqrt{6-x}+2)}=\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}$.

2) $f(x)$ is between $-1$ and $2$. Hence $xf(x)$ is between $-x$ and $2x$ -- two continuous functions with a common value at $x=0$.[/quote]
in this case limit is equal 1/2. by the Hopital's   we give $\frac{\sqrt{3-x}}{\sqrt{6-x}}=1/2$


ratio
geometry
trigonometry
area of a triangle
Heron\u0027s formula

Let $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD,PE,PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen, \[ \frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}. \]

Only a disguise of the [b]Viviani's Theorem[/b]
Hint
[hide][PAB]+[PBC]+[PCA]=[ABC][/hide]

[hide="Hint"]Try proving the $PD+PE+PF$ is const and then look at the case when it's easiest to calculate $PD+PE+PF$ in terms of $AB+BC+AC$ ;) [/hide] ;)

[hide="Not a Hint"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=861&sid=f68fdee8d736e17cede15e1adf7495c5[/img]

Let a side of the triangle be $s$. Note that because $2[\triangle ABC]=2[\triangle APB]+2[\triangle APC]+2[\triangle BPC],$ $\frac{\sqrt3}{2}s^2=s(PD+PE+PF).$ Dividing both sides by $s$, the sum of the perpendiculars from $P$ equals $PD+PE+PF=\frac{\sqrt3}{2}s.$ (It is independant of point $P$) Because the sum of the sides is $3s$, the ratio is $\frac{\frac{\sqrt3}{2}}{3s}=\frac{1}{2\sqrt3}.$ [/hide]

[hide="solution"]

Since equilateral, let AB = AC = BC = a. Also, let PD = d, PE = e, and PF = f.

Therefore, using Heron's formula we have:

a(d+e+f)/2 = ((3a/2)(a/2)(a/2)(a/2)) ^ 1/2  (one)

This equation will ultimately give:

a/(d+e+f) = (2/3)*(3 ^ 1/2)  (two)

Since we are looking for (d+e+f)/(a+a+a), multiply (two) by 3 and take the reciprocal.

Q.E.D  :D

[/hide]

[hide]Draw $\Delta ABC$ with point $P$ in the centre. Draw perpendiculars to each of the three sides with $D$ on $AB$, $E$ on $BC$, and $F$ on $AC$. Draw $AP$, $BP$, $CP$. We find the area of $ABC$ in terms of the three triangles $ABP$, $BPC$, $APC$. We have:

$[ABC] = \frac{1}{2}(AB*DP+AC*PF+BC*EP)$

But $[ABC] = AB*AB*\sin 60$. So, we have:

$AB*DP+AC*PF+BC*EP = AB*AB*\sin 60$

Let $AB = AC = BC = s$.

$DP+PF+DE = s*\frac{\sqrt{3}}{2}$

Dividing by $3s$,

$\frac{DP+PF+DE}{AB+BC+CA}= \frac{\sqrt{3}}{6}=\frac{1}{2\sqrt{3}}$[/hide]

By Viviani's theorem, this point's "power" will be constant for anywhere inside the triangle. Then we can choose the point P=A, which means PD+PE+PF=AD, where D is the foot of altitude from A. That is equivalent to $s\sqrt{3}/2$, and when divided by 3s, gives $1/(2\sqrt{3})$. $\blacksquare$

[hide=solution]
Let the side length of the equilateral triangle be $s$. Vivani's Theorem tells us that $PD + PE + PF$ will equal the length of the altitude of the triangle which is $\frac{\sqrt{3}s}{2}.$ Since $ABC$ is equilateral, we have $AB = BC = CA = s,$ so $AB+BC+CA = 3s.$ Hence, we have $$\frac{PD+PE+PF}{AB+BC+CA} = \frac{\frac{\sqrt{3}s}{2}}{3s} = \frac{\sqrt{3}}{6} = \frac{1}{2\sqrt{3}}.$$ $\square$
[/hide]

Area of ∆ABC=√3/4 AB²( as it is equilateral)
Area of ∆ABC=Area of ∆BPC+Area of ∆APC+
Area of ∆APB
√3/4 AB²=(1/2){PD+PE+PF)
PD+PE+PF=√3/2 AB
(PD+PE+PF)/ 3AB=1/2√3
PD+PE+PF / AB+ AC+BC= 1/2√3


calculus
integration
trigonometry
limit
calculus computations

hope i haven't asked this one before  :maybe:

show  :D

${ \int_0^\infty \; \frac{e^{-ay} - e^{-by} }{y\;\; \text{cosec}\, my} \;\; dy \;\; = \; \; \boxed {\tan^{-1}\;\frac{b}{m} - \tan^{-1} \;\frac{a}{m} } }$

[b]no[/b] techniques from complex analysis.

what will be if I define:
\[ I(\epsilon) \equal{} \int_{0}^{\infty} \frac {\cos (my) ( e^{ \minus{} ay\epsilon} \minus{} e^{ \minus{} by\epsilon})}{y}dy
\]
calculate $ I'(\epsilon)$ , hence find $ I(\epsilon)$ and also $ \lim_{\epsilon \to 1} I(\epsilon) \equal{} your integral$
here is [b]good[/b] question:
Can I differentiate it by $ \epsilon$ ?  indeed, I can  :roll:

$ \frac {e^{ \minus{} \alpha y} \minus{} e^{ \minus{} \beta y}}{y} \equal{} \int_{\alpha }^{\beta }{e^{ \minus{} xy}\,dx}.$ The integral becomes $ \int_{\alpha }^{\beta }\!{\int_{0}^{\infty }{e^{ \minus{} xy}\sin (my)\,dy}\,dx} \equal{} \int_{\alpha }^{\beta }{\frac {m}{m^2 \plus{} x^{2}}\,dx},$ and the conclusion follows.

yes , it's true.
some comment:
you also need to explain why you can change integrals $ \int_{0}^{\infty} \int_{\alpha}^{\beta} \to     \int_{\alpha}^{\beta}  \int_{0}^{\infty}$

in this case it's simple. it's enough to show that integral :
$ \int_{0}^{\infty} e^{\minus{}xy}\sin (my) dy$ uniform converge on $ [\alpha,\beta]$ . it's easy by Weierstrass criterion. since $ \sin$ is bounded and $ e^{\minus{}..}$ converge


How would I make the two axes (x and y) and the graph of y=f(x)??

See Dojo's blog - you may have to go back one or two pages. He shows the code for making graphs in Asymptote.

[asy]import graph;Label f;f.p=fontsize(6);xaxis(-8,8,Ticks(f, 8, 0.5));yaxis(-8,8,Ticks(f, 8, 0.5));real f(real x){return x^3;}draw(graph(f,-2,2));[/asy]

Experimenting a bit:

[asy]import graph;
Label f;
f.p=fontsize(6);
xaxis(-8,8);
yaxis(-8,8);
real f(real x){return x-1;}
draw(graph(f,-2,2));[/asy]

[asy]import graph;
Label f;
f.p=fontsize(6);
xaxis(-8,8);
yaxis(-8,8);
real f(real x){return x-1;}
draw(graph(f,-2,2));
real f(real x){return x^2;}
draw(graph(f,-2,2));[/asy]

See here:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213540]A thread[/url]
Or
[ur=http://www.artofproblemsolving.com/Forum/weblog.php?ppage=5&w=1157&start=5l]My Blog[/url]


I just looked at the forums in Safari, and found that the expand/collapse buttons for forums didn't work.

I've seen this before, just after the feature was introduced; my problem was an old cached version then. Since I haven't used Safari to visit AOPS in a long time (and it worked back then), I don't think that's the problem this time.

Unfortunately we don't provide support for Safari (especially for the little visual / javascript tricks). Less than 0.5% of the visitators of the site use Safari (around 70% use IE, 25% Firefox and 5% other browsers, Opera having a high cut on this).

If anyone wants to debug Safari for that javascript error, and they find the culprit I'll be glad to issue a fix. Otherwise, I'm afraid it's just going to be that way.

Update: it seems to be a bug in the Firefox IE7 too, so I presume it's a small glitch here too. I'm looking into it.

Problem fixed on IE, Firefox and Opera. Is it working in Safari now? (to test that it works, you will have to refresh the page to get the new javascript -- with F5).


I'm having difficulty with problem 3 on this page:
http://mathcircle.berkeley.edu/mpgeo/node7.html

In $ \bigtriangleup ABC$, $ D$ is the midpoint of $ \overline{BC}$ and $ E$ is the trisection point of $ \overline{AC}$ nearer$ A$. Let $ G$ = $ \overline{BE} \cap \overline{AD}$. Find $ AG: GD$ and $ BG: GE$.
Solution: Draw the figure! Assign weight 2 to $ A$ and weight 1 to each of $ B$ and $ C$. Then $ 2A\plus{}1B \equal{} 3E$ and $ 1B\plus{}1C\equal{}2D$. Note that the center of mass of the system is $ 2A \plus{} 1B \plus{} 1C\equal{} 3E \plus{}1C \equal{} 2A \plus{} 2D \equal{} 4G$. From this we can see that $ AG: GD \equal{} 2: 2 \equal{} 1: 1$ and $ BG: GE \equal{} 3: 1$.

1. What is the inspiration for assigning weights of 2 to $ A$ and 1 to $ B$ and $ C$ and how does choosing those values help us solve the problem?

2. Why would we have noticed that G happens to be the location of the center of mass?

3. In mass point problems, is it a good idea to find the weights of all of your relevant points first?

[quote]Why would we have noticed that G happens to be the location of the center of mass?[/quote]
Why should G be the center of mass?The center of mass of a triangle lies on the intersection of medians?

It is a consequence of the problem that G is located at the centre of mass.

Essentially, by the 3E+1C=2A+2D which follows by expanding out E and D, we see that G lies on both lines, and has weight 4, and so we readily find the ratios as well. This is how we solved the problem. Now, they both expand to 2A+B+C, which is the center of mass for the whole thing.


quadratics
modular arithmetic
algorithm
search
abstract algebra
number theory
Diophantine equation

I know following theorem ([1])
The Diophantine equation $y^{2}=x^{3}+k$ has no solutions if $k$ has form $(4n-1)^{3}-4m^{2}$, where $m,n$ are integers such that no prime $p\equiv-1\pmod{4}$ divides $m$.

[1] Tom M. Apostol, Introduction to Analytic Number Theory.

[quote="Continuous"]Is there a general statement about the solution set for arbitrary $d$?[/quote]
For every $d\in\mathbb{Z}$ there are only finitely many integer solutions.

[quote="Continuous"]I'm looking for a method for solving diophantine equations of the form $y^{3}=x^{2}+d$ with $d\in\mathbb Z$.
For special cases (e.g. $d=11$) I can bash out this equation by calculations in quadratic fields. But it looks to me more like a happy accident.
Is there a general statement about the solution set for arbitrary $d$?[/quote]

Actually this is a theorem which I love.

$y^{3}= x^{2}+d$
Then, the elliptic curve $E(\mathbb{Q})$ is finitely generated, http://en.wikipedia.org/wiki/Mordell_theorem

So it has a FINITE number of solutions for any $d\in \mathbb{Z}$.

I think, im not sure about this, that in the book "Rational Points on Elliptic Curves" by Tate and Silverman some algorithms are given.

Thanks for your answers. As far as I know, it is still unknown that the algorithm we have for computing the group of an elliptic curve over Q terminates. Only the torsion group of E(Q) can be easily computed.
Actually I'm looking for an approach without using conjectured theorems about elliptic curves. For instance $y^{3}=x^{2}+5$ can be solved by considering this factorization in $\mathbb Q(\sqrt{-5})$:
$y^{3}=(x-\sqrt{-5})(x+\sqrt{-5})$. The ideals $(x-\sqrt{-5})$ and $(x+\sqrt{-5})$ are coprime in $\mathbb Z(\sqrt{-5})$. So $(x-\sqrt{-5})=(\alpha)^{3}$  since the order of the class group of $\mathbb Q(\sqrt{-5})$ is coprime to 3. This gives the solutions for $\alpha$ and hence also for x and y in $\mathbb Z$.
But how can I treat this equation with other values of d (e.g. for squares)? This method only work in few cases.
Maybe it is not possible (or only conjectured) in the general case.

[quote="Continuous"]As far as I know, it is still unknown that the algorithm we have for computing the group of an elliptic curve over Q terminates.[/quote]
I have heard that there are also algorithms to compute the rank of a given elliptic curve. I don't know much about elliptic curves so I don't know what do you mean by the group of an elliptic curve.

look at google, wikipedia, etc for definition.

[quote="puuhikki"]
I have heard that there are also algorithms to compute the rank of a given elliptic curve. I don't know much about elliptic curves so I don't know what do you mean by the group of an elliptic curve.[/quote]
Please see at
[1] Advanced Topics in the Arithmetic of Elliptic Curves (Graduate Texts in Mathematics)
[2] The arithmetic of elliptic curves (Graduate Texts in Mathematics)
Both by J. Silverman.

[quote="puuhikki"]
I have heard that there are also algorithms to compute the rank of a given elliptic curve.
[/quote]
Unanswered question in mathematics. Only specific cases can be computed. The general case is the Birch-Swinnerton-Dyer Conjecture and open problem worth 1,000,000,000 for its solution.
[quote]
I don't know much about elliptic curves so I don't know what do you mean by the group of an elliptic curve.[/quote]
By that I mean the "Weierstrass Group Law". (Search for it).

For example, given the field $\mathbb{Q}$ and the elliptic curve $y^{2}=x^{3}+1$ then we define $E(\mathbb{Q}) = \{ \infty\}\cup S$ where $S=\{(x,y)| y^{2}=x^{3}+1 , x,y\in \mathbb{Q}\}$ (it can be empty!). The $\infty$ symbol here is "the point at infinity" and it plays the role of the identity element for this group $E(\mathbb{Q})$ of course the question is "what is the binary operation". Well, that is a little hard to explain. I would recommend [u]Elliptic Curve: Number Theory and Cryptography[/u] he (Washington) explain near the beginning of the book what this binary operation is. So under this binary operation (which I would explain it is just complicated and long) the set $E(\mathbb{Q})$ for an abelian group.

Now, the more general case is when $F$ is any arbitray field and the elliptic curve is simply the equation $y^{2}= x^{3}+Ax+B$ (with $4A^{3}+27B^{2}\not = 0$) where $A,B\in F$. And $S=\{(x,y)|y^{2}=x^{3}+Ax+B , x,y \in F\}$

[quote="jmerry"]Another nice source is "Rational Points on Elliptic Curves", by Silverman and Tate. That's UTM rather than GTM- you don't need much background to read and understand it.[/quote]

The problem with that book is that it doesn't proof much of its results.

[quote="sylow_theory"][quote="puuhikki"]
I have heard that there are also algorithms to compute the rank of a given elliptic curve.
[/quote]
Unanswered question in mathematics. Only specific cases can be computed.[/quote]
Nope. There are such algorithms but they are very slow so therefore they are unpractical. They do not require Swinnerton-Dyer conjecture.


logarithms
inequalities
function
inequalities proposed

Let $x,y,z >0$ such that    $x(1-y)+y(1-z)=z(x-1)$

prove that  $x^{\dfrac{1}{y}}\cdot y^{\dfrac{1}{z}}\cdot z^{\dfrac{1}{x}}\geq 1$

Basically the condition tells us that $x + y + z = xy + yz + zx$.

I am wondering why you didn't state the problem that way, it looks nicer, IMHO.

Let $\frac{1}{x}=a, \frac{1}{y}=b, \frac{1}{z}=c,$  $a,b,c>0.$
We have $x^{\frac{1}{y}}\cdot y^{\frac{1}{z}}\cdot z^{\frac{1}{x}}\geq 1 \Leftrightarrow$
$\Leftrightarrow a^{b}\cdot b^{c}\cdot c^{a}\leq1\Leftrightarrow b\ln{a}+c\ln{b}+a\ln{c}\leq 0\Leftrightarrow$
$\Leftrightarrow \frac{b\ln{a}+c\ln{b}+a\ln{c}}{a+b+c}\leq \ln{\frac{ab+bc+ca}{a+b+c}}.$
The last one is Jensen Ineq for concave function $f(x)=\ln{x}$ on $(0,\infty )$ and hence the conclusion.


Given that the function $f$:

$f(x) = \frac {cx}{2x+3}$, $x \neq - \frac {3}{2}$,

satisfies $f(f(x)) = x$ for all real numbers x except $- \frac {3}{2}$, then $c$ is what?

 :P

WHAT?

This is not pre-olympiad material. I demand it back to the Intermediate Forum.

Haven't posted in a while, so I thought I'd contribute!

f(f(x))=x
(c*(cx/(2x+3))/(2(cx/(2x+3))+3)=x
This yields...

(c^2x-2cx^2)=6x^2+9x
c^2=9 and -2c=6
c=-3

 :D


vector
Euler
real analysis
real analysis unsolved

I have a problem that I couldn't figure out.

In a 2D velocity vector field $ \vec v(x,y)$, all $ \vec v$ eventually point to a region $ \Omega(x,y)$, i.e. $ \Omega(x,y)$ region has the lowest potetial, so you will end up going into $ \Omega(x,y)$ no matter where you start. How to calculate the total time of traveling from a given point $ (x_0,y_0)$ into $ \Omega(x,y)$ (touching $ \Omega(x,y)$ boundary).

I don't how to solve this mathamatically, and my final goal is to solve this problem numerically (could aslo with some level of approximation). Thanks for your help!

Why don't you just simulate the motion (using Euler or Runge-Kutta)? The problem is that if you know the trajectory, you can easily find the time by integrating the inverse velocity over it but I do not see any way to find the trajectory without solving the corresponding ODE numerically (unless it is of some special type that can be solved analytically).


hello, by rationalizing the given expression you will get
$ \frac {( \minus{} a \plus{} 1 \plus{} \sqrt {1 \plus{} a^2})(b \minus{} 1 \plus{} \sqrt {1 \plus{} b^2})}{2b}$ for $ b\neq 0$.
Sonnhard.


You enter a deserted house late at night.  Inside there is an oil lamp, a gas fire and a stove full of wood.  You have one match however, so which do you light first?

[hide="Click to reveal answer"]The match[/hide]

This one os good fun[hide] the match[/hide][/u]

lol great riddle... i'll hafta remember that for future reference  :D


kinda tough to do it by hand but here you go:

[url]http://www.1728.com/cubic2.htm[/url]

[quote="helpmeplz"]I dont just want the answer.. I wanna see a step by step breakdown of how you do it.

x^3+ 3x^2 -4x-12
________________
x^2 + 5x+ 6[/quote]
I assume you mean solve the equation \[x^{3}+3x^{2}-4x-12=0.\] Well, we can just factor it by doing the following: \[x^{2}(x+3)-4(x+3)=0\]  \[(x^{2}-4)(x+3)=0\]  \[(x+2)(x-2)(x+3)=0\] $x=-2$ or $x=2$ or $x=-3$

Another way to do this is to "guess" roots, and then after you find some, factor them out (through synthetic division or long division).

By the way, if you want a signature, then just enter it [url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=editprofile&sub=sig]here[/url] and click "Submit."

EDIT: Oh, wait a second. Never mind about the signature.  :oops:  He/she was asking about a polynomial division problem. I should've realized that.  :blush:

But anyway, one way you can do it is to factor them both (as I did) and cancel out all common factors. (However, this doesn't work if you're looking for a remainder too.  :wink: )

Here's a website explaining how to do general polynomial long division (which happens to be very similar to regular long divison):
[url]http://www.sosmath.com/algebra/factor/fac01/fac01.html[/url]

Actually the sum of the roots iis $-\frac{b}{a}$ :wink:

[quote="Totally Zealous"][quote="helpmeplz"]I dont just want the answer.. I wanna see a step by step breakdown of how you do it.

x^3+ 3x^2 -4x-12
________________
x^2 + 5x+ 6[/quote]

For the second, $x^{2}+5x+6=(x+2)(x+3)$. You need 2 factors that add to 5 and multiply to 6.
Generally, for $ax^{2}+bx+c$, the product of the roots is $\frac{c}{a}$ and the sum of the roots is $\frac{b}{a}$.

Also, this belongs in Classroom Math.[/quote]

I did not mean to ask how to factor those two numbers. I was asking how you divide this polynomial.

I hope this helps:

http://www.purplemath.com/modules/synthdiv.htm

It demonstrates polynomial long division and Synthetic division, both great ways to divide. I prefer synthetic division-it is more confusing at the beginning, but it is [i]faster[/i].

just use that one theorem (i forgot what it was called  :blush:):

the possible factors of a polynomial are within all the factors of $c$, $a$, and the factors of $c$ over factors of $a$

randomly pluck out and test one by one in a synthetic division

or

you can simply take a graphing calculator and use the factor command  :lol:

[quote="helpmeplz"]I dont just want the answer.. I wanna see a step by step breakdown of how you do it.

x^3+ 3x^2 -4x-12
________________
x^2 + 5x+ 6[/quote]
[hide]$\frac{x^{3}+3x^{2}-4x-12}{x^{2}+5x+6}$

$\frac{x^{2}(x+3)-4(x+3)}{(x+2)(x+3)}$

$\frac{(x-2)(x+2)(x+3)}{(x+2)(x+3)}$

$\therefore \frac{x^{3}+3x^{2}-4x-12}{x^{2}+5x+6}=x-2$

How do you do therefore symbol in Latex?[/hide]

[quote="i_like_pie"][quote="helpmeplz"]I dont just want the answer.. I wanna see a step by step breakdown of how you do it.

x^3+ 3x^2 -4x-12
________________
x^2 + 5x+ 6[/quote]
[hide]$\frac{x^{3}+3x^{2}-4x-12}{x^{2}+5x+6}$

$\frac{x^{2}(x+3)-4(x+3)}{(x+2)(x+3)}$

$\frac{(x-2)(x+2)(x+3)}{(x+2)(x+3)}$

Therefore, $\frac{x^{3}+3x^{2}-4x-12}{x^{2}+5x+6}=x-2$

How do you do therefore symbol in Latex?[/hide][/quote]
To do the "therefore" symbol in $\LaTeX$, simply use "\therefore"
This produces $\therefore$

Many symbols in $\LaTeX$ are typeset in this way; for example, "\alpha" produces $\alpha$, "\prime" produces $\prime$, etc.


Let $ ABC$ be an acute triangle with $ AB > AC$ and its height is $ AD (D \in BC)$. $ O$ is the center of the circumcircle (ABC), $ H$ is the orthocenter of the triangle ABC. Let E is contained on AC such that $ \hat{EDO} \equal{} 90^o$. Prove that  $ \hat{DHE} \equal{} \hat{BCA}$

For every $ E \in ED$ $ \angle{EDO}\equal{}90^\circ$, but $ \angle{DHE}$ is not constant....

Sorry, my post made a mistake. $ E\in AC$

[quote="mr.danh"][color=darkred] Let $ ABC$ be an acute triangle with $ c > b$ . Let $ w = C(O)$ be the its circumcircle and $ H$ be the its orthocenter.

Denote $ D\in AH\cap BC$ and define the point  $ E\in AC$ such that $ DE\perp DO$. Prove that  $ \widehat{DHE}\equiv \widehat{BCA}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \{\begin{array}{c} \{A,L\} = w\cap AH \\
\ F\in DE\cap BL\end{array}$ . Apply the [b]butterfly property[/b] to the point $ D\in BC\cap AL$ in the circle $ w$ :

$ OD\perp EF\implies DE = DF$ . Thus, $ \{\begin{array}{c} DE = DF \\
\ DH = DL\end{array}\|$ $ \implies$ $ HE\parallel LB$ $ \implies$ $ \widehat {DHE}\equiv\widehat {BCA}$ .

[b][u]Remark.[/u][/b] See and my post from[/color] [url=http://www.mathlinks.ro/viewtopic.php?t=204275][color=darkred][b]here .[/b][/color][/url]


inequalities
induction
LaTeX
inequalities unsolved

Let $ x_1, x_2, ..., x_n$ be arbitrary real numbers. Prove the inequality

$ \frac{x_1}{1 \plus{} (x_1)^2} \plus{} \frac{x_2}{1 \plus{} (x_1)^2 \plus{} (x_2)^2} \plus{} ... \plus{} \frac{x_n}{1 \plus{} (x_1)^2 \plus{} (x_2)^2 \plus{} ... \plus{} (x_n)^2}<\sqrt{n}$

I solved it by induction and I 'll post my solution later......Can you find any other ways?

Thank you in advance

I'm being very unhelpful here, but I think I saw that this inequality was presented first by Polya. The hint was to use induction. If i can find that book I'll post the solution...

Well I think it reminds me of a BMO... :wink:


Let a,b,c are reals such that a,b,c>0 and a.a+b.b+c.c=3.prove that 5(a+b+c)+3/(abc)>18 or=18.

Sorry for looking dumb but what does a.a+b.b+c.c=3 mean?

[quote="duy_nguyen"]Let $ a,b,c$are reals such that $ a,b,c>0$ and $ a^{2}+b^{2}+c^{2}=3.$ prove that $ 5(a+b+c)+\frac{3}{abc}\geq18.$ [/quote]
It's known Vasc's inequality.
By the way, for all $ 0\leq k\leq\frac{13+\sqrt{329}}{6}$ the following inequality holds:
$ k(a+b+c)+\frac{3}{abc}\geq3(k+1),$
where $ a,$ $ b$ and $ c$ are positive numbers such that $ a^{2}+b^{2}+c^{2}=3.$


vector calculus integration rotation function parameterization algebra

I am reading a document about Green formula.But I don't know about theory concerned.I will indicate after.... :
1)If boundary U belong C1 then along bdr U is defined the outward pointing unit normal vector .Please define for me :outward pointing unit normal vector  ? it calculate ?
2)In Green formula ,we calculate integral surface on (boundary U).Definition of integral surface  ?
(U is open set in Rn)

I suggest working on the simpler case $n=2$ first. Suppose that the boundary of $U\subset\mathbb{R}^{2}$ is parametrized by a $C^{1}$-smooth simple closed curve $\Gamma\colon [0,1]\to \mathbb{R}^{2}$ that is oriented counterclockwise. Then $\vec\tau=\frac{\Gamma'(t)}{|\Gamma'(t)|}$ is the unit tangent vector to $\partial U$ in the counterclockwise direction. To obtain the outward unit normal vector $\vec n$, rotate $\vec\tau$ by $\pi/2$ degrees clockwise. Formally: if $\vec\tau=(a,b)$, then $\vec n=(b,-a)$ .
For the second part, you can integrate a function $f(x,y)$ over the boundary of $U$ using the parametrization $\Gamma$: \[\int_{\partial U}f = \int_{0}^{1}f(\Gamma(t))|\Gamma'(t)|dt\] This is basically the change of variables formula.

Wouldn't $|\partial U| = 0$ so $\int_{\partial U}f = 0$? Or is the integral not the standard riemann integral?

Good point: it isn't. The standard Riemann integral - $\int_{a}^{b}f(x)dx$ - depends on the orientation of the region of integration (changes sign when $a$ and $b$ are exchanged). In higher dimensions there is a choice of integrating either a function $f$ or a vector field $\vec F$. Integrals of functions do not depend on the orientation of their domain of integration (note $|\Gamma'(t)|$ in the above formula instead of $\Gamma'(t)$); integrals of vector fields do. So, $\int_{a}^{b}f(x)dx$ is actually an integral of a one-dimensional vector field $\vec f(x)=(f(x))$.

(An aside: in multivariable calculus, the Fundamental Theorem applies only to integrals of vector fields, not of scalar functions.)

With n=2 or 3 It is simp .Can you show me with n >3?.

The form of computations depends on how $U$ is defined. For example, suppose that $U=\{\vec x\in\mathbb R^{n}\colon \rho(\vec x)<0\}$, where $\rho\in C^{1}(\mathbb R^{n})$ and $\nabla\rho\ne \vec0$ on $\partial U$. (This $\rho$ is called a defining function of $U$.) In this situation the outward unit normal vector is simply $|\nabla\rho|^{-1}\nabla\rho$. However, this setup makes it difficult to integrate over $\partial U$.

Suppose instead that $\partial U$ is provided with parametrization $x_{i}=x_{i}(t_{1},\dots,t_{n-1})$, where $t_{1},\dots,t_{n-1}$ are parameters. Now the surface integral of a function can be computed as in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105127]this thread[/url]. To find a normal vector to $\partial U$, compute this generalized cross product: \[\begin{vmatrix}\vec e_{1}& \dots & \vec e_{n}\\ \partial x_{1}/\partial t_{1}& \dots & \partial x_{n}/\partial t_{1}\\ \dots & \dots & \dots \\ \partial x_{1}/\partial t_{n-1}& \dots & \partial x_{n}/\partial t_{n-1}\\ \end{vmatrix}\] Is this an inner or outer normal? You'll have to check in each concrete example. And normalize the vector, if you want.

[When I teach flux integrals, I avoid normalizing a normal  :) . You don't actually need the unit normal - what you need is "a normal of right length", which you get from the above determinant, up to orientation.]

Thanks mlok.But ours Green theorem is:$u \in C^{1}\left({\overline U }\right)$ then $\int_{U}{\frac{{\partial u}}{{\partial x_{i}}}dx}= \int_{\partial U}{uv^{i}dS}$ .$v$ is outward pointing unit normal vector.
Where $U \subset R^{n}$ be  arbitrary open and bounded, $\partial U \in C^{1}$.
So we need a general definition so that prove it.You can explain for me? Thanks you.


geometry
parallelogram
trigonometry
geometric transformation
rotation
homothety
ratio

Let a parallelogramm $ ABCD$ and the equilateral triangles $ ACM , BDN$. Prove that  $ MN$ is perpendicular to at least one of the the sides of the parallelogram.

 Babis

[hide]Consider the complex numbers $ a\equal{}0,b\equal{}1,c\equal{}1\plus{}q,d\equal{}q,\omega\equal{}\cos\frac{\pi}{3}\plus{}i\sin\frac{\pi}{3}$. Then $ m\equal{}\omega(1\plus{}q)$ and $ n\equal{}q\plus{}\omega(1\minus{}q)$ and $ m\minus{}n \equal{} (2\omega\minus{}1)q$. However $ 2\omega \minus{}1 \equal{} i\sqrt{3}$ so clearly $ MN$ and $ BC$ are orthogonal. (The case where the triangles are constructed by other angles is treated similarly.)[/hide]

[quote="bertram"][hide]Consider the complex numbers $ a \equal{} 0,b \equal{} 1,c \equal{} 1 \plus{} q,d \equal{} q,\omega \equal{} \cos\frac {\pi}{3} \plus{} i\sin\frac {\pi}{3}$. Then $ m \equal{} \omega(1 \plus{} q)$ and $ n \equal{} q \plus{} \omega(1 \minus{} q)$ and $ m \minus{} n \equal{} (2\omega \minus{} 1)q$. However $ 2\omega \minus{} 1 \equal{} i\sqrt {3}$ so clearly $ MN$ and $ BC$ are orthogonal. (The case where the triangles are constructed by other angles is treated similarly.)[/hide][/quote]

 Thanks for the solution ! I have solved the problem with similar triangles , but this solution is new to me.

 Babis

A rotation of -90° of center $ O$ and then a homothety of ratio $ \sqrt3$ takes the triangle $ AOB$ into the triangle $ MON$. The rest is...silence ! :wink:


How many four-digit positive integers have at least one digit that is a 2 or a 3?

There is a total of 9000 (which is $ 9 \cdot 10 \cdot 10 \cdot 10$) 4 digit numbers. Out of these, $ 7 \cdot 8 \cdot 8 \cdot 8\equal{}3584$ don't fit our criteria. So $ 9000\minus{}3584\equal{}5416$ of them do.


[quote="zaizai-hoang"]Let $a,b,c$ are positive real number. Prove that:
$\frac{27}{64}(a-b)^{2}(b-c)^{2}(c-a)^{2}\ge \frac{3(a^{3}+b^{3}+c^{3})^{2}}{27}$

 :D welcome :P
[b]Problem 2:[/b]
Let $a,b,c$ are positive real number. Prove that:
$\frac{27}{64}(a-b)^{2}(b-c)^{2}(c-a)^{2}\ge \frac{(a^{2}+b^{2}+c^{2})^{3}}{27}$[/quote]
What if $a=b$ $?$


a) A [size=150]good[/size] number is the sum of a two-digit number, with distinct digits, and its reverse. For example, 110 = 37 + 73 is good. How many good numbers are perfect squares?

b) How many numbers less than 10000 have the product of their digits equal to 84?

to make sure its of intermediate level I'm not posting the choices. both of the questions above are multi-choice.

[hide=""mohd_ahsan""]Indeed!!! Very good  ;)  As long as there are more problems that keeps your mind going, there is never a boring moment. For example, Uncle Erdos and J.R. Chen, both exceptional mathematicians, never stopped thinking mathematics, even during sleep too![/hide]

WLOG, the two numbers are ab and ba, where the letters a, b represent digits. When added, we have 11(b+a). Thus, every good number is divisible by 11. The maximum value of such a number is 99+99=198, so the only possibility for a good square number is 121, which checks  out as 56+65.

[hide]a - A good number can be expressed as ab+ba, where a and b are the digits of the number.  This can be simplified to 10*(a+b)+(a+b) = 11(a+b).

Now, all we have to is find the number of square numbers from 11(the smallest good number) to 198(the largest good number) that are divisible by 11.  The only number that works is 121, making the answer 1.

b - There are 5 different sets of numbers we can use to make numbers in which the product of their digits is 85 - {2, 2, 3, 7}, {1, 4, 3, 7}, {1, 2, 6, 7}, {4, 3, 7}, and {2, 6, 7}.  We can just find the number of combinations there are for each set, 12 for the first, 24 for the next two, and 6 for the last two, making a total of 72.[/hide]


If the set of vectors $ {{x_1\cdots x_{n + 1}}}$ is linearly independent, show that $ {x_1\cdots x_n}$ is also LI.

This is obvious intuitively. I am just having trouble putting it onto paper. Can someone point me in the right direction?

Use the definition of linear independence.  If you get stuck, show us your initial steps, and where you got stuck.

You could also go with the contrapositive: if $ x_1,\dots,x_n$ are linearly dependent, $ x_1,\dots,x_n,x_{n\plus{}1}$ are also linearly dependent. Straight from the definition, but you're writing down explicit linear combinations rather than saying they don't exist.

Well, if $ {x_1\cdots n_{n\plus{}1}}$ in LI, then

$ a_1x_1\plus{}\cdots \plus{}a_{n\plus{}1}x_{n\plus{}1} \equal{} 0$ ==> $ a_i\equal{}0 for i in [1,n]$

therefore...

I need to somehow state that same thing for the set of vectors minus the last one, but I have no idea how. I can prove it by saying that if $ {x_1\cdots n_{n}}$ is LD then so must $ {x_1\cdots n_{n\plus{}1}}$ be LD, I have already done the proof of that and so I can always use the contrapositive, but I can't figure out how to prove it alone.

I get stuck right after the definition.

Using the contrapositive is perfectly fine and logical, but since you say you want a direct proof...

You wrote down the definition for $ x_1$, ..., $ x_{n \plus{} 1}$ to be independent.  Even better, write down what it would mean for $ x_1$, ..., $ x_n$ to be independent.  Next, how do you prove such an implication?  Remember that any linear combination of $ x_1$, ..., $ x_n$ is also a linear combination of $ x_1$, ..., $ x_{n \plus{} 1}$, because we can always add in $ 0x_{n \plus{} 1}$.

Yeah I know the contrapositive works but I am new to proofs so it's always good to prove things different ways to get a feel for different methods.

Can you simply tack on the $ 0x_{n\plus{}1}$ onto the end of the definition of what it would mean for $ x_1\cdots x_n$ to be LI and then say that since the first set is LI by assumption, $ x_1\cdots x_n$ must be as well? For some reason that seems like circular logic to me ^_^

Yes, that method is correct, though I would have to see your full proof to make sure you have implemented the idea correctly.  Didn't you use the same trick of adding $ 0x_{n\plus{}1}$ to show that if $ x_1$, ..., $ x_n$ is dependent, then $ x_1$, ..., $ x_{n \plus{} 1}$ is dependent?

Yes I did

OK:

1) If $ {x_1\cdots x_{n \plus{} 1}}$ is LI, then:

1a) $ c_1x_1 \plus{} \cdots \plus{} c_{n \plus{} 1}x_{n \plus{} 1} \equal{} 0$ ==> $ c_i \equal{} 0$ for $ 1 < \equal{} i < \equal{} n \plus{} 1$

2) If $ {x_1\cdots x_{n}}$ is LI, then:

2a) $ c_1x_1 \plus{} \cdots \plus{} c_{n}x_{n} \equal{} 0$ ==> $ c_i \equal{} 0$ for $ 1 < \equal{} i < \equal{} n$

But 2a can be written as $ c_1x_1\plus{}\cdots \plus{}c_{n}x_{n}\plus{}0x_{n\plus{}1}\equal{}0$, so by 1), $ c_1$ through $ c_n$ must also be zero.

Therefore, $ {x_1\cdots x_{n}}$ is LI.

Really? For some reason that isn't sitting well with me, although I couldn't really say why.

Haha don't worry I don't doubt you, it just seems odd for some reason.

Thanks for the help :)


[quote="GameBot"]A palindrome is a number that reads the same forwards and backwards, such as 3003. How many positive four-digit integers are palindromes?[/quote]

There are $ 9$ choices for the first digit, because $ 0$ can't be the first number,
Also, there are $ 10$ choices for the second digit, because $ 0$ can be included.
If its a palindrome, then digit $ 3$ has to be the same as digit $ 2$ and $ 4$ is the same as $ 1$.

So by this we have $ 9*10*1*1\equal{}90$

I did this problem a different way and got 99.I counted all of the numbers from 01 to 99. I thought this worked because for every 2-digit number you could flip it ,multiply by 100 ,and add the orignal to ge a palindrome. Why dosen't this work?

(Sorry for revivnig a semi-old topic)

Because 10, 20, 30, 40, 50, 60, 70, 80, and 90 don't work, so you have to subtract 9 and get 90.


The question is: prove that $ e^{2BA}\equal{}I$. Unfortunately it's false.
$ AB$ is diagonalizable and similar to $ diag(Z,\lambda_1,\cdots,\lambda_k)$ where $ Z$ is the $ 0$ matrix of dimension $ r$ and $ \lambda_j$ is in the form $ 2li\pi$ with $ l$ a non zero integer.
$ spectrum(BA)\equal{}\{0,\cdots,0,\lambda_1,\cdots,\lambda_k\}$.
Now we use a result  by Flanders:
The eigenvalue $ \lambda_j$ of $ AB$ and $ BA$ has the same Jordan structure in both cases.
The eigenvalue $ 0$ of $ AB$ and $ BA$ has not the same structure in both cases but each  nilpotent jordan block of $ AB$ can be associated to a nilpotent jordan block of $ BA$ s.t. their dimensions differ at most by one (and conversely).
Thus $ BA$ is similar to $ diag(N_1,\cdots,N_s,0,\cdots,0,\lambda_1,\cdots,\lambda_k)$  where $ N_j\equal{}\begin{pmatrix}0&1\\0&0\end{pmatrix}$.
$ e^{2BA}$ is similar to $ diag(M_1,\cdots,M_s,1,\cdots,1)$ with $ M_j\equal{}\begin{pmatrix}1&2\\0&1\end{pmatrix}$.

[quote="QuyBac"]Let $ A,B \in M_n(C)$ Such that $ e^{AB} \equal{} I_n$ Prove that $ Ch(BA) \equal{} I_n$
Here $ I_n$ is identity matrix .
$ Ch(X) \equal{} \frac {e^X \plus{} e^{ \minus{} X}}{2}$[/quote](Edit)
  Put $ C \equal{} \sum_{k \equal{} 1}^{ \plus{} \infty}{\frac {(BA)^{k \minus{} 1}B}{k!}} \equal{} \sum_{k \equal{} 1}^{ \plus{} \infty}{\frac {B(AB)^{k \minus{} 1}}{k!}}$
Since : $ e^{AB} \equal{} I_n \plus{} AC$,$ e^{BA} \equal{} I_n \plus{} CA$ From $ e^{AB} \equal{} I_n$ We have : $ AC \equal{} 0$ and
  $ (e^{BA} \minus{} I_n)^2 \equal{} (CA)^2 \equal{} C(AC)A \equal{} 0$ or $ Ch(BA) \equal{} 0$
I believe [b]QuyBac [/b]...  :play_ball:

ZengZeng you are right.
First $ ch(X)\equal{}\dfrac{e^X\plus{}e^{\minus{}X}}{2}$.
Second my mistake is here: in fact $ Ch(BA)\equal{}I$ iff $ (BA\minus{}I)^2\equal{}0$.
Clearly my post implies this result but the proof of Zengzeng is prettier.

add.
$ (e^{BA} \minus{} I_n)^2 \equal{} e^{2BA} \minus{} 2e^{BA} \plus{} I_n \equal{} 0$ or $ e^{2BA} \plus{} I_n \equal{} 2e^{BA}$ or $ e^{BA} \plus{} e^{ \minus{} BA} \equal{} 2I_n$ or $ Ch(BA) \equal{} I_n$
 [b]QuyBac have a mistake !![/b]
 See if $ e^{AB}\equal{}aI_n ,(a \neq 1)$ since $ e^{BA}\equal{}aI_n$

Read in my last post $ ch(BA)\equal{}I$ iff $ (e^{BA}\minus{}I)^2\equal{}0$

zengzeng uses a trick that is valid for any analytic function $ f$ s.t. $ f(AB)$ and $ f(BA)$ are defined: $ Af(BA)\equal{}f(AB)A$. Here we obtain: $ Ae^{BA}\equal{}A$ or $ A(e^{BA}\minus{}I)\equal{}0$.
Thus $ (BA)(e^{BA}\minus{}I)\equal{}0$ therefore $ (e^{BA}\minus{}I)^2\equal{}0$.   QED

[quote="zengzeng"]add.
$ (e^{BA} \minus{} I_n)^2 \equal{} e^{2BA} \minus{} 2e^{BA} \plus{} I_n \equal{} 0$ or $ e^{2BA} \plus{} I_n \equal{} 2e^{BA}$ or $ e^{BA} \plus{} e^{ \minus{} BA} \equal{} 2I_n$ or $ Ch(BA) \equal{} I_n$
 [b]QuyBac have a mistake !![/b]
 See if $ e^{AB} \equal{} aI_n ,(a \neq 1)$ since $ e^{BA} \equal{} aI_n$[/quote]
Thanks you Zeng Zeng !!! I Edited  ! :(


linear algebra
matrix
linear algebra solved

Suppose [b]A[/b] is an [i](n x n)[/i] matirx which satisfies  [b]A.A = 0[/b] matrix[i](nxn)[/i]

I will type identity matrix as [b]In[/b].
Show that [b]In + A[/b] is invertible , with inverse [b]In - A[/b] .

Please help me with this thanks .

[quote="blahblahblah"]You want to show that $(I+A)(I-A)=I$.  See if you can justify the below calculations.

$(I+A)(I-A)=I^2-IA+AI-A^2=I-IA+AI-A^2=I-A+A-A^2=I$[/quote]

I tried this but i cant get the $I-A+A-A^2=I$

Okay.  You know $A^2=0$ (that's given) and $IA=AI$, since the identity commutes with everything.

Oh yeah . That was silly of me i forgot that $A^2=0$. Thanks a lot


Έστω ΑΒΓ ένα όχι ισόπλευρο τρίγωνο και P ένα εσωτερικό του σημείο ώστε οι τρεις σεβιανές που περνούν από το Ρ να έχουν ίσα μήκη και ίσα με K<min(AB,BC,CA) .Αποδείξτε ότι υπάρχει και άλλο διαφορετικό σημείο Q ώστε άλλες τρεις σεβιανές που περνούν από το Q  να έχουν ίσα μήκη .

Ελπίζω να ασχοληθείτε  ;)

σεβιανές? τι σημαίνει αυτό σιλουανέ?

ενημερώνω τους αγπαητούς φίλους που δεν το γνωρίζουν οτι σεβιανές ειναι οι συντρέχουσες ευθείες ετσι αγαπητέ σιλουανέ ? Διορθωσε με αμεσα αν κάνω λάθος η παράλειψη

ένα ευθύγραμμο τμήμα που ενώνει μία κορυφή του τριγώνου με ένα σημείο της απέναντι πλευράς

μετα απο συζητηση με τον Σιλουανο βαζω λυση σε αυτο το παλιο αλυτο και κατα την γνωμη μου αρκετα απαιτητικο θεμα.
[hide]Λοιπον, εστω οτι οι αρχικες σεβιανες που περνανε απο το $ P$ ειναι οι $ AD,BK,CL$.
Τοτε παρατηρουμε οτι οι συμμετρικες των σεβιανων της εκφωνηση (εστω $ AD',BK',CL'$) ως προς τα αντιστοιχα υψη (εστω $ AH,BP,CU$) ειναι και αυτες προφανως ισου μηκους k αλλα και ειναι εσωτερικες στο τριγωνο (αλλιως $ k \geq min(AB,BC,CA)$ και ατοπο).

 συνεπως αρκει να δειξουμε οτι διερχονται απο το ιδιο σημειο εστω Q.

Αρκει δηλαδη απο ceva,
$ \frac{BD'}{D'C}\frac{CK'}{K'A}\frac{AL'}{L'B}\equal{}1$

Ομως η δυναμη σημειου του Β ως προς τον $ (A,AD\equal{}k)$ κυκλο ειναι ιση με $ AB^2\minus{}AD^2\equal{}AB^2\minus{}k^2$ δηλαδη ειναι ιση με την δυναμη του A ως προς τον $ (B,BK\equal{}k))$!!

αρα $ (KA)(K'A)\equal{}(BD)(BD')$ και ομοια $ (LB)(L'B)\equal{}(CK)(CK')$ , $ (DC)(D'C)\equal{}(AL)(AL')$

Συνεπως $ \frac{BD'}{D'C}\frac{CE'}{E'A}\frac{AF'}{F'B}\equal{}\frac{KA}{BD}\frac{LB}{CK}\frac{DC}{AL}\equal{}1$ και τελος.
Η τελευταια ισοτητα ειναι προφανης απο ceva για το P.
To Q ειναι διαφορο του P αφου σε διαφορετικη περιπτωση θα ειχαμε το P να ειναι το ορθοκεντρο ενος τριγωνου με ισα υψη δηλαδη ενος ισοπλευρου τριγωνου-ατοπο.
 :) [/hide]

ΥΓ ρε σιλ μου ειπες οτι ηταν στο forum η ασκηση αλλα απο το 2006? που το θυμοσουν!!??  :P

Είναι και αυτό μια δουλειά ενός moderator  :P . Λοιπον εγώ παραθέτω ακόμη ένα άλλο λινκ για λόγους μαθηματικής πολυφωνίας  :roll:
http://www.mathlinks.ro/viewtopic.php?search_id=132512562&t=16653


What is Droz Farny circle? My book define Droz Farny circle as follow but I think It is wrong.
$ P,Q$ are isogonal conjugate point wrt triangle $ ABC$ and triangle $ DEF$ is pedal triangle of $ P$. Let $ A_1,A_2$ be point on $ BC$ such $ DA_1 \equal{} DA_2 \equal{} PQ$ and define $ B_1,B_2,C_1,C_2$ as similarly. Then $ A_1,A_2,B_1,B_2,C_1,C_2$ are on circle with center $ P$ and this circle called Droz Farny circle.

The result you are asking for is due to R. Honsberger and has the following statement:

[color=blue][b]Theorem[/b][/color]: [color=blue]Let $ P$ and $ Q$ be any pair of isogonal conjugates of a triangle $ ABC$, and let $ D$, $ E$, and $ F$ be the feet of the perpendiculars to the sides from one of the points (say, $ P$), and let circles with centers $ D$, $ E$, and $ F$ be drawn to pass through $ Q$. Then the three pairs of points on the sides of $ ABC$ which are determined by these circles always lie on a circle with center $ P$, and the two circles constructed in this way are congruent.[/color]

See also here: [url]http://mathworld.wolfram.com/Droz-FarnyCircles.html[/url].


geometry
rectangle
analytic geometry
ratio
function
geometry proposed

Hi!

Let $R$ be a rectangle with sides $a$ and $b$. We assume that $R$ can be partionned into $n$ squares with sides $x_1,\cdots,x_n$ (respectively).

Prove that for every $i\in \{1,\cdots,n\}$:
\[ {a\over b}\in {\mathbb Q}\qquad\hbox{and}\qquad {{x_i}\over{a}}\in{\mathbb Q}. \]

I'll prove the weaker part (i.e. $\frac ab\in\mathbb Q$). The second assertion (which clearly implies the first) has been discussed before on the forum. Look through lhvietbao's messages in the "Combinatorics" section. I'd  try to solve that too, but it's hard not to imitate his proof :).

Assume all our rectangles have sides parallel to the coordinate axes. Given such a rectangle $\mathcal R$ with height $u$ and width $v$, define $\tilde f(\mathcal R)=uf(v)-vf(u)$, where $f: \mathbb R\to\mathbb R$ is an additive function. It's clear that $\tilde f$ is additive, in the sense that if $(\mathcal R_i)_{i=1}^n$ tile the rectangle $\mathcal R$, then $\sum_{i=1}^n\tilde f(\mathcal R_i)=\tilde f(\mathcal R)$. If $\frac uv\in\mathbb Q$, then $\tilde f(\mathcal R)=0$. This means that $\tilde f(\mathcal R)=0$ whenever $\mathcal R$ can be tiled with rectangles with rational height-to-width ratio (which is equivalent to the fact that it can be tiled with squares).

Assume now that our rectangle $\mathcal R$ has width $b=1$ and height $a$. We have $\tilde f(\mathcal R)=af(1)-f(a)=0\ (*)$. Choose the additive function $f$ such that $f(x)=0\iff x\in\mathbb Q$ (if $(a_i)_{i\in I}$ is a basis of $\mathbb R$ over $\mathbb Q$ with $a_{i_0}=1$, set $f(a_i)=a_i$ if $i\ne i_0$ and $f(a_{i_0})=0$, and extend to the whole real line by linearity). Now $(*)$ implies that $f(a)=0\Rightarrow a\in\mathbb Q$, and we're done.

Edit:

For the second (stronger) part, the approach outlined [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80695]here[/url] should work. It shows that whenever we tile a rectangle with squares, the associated electrical network determines the lengths of the sides of the squares uniquely, and since the linear systems involved have rational coefficients, the solution we get must have rational solutions as well.

As I said, there's also a different (complete, unlike this one, which is just a sketch :)) solution posted on the forum by lhvietbao.


Let ABC be a triangle in which $B=60+A,\ A\ne 30,\ D\in BC,\ AD\perp BC,\ E\in AC,\ AE=2\cdot BD$ thus: $if\ B\in (DC),\ then\ A\in (EC);\ if\ not\  B\in(DC),\ then\ E\in (AC).$ Ascertain m(<BEC).

Denote a = BC, b = CA , c = AB the sides and $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$ the internal angles of the triangle $\triangle ABC$. Since $AD \perp BC$, the line AD is the triangle altitude from the vertex A and

$BD = AB \cos \beta = c \cos \beta$

Let the perpendiculer bisector of the side AB intersect the altitude AD at a point P. Let the line BP intersect the line CA at a point E'. The triangle $\triangle ABP$ is isosceles with PA = PB. Its base angles $\angle PAB = \angle PBA$ are obtained from the right angle triangle $\triangle ABD$:

$\angle PAB = \angle PBA \equiv \angle DBA = 90^\circ - \beta$

From the triangle $\triangle ABE'$,

$\angle AE'B = 180^\circ - (\angle E'AB + \angle ABE') \equiv 180^\circ - (\angle PAB + \angle ABC) =$

$= 180^\circ - (90^\circ - \beta + \alpha) = 180^\circ - (90^\circ - 60^\circ) = 150^\circ$

$\angle BE'C = 180^\circ - \angle AE'B = 180^\circ - 150^\circ = 30^\circ$

Using the sine theorem for this triangle,

$\frac{AE'}{c} =\frac{AE'}{AB} = \frac{\sin \widehat{PBA}}{\sin \widehat{AE'B}} = \frac{\sin (90^\circ - \beta)}{\sin 150^\circ} = 2 \cos \beta$

$AE' = 2c \cos \beta = 2BD$

Consequently, the points $E' \equiv E$ are identical and the angle $\angle BEC \equiv \angle BE'C = 30^\circ$.

If we consider BD, AE as directed segments, it is clear that if $\beta < 90^\circ,\ \cos \beta > 0$, both the points D, E lie inside of the segments BC, CA, respectively, and if $\beta > 90^\circ,\ \cos \beta < 0$, both the points D, E lie outside of the segments BC, CA, respectively. $\beta = 90^\circ,\ \cos \beta = 0$ is excluded by the condition $\alpha = \beta - 60^\circ \neq 30^\circ$.


geometry
circumcircle
trigonometry
complex numbers
geometry proposed

P is a point on the incircle $\mathcal{C}$ of an equilateral triangle $\triangle ABC$. Show that
$AP^2 + BP^2 + CP^2$ is constant. In fact, if each side of the equilateral triangle has length $2a$, then $AP^2 + BP^2 + CP^2=5a^2$.

I made two proofs of this result: one trigonometric and the other one using complex numbers. The latter is amazingly simple!
I'm looking for a synthetic proof, and I guess it has something to do with Ptolomy's Theorem. Anyway,using the main result I found the following.
If $A'$, $B'$ and $C'$ are such that $AP\cap\mathcal{C}=\{P,A'\}$, etc. what I'm trying to prove is equivalent to
\[ AP\cdot A'P+BP\cdot B'P+CP\cdot C'P=2a^2  \]

Thanks in advance...

try using Steiner's theorem (this one from physics) - it's quite a nice application of center of mass method  :)

hpoincare,

I didn't understand.

You proved that $AP\cdot A'P+BP\cdot B'P+CP\cdot C'P=2a^2 \ \ (1)$

and you want to use this to prove the theorem?

Or you need a proof for $(1)??$

By the way, check the related links

http://www.mathlinks.ro/Forum/viewtopic.php?t=58199
http://www.mathlinks.ro/Forum/viewtopic.php?t=59086
http://www.mathlinks.ro/Forum/viewtopic.php?t=849

and one with circumcircle: http://www.mathlinks.ro/Forum/viewtopic.php?t=50605

The last one uses Ptolemy's Theorem

Hey thanks for the links!

I just said that I proved that

$PA^2+PB^2+PC^2=5a^2$ (*)

if and only if

$PA'\cdot PA+PB'\cdot PB+PC'\cdot PC=2a^2$

I made two proofs of (*) one using complex numbers and the other using trigonometry, so I was looking for a synthetic proof.
The generalization is great but you are using in the proof the scalar product, so thats Law of Cosine...
Anyways, thank you very much!

[quote="hpoincare"]please explain in more detail...
thanks[/quote]

give weights $m$ to vertices $A,B,C$ and use Steiner Theorem to count moment of inertia (I think this is its english name  :) ) for point $P$ by using moment of inertia of $O$ - center of circle. I know this explanation isn't very good but my physics english is awfull - hope it helps


Euler
geometry
circumcircle
geometry unsolved

Give a cute straingle $ \triangle ABC$ . I is Tolicelli of it . Prove Euler line of  $ \triangle IBC$ , $ \triangle IAB$ , $ \triangle IAC$
current ( cut at a point)

[quote="shortlist"]Who can help me?
I thoungt it pay for much times[/quote]

1) You posted a geometry problem in an algebra forum so less interested readers.

2) you posted in a "proposed and own" forum, so you claimed that you have the solution and ask for a second one. So nobody is on hurry to answer you.

Maybe a better  choice would be to post in "geometry unsolved problems" :)

The notation and wording is inconvenient. Try to use $ I$ for the incenter.

[quote] $ T$ is Toricelli point of $\triangle ABC.$ Prove Euler lines of  $ \triangle TBC,$ $ \triangle TAB,$ $ \triangle TAC$ concur.[/quote]
Let $ \triangle A'BC , \triangle B'CA$ and $ \triangle C'AB$ be the three equilateral triangles erected outwardly on $ BC,CA,AB.$ Let $ X,Y,Z$ be their circumcenters, thus $ T \equiv (X) \cap (Y) \cap (Z)$ and $ T\equiv AA' \cap BB' \cap CC'.$ Let $ G_1,G_2,G_3$ denote the centroids of $ \triangle TBC, $ $\triangle TCA ,$ $ \triangle TAB$ $ \Longrightarrow$ $ XG_1,YG_2$ and $ ZG_3$ are the Euler lines of $ \triangle TBC, \triangle TCA , \triangle TAB.$ If $ M$ is the midpoint of $ BC$ and $ G$ is the centroid of $ \triangle ABC,$ we get

$ \frac {MX}{MA'} \equal{} \frac {MG_1}{MT} \equal{} \frac {MG}{MA} \equal{} \frac {1}{3}$

It follows that $ X,G_1,G$ are collinear on a parallel line to $ AA'.$ Hence, Euler lines of $ \triangle TBC,$ $ \triangle TCA ,$ $ \triangle TAB$ concur at the centroid $ G$ of $\triangle ABC.$


linear algebra
matrix
inequalities
vector
induction
abstract algebra
special factorizations

Let $A=\begin{pmatrix}1&\frac12&1\\ \frac12&\frac13& 1\\1&1&x\end{pmatrix}.$

Suppose that $A$ is (at least) positive semidefinite.

Prove that $x\ge 4.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77089]The source of this problem.[/url]

Here's a more challenging version: Let $A$ be the following $n\times n$ matrix.

$A=\begin{pmatrix}1&\frac12&\cdots&\frac1{n-1}&1 \\ \frac12&\frac13&\cdots&\frac1n&1 \\ \vdots&\vdots&\ddots&\vdots\\ \frac1{n-1}&\frac1n&\cdots&\frac1{2n-3}&1\\ 1&1&\cdots&1&x\end{pmatrix}$

There is a number $C_n$ such that if $A$ is (at least) positive semidefinite, then $x\ge C_n.$ $C_3=4,$ assuming the original problem is correct.

Calculate $C_4,C_5,\dots.$

What do u mean by semidefinite positive ?
Sorry i'm french, i don't know yet the right english terms to use.
Is this when sp(A) is included in the set of positive reals or positive reals <>0 ?

Yes that's what I mean. All eigenvalues of $A$ are $\ge 0,$, or, equivalently, $x^TAx\ge0$ for all column vectors $x.$

Positive definite would mean that all eigenvalues are positive, or that $x^TAx>0$ for all $x\ne 0.$ Saying "semidefinite" instead of "definite" simply relaxes that $>$ to a $\ge.$

[quote="Kent Merryfield"]In retrospect, I started this out in the wrong place, so I've moved it now. Does anyone want to get concrete and just do the specific cases of $n=3,\,n=4,$ and maybe $n=5?$[/quote]Let $A(x)$ be the matrix in question. The top left $(n-1) \times (n-1)$ matrix is the infamous Hilbert matrix $H_{n-1}$, so it's (barely) positive definite. Then $A(x)$ is positive semidefinite (with one zero eigenvalue) for the single $x$ for which $\det A(x) = 0$, and positive definite for all larger $x$.

Thus let's find an eigenvector $(a,\alpha)^T$ of $A(x)$ with eigenvalue zero, where $a \in \mathbb{R}^{n-1}$ and $\alpha$ is a real number. This leads to the conditions $H_{n-1}a + \alpha e = 0$ and $e^Ta + \alpha x = 0$, resulting in $x = e^T H_{n-1}^{-1}e$. Here $e = (1, \dots, 1)^T \in \mathbb{R}^{n-1}$. Thus $x$ = sum of all entries of $H_{n-1}^{-1}$, which are known to be products of binomial coefficents. It looks like $x = (n-1)^2$. That's probably well-known.

[quote="hpe"]
Thus let's find an eigenvector $(a,\alpha)^T$ of $A(x)$ with eigenvalue zero, where $a \in \mathbb{R}^{n-1}$ and $\alpha$ is a real number. This leads to the conditions $H_{n-1}a + \alpha e = 0$ and $e^Ta + \alpha x = 0$, resulting in $x = e^T H_{n-1}^{-1}e$. Here $e = (1, \dots, 1)^T \in \mathbb{R}^{n-1}$. Thus $x$ = sum of all entries of $H_{n-1}^{-1}$, which are known to be products of binomial coefficents. It looks like $x = (n-1)^2$. That's probably well-known.[/quote]

Could you please explain why$H_{n-1}a + \alpha e = 0$ and $e^Ta + \alpha x = 0$?

It's funny that no one has provided the simple solution to this problem. Completing the square we have that
$x^TAx = (x_1 + \frac12 x_2 + x_3)^2 + \frac{1}{12}(x_2+6x_3)^2 + (x-4)x_3^2$

This approach is obvious if you've seen the law of inertia (or so we call it in swedish). To prove the more challenging version I believe you only need to use induction on completing the last square, but I'm too lazy.

[quote="spix"]Could you please explain why$H_{n-1}a + \alpha e = 0$ and $e^Ta + \alpha x = 0$?[/quote]Let $e$ be the column vector with all ones, and let $b = (a^T,\alpha)^T$ be a vector in the kernel where $a$ is an $n-1$-vector and $\alpha \in \mathbb{R}$.Then
\[ A(x)b = \begin{pmatrix} H_{n-1} &\ e \\ e^T &\ x \end{pmatrix}\begin{pmatrix}a \\ \alpha\end{pmatrix} = \begin{pmatrix} H_{n-1}a + \alpha e \\ e^T a + \alpha x \end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} \]
by block multiplication. Therefore $H_{n-1}a + \alpha e = 0$ and $e^Ta + \alpha x = 0$. It follows that $a = -\alpha H_{n-1}^{-1}e$ and therefore $\alpha x = - e^Ta = \alpha e^TH_{n-1}^{-1}e$. Now $\alpha \ne 0$, since $H_{n-1}$ is non-singular. Therefore $x = e^TH_{n-1}^{-1}e$.


Reduced to lowest terms, $ \frac {a^2\minus{}b^2}{ab}\minus{} \frac {ab\minus{}b^2}{ab\minus{}a^2}$ is equal to:

$\textbf{(A)}\ \dfrac{a}{b} \qquad
\textbf{(B)}\ \dfrac{a^2-2b^2}{ab} \qquad
\textbf{(C)}\ a^2 \qquad
\textbf{(D)}\ a-2b \qquad
\textbf{(E)}\ \text{None of these}$

[hide="Solution"]

$ \frac{a^2\minus{}b^2}{ab} \minus{} \frac{b(a\minus{}b)}{a(b\minus{}a)} \equal{} \frac{a^2\minus{}b^2}{ab} \plus{} \frac{b}{a} \equal{} \frac{a^2\minus{}b^2\plus{}b^2}{ab} \equal{} \frac{a}{b}$. A. [/hide]

Another way to do this is by taking a and b to be two simple integers and trying all possible ways. For example, if a = 1 and b = 2 then we can simplify the main expression to 1/2. Checking A, B, C, and D we find that A is the only one that results in 1/2 also and so must be the simplified expression.


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Let $ k$ be a field and $ k[X]$  the polynomial ring in one variable.
Let $ P \equal{} (X)$, the ideal generated by $ X$ and let $ k[X]_P$ be the localization of $ k[X]$ with respect to $ P$ (i.e. $ S^{ \minus{} 1}(k[X])$, where $ S \equal{} k[X] \minus{} P$)
What is the dimension (sup of lengths of prime chains) of $ k[X]_P$?

As localisation preserves PID, it is one-dimensional.


if a,b,c,d are positive real numbers such that a>=b+c+d,prove that

                         (a+b)(b+c)(c+d)(d+a)>=20abcd.

Please don't use the Round Table to post math problems.  Use the math forums.  Thanks.


Find the maximum value of $ 3sint+2cost$

Answer

let $ 3\sin{t}+2\cos{t}\equiv R\sin{(t+\alpha)}$ where $ R > 0$

(Obviously the amplitude, $ R$, gives the max value)

$ \Rightarrow 3\sin{t}+2\cos{t}\equiv R\sin{t}\cos{\alpha}+R\cos{t}\sin{\alpha}$

$ \Rightarrow R\cos{\alpha}= 3$ and $ R\sin{\alpha}= 2$

$ \Rightarrow R^{2}(\cos^{2}{\alpha}+\sin^{2}{\alpha}) = 13$

$ \Rightarrow R^{2}= 13 \Rightarrow R = \sqrt{13}$

$ \Rightarrow$ Max value $ = \sqrt{13}$


If $\log_MN=\log_NM$, $M\ne N$, $MN>0$, $M\ne 1$, $N\ne 1$, then $MN$ equals:

$\text{(A)} \ \frac12 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 10 \qquad \text{(E)} \ \text{a number greater than 2 and less than 10}$

[hide]By the logarithm chain rule, we know that $(\log_NM)(\log_MN)=\log_NN=1$. Using this equation with the one provided, we see that $(\log_MN)^2=1$. Therefore, $\log_MN=\pm 1$. If it equals $1$, we get $M=N$, which is useless. If it's $-1$, we see get $M=\frac{1}{N}\implies MN=1$.[/hide]

[hide="Solution"]$\log_M{N}=\frac{1}{\log_M{N}}$, so $(\log_M{N})^2=1 \implies \log_M{N}=\pm 1$. there are thus two cases:
1) $M^1=N$, $MN$ is not ascertainable.

2) $M^{-1}=N \implies MN=\boxed{1}$[/hide]


Find all polynomials with real coefficients satisfying $f(x^{2})=f(x)f(x-1)$.

Use lemma : Let $P,Q,H\in\mathbb{R}[x]$ such that $\min\{\deg{P},\deg{Q}\}\geq 1$ and $\deg{P}+\deg{Q}=\deg{H}$. If $\deg{P}>\deg{Q}$ or $\deg{P}=\deg{Q}$ but $\deg{P}+\deg{Q}=\deg{Q}=\deg{P}$ then for any positive integer $n$ have at most $f\in\mathbb{R}[x]$ such that $\deg f=n$ and $f(P(x)).f(Q(x))=f(H(x))\forall x\in\mathbb R$.

@N.T.TUAN: Can you give a solution of your lema please?
Also I do not understand this part of the lema [quote]$\deg{P}>\deg{Q}$ or $\deg{P}=\deg{Q}$ but $\deg{P}+\deg{Q}=\deg{P}=\deg{Q}$[/quote]
You mean that you can only have $\deg{P}=\deg{Q}$ only when both are $0$?

Ok ...but i think  that you can do the problem directly: you take a to be a zero of f. If a=0 then you can easily get that f has an infinity of roots so f is 0. If |a| is different from 1, you get again a contradiction. So it remains |a|=1 for any root. but (a+1)^2 is also a root so |a+1|=1. If you draw the 2 circles and you write their equations you get that a is one of the roots of the polynomial x^2+x+1. As the coefficients of f are real, it follows that f=(x^2+x+1)^n.

[quote="ciprian"]
You mean that you can only have $\deg{P}=\deg{Q}$ only when both are $0$?[/quote]
No, $P=b_{n}x^{n}+...$ and $Q=c_{n}x^{n}+...$ with $b_{n}+c_{n}\not = 0$.

[quote="ciprian"]Ok :) What about the solution to the lema N.T.TUAN?[/quote]
Because prove of this lemma  is long and not of mine, therefore I only post this prove if you can't prove it, later endeavour :)


ratio
geometry
rectangle
integration
calculus
calculus computations

Can i get some help with this problem? i don't really understand what to do, it looks related to area between curves but it's kinda confusing to me :oops:

The figure below shows part of the curve $y=x^3$ and a rectangle with two vertices at $(0,0)$ and $(c,0)$. what is the ratio of the area of the rectangle to the shated part of it above the cubic?

a) 4:3
b) 4:1
c) 3:2
d) 3:1
e) 5:4

[img]http://img91.imageshack.us/img91/2659/pic1pe.jpg[/img]

hello, the area of the rectangle is $ A_r\equal{}cf(c)\equal{}c^4$.The area under the curve is given by $ A_c\equal{}\int_{0}^cx^3dx\equal{}\frac{c^4}{4}$. So our ratio is
$ \frac{c^4}{c^4\minus{}\frac{c^4}{4}}\equal{}\frac{c^4}{\frac{3}{4}c^4}\equal{}\frac{4}{3}$.
Sonnhard.


1. Mewto55555
2. gogators
3. cyberspace
4. KING1
5. (your username here)

meewhee009, if you want to signup, please hurry, I think the cutoff will be 9/6/09

1. Mewto55555
2. gogators
3. cyberspace
4. KING1
5. EggyLv.999

I deleted the previous for a reason. Don't make this again. Put it in the SSBB thread or PM each other among yourselves.


I'll make it easier: There are 3 starters, 10 main dishes, and 5 desserts to choose from for dinner. 4 people go to the restuarant. How many different combinations of the dinner are there?

P.S. Sorry for making the wording strange...

[hide]I'm guessing it would be $3*3*3=27$ but i don't fully understand the question[/hide]

[quote="pascal_1623"]If there is the person who hates ice cream and Bobby who just gets the full entree, how many different combinations are there?
(The person who hates ice cream doesn't order it.)[/quote]

I don't get what that part means.

The wording is kind of confusing, but here is my best shot:

[hide="best shot"]It appears as if there are 4 choices for spaghetti, 3 choices for sauce and 4 choices for icecream.
Person who hates icecream only has $4\cdot3=12$ choices to order.
Bobby has $4\cdot3\cdot4=48$ choices to order.
Multiply them together to get the total number of combinations $12\cdot48=\boxed{\boxed{576}}$[/hide]


complex numbers
geometry
perpendicular bisector

I have no idea how to even start this problem....

If p and q are unequal complex numbers, the equation of the perpendicular bisector of the segment pq is |z - p| = |z - q|
[size=59]............................................................................[/size]_[size=59]..........[/size]       _
By squaring, reduce this to the form Bz + Bz + C = 0 where B = p – q which does not = 0.

Also show, conversely, that the latter equation with B not equal to 0 can be reduced to the former with p not equal to q.

And sorry I do not seem to know how to use the symbols on here, so I did the best II could

Thank you for your help...

use the fact that for every complex number $ w$, $ |w|^2\equal{}w\overline{w}$

I am still not sure.  I see that, but I cannot figure what to do with it.


the problems confuse me quite a while...

for the inequality (1/a^3(b+c) +(1/b^3(a+c) +(1/c^3(b+a)>= 3/2, why can we make a conclusion that LHS can never be equal to RHS because it's not homogenous?

and for a homogenous inequality, why we can normalize it by adding a condition that is designed by ourselves? :?:

it would be great if someone can explain these qts to me wif example! :D

[b]Normalisation[/b]

Essentially, we can apply a transform to get the variables into a form which suits us because any transform does not affect the value of an expression.

Consider this simple example

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$. Let $ a \plus{} b \plus{} c \equal{} k$. Then $ \frac {a}{k} \plus{} \frac {b}{k} \plus{} \frac {c}{k} \equal{} 1$

Well, if we substitute this into out original expression

$ \frac {\frac {a}{k}}{\frac {b}{k}} \plus{} \frac {\frac {b}{k}}{\frac {c}{k}} \plus{} \frac {\frac {c}{k}}{\frac {a}{k}}$

But this all cancels and we are left with out original expression. So the transform is valid.

A good example given in Engel.

Let $ f(x,y,z) \equal{} x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} xz$

If $ f(x,y,z) \ge 0$ then $ f(kx,ky,kz) \equal{} k^2f(x,y,z) \ge 0$. So by constructively choosing $ k$, this problem becomes much simpler (than it already was)

Some standard exercises:

Show that $ x^4\plus{}y^4\plus{}z^4 \ge x^2y^2\plus{}y^2z^2\plus{}z^2x^2$

actually i am still confused...

but i think it would be better if you solve x^4+y^4+z^4>=x^2y^2+y^2z^2+x^2z^2 with normalization and tell why you can use normaliation and how to custruct it on that case. then i think i can understand wt normalization is. :lol:

Engel's Problem Solving Strategies (a book)

Anyway, for the exercise, I'll give you a helping hand, then you can try and give it a go yourself.

Let $ f(x,y,z) \equal{} x^4\plus{}y^4\plus{}z^4\minus{}x^2y^2\minus{}y^2z^2\minus{}z^2x^2$.

Then look at $ f(kx,ky,kz)$ and tell me what happens.

i think i can partly get the idea now.

for any x,y,z we have summation{x^2y^2}=k, we can take x=k^(1/4)x_1, y= x=k^(1/4)y_1, z=k^(1/4)z_1, and if we substitute these three equation into summation{x^2y^2}=k, we have summation{ (x_1)^2(y_1)^2}=1

for the problem, summation{x^4}>=summation{x^2y^2}, as it's homogenous, we can pull the k out from each side, then it remains to prove that

summation{(x_1)^4}>=1, with the condition summation{ (x_1)^2(y_1)^2}=1 .

for conveneince, we may just assume that for every homogenous equation with n variables, we have summation {a_1)=1, and it changes the structure and simplifies the question. am i right??


i guess this is very easy for this forum, but sorry I am new to functions :(  , and i cant get these problems :(

if f(x) is a polynomial satisfying f(x)f(1/x)= f(x) + f(1/x) and f(3)=28 then find f(4)

if u know that there is only one answer u could find a funtion like $f(x)=x^3+1$ and say $f(4)=65$, but i guess that is not a quite satisfiying answer ...

Im very new to functions aswell are there some rules?

[quote="Hikaru79"]Is the degree of the function given? I do not think the problem is solvable if it's not, but I could be very wrong  :blush:[/quote]

The problem is solvable. I may ask you to solve the problem.
Actually,
if the function as is given by eminem is true, then function is
either x^n +1 or x^n -1.
Here we can take n=3 and + sign. The answer is 65.

[quote="varun"]
if the function as is given by eminem is true, then function is
either x^n +1 or x^n -1.
Here we can take n=3 and + sign. The answer is 65.[/quote]

can u prove that?

Well I don't remember the solution. But I will try to find it out and post the solution by tomorrow.

Hello

Let g(x)=f(x)-1 ;Since f is a polinom : g(1/x)=f(1/x)-1 because the additional "-1" only changes the constant number of the function
So g(x)*g(1/x)=1
Suppose deg(g)=n and the least degree in g is m (m may equal 0) :
g(x)=an*x^n +...+ am*x^m
g(1/x)=an*(1/x)^n +...+ am*(1/x)^m
so deg( h(x)=g(x)*g(1/x) )=n-m
But since h(x)=1 we can say n-m<=0 ;otherwise
h(x)=an*am*x^(n-m) +...
if x grow enough h(x) will grow a lot - bigger than any certain number .
So n-m<=0
But m cannot be bigger than n (as defult), so n=m
therfore g is like: "ax^n"
and a=+-1 because g(x)*g(1/x) = a*x^n * a*(1/x)^n = a^2 =1
and it means g(x)=+-x^n       (+- means + or -)
g(3)=f(3)-1=27=+-3^n
therfore n=3 and g(x)=x^3
So f(x)=x^3+1

f(4)=64+1=65

Sorry for my poor English
Good Luck
Khoda Hafez (a Persian Farewell)


inequalities
LaTeX
search
real analysis
real analysis unsolved

Show that for a sequence of positive reals $ (a_{n})_{n \equal{} 1}^{\infty}$ we have $ \limsup\left(\frac{1\plus{}a_{n\plus{}1}}{a_{n}}\right)^{n}\geq e$.
(Also - how does one get Latex to put big brackets around expressions such as the one in my above post?)
EDIT;Fixed the brackets thing. Thanks JBL.

[quote="EBz89"]
(Also - how does one get Latex to put big brackets around expressions such as the one in my above post?)[/quote]

\left( .... \right)  expands the ( and ) to a larger size when the expression between them is large.  This also works for other delimiters.  You need a \right to accompany each \left and vice-versa.  Another option is to use size commands like \Big.


geometry
tangent circles
Inequality
geometric inequality
IMO Shortlist
IMO Longlist

Given two internally tangent circles; in the bigger one we inscribe an equilateral triangle. From each of the vertices of this triangle, we draw a tangent to the smaller circle. Prove that the length of one of these tangents equals the sum of the lengths of the two other tangents.

We use Casey's theorem on the system formed by the three vertices of the triangle and the small circle. Let $A,B,C$ be the vertices of the triangle, and let $t_a,t_b,t_c$ be the lengths of the tangents from $A,B,C$ (respectively) to the small circle. Then, if we assume $t_a\ge t_b\ge t_c$, we get $t_a\cdot BC=t_b\cdot CA+t_c\cdot AB\ (*)$. Since $BC=CA=AB$, we can simplify $(*)$ by $BC=CA=AB$ to get $t_a=t_b+t_c$.

i think there is a mistake in the solution . it should be $t_a + t_c$ = $t_b$


Find [tex]$\displaystyle \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta + \tan \theta}$[/tex].

I can solve this by writing [tex]$\tan \theta = \frac{\sin \theta}{\cos \theta}$[/tex], simplifying, and then using l'Hospital's Rule. But is there a way to evaluate this limit without l'Hospital?

(hehe, [tex]\TeX[/tex] is fun.)

Does L'Hospital say that [tex]\displaystyle \lim_{\theta \to 0}~\sin~\theta = \theta[/tex]?

L'Hospital's Rule (or at least the case of it that applies here):

If [tex]$\displaystyle \lim_{x \rightarrow a} f(x)=0$[/tex] and [tex]$\displaystyle \lim_{x \rightarrow a} g(x)=0$[/tex], then [tex]$\displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)}=\displaystyle \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}[/tex]

By the way, how did you get the [tex]x \rightarrow a[/tex] to go under the [tex]\lim[/tex]?

[i]Edit: Thanks for the tip, nr1337.[/i]

\displaystyle on the same line as the limit:
[tex]
\displaystyle \lim_{\theta \to 0}
[/tex]

Not having taken calc (:-(), but knowing some basic stuff, is the answer [hide]1/2[/hide]?  Most likely I have no clue what I'm doing, but...

Yeah, it's [hide]1/2[/hide], but I'm not sure how to get to it without l'Hospital.

Here's how to get it without L'hopital:

[hide]Rewrite it as [tex]\displaystyle \lim_{\theta\to0}\frac{\tfrac{1}{2}\sin2\theta}{\theta\cos\theta+\sin\theta}[/tex] and then use Power Series.[/hide]

I'm not sure what those are...I'm only through about the first 1/4 of calculus. This problem was in an early chapter of the textbook, even before l'Hospital. The section of the chapter that this problem was in covered stuff like [tex]$\displaystyle \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1$[/tex].

Well, thinking geometrically with the unit circle, [tex]\displaystyle \lim_{\theta \to 0}~\tan\theta = \theta\mbox{, and }\lim_{\theta \to 0}~\sin\theta = \theta[/tex]

Of course, this is for theta in radians, and I might be wrong.

Edit:

Also, remember that tan :theta: = sin :theta:/cos :theta:, and cos :theta: should be 1 (I think)

[tex] \displaystyle {\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta + \tan \theta}} [/tex] = [tex] \displaystyle {\lim_{\theta \rightarrow 0} \frac{\frac {\sin \theta}{\theta}}{1 + \frac {\tan \theta}{\theta}}} [/tex] = [tex] \displaystyle {\frac {1}{1 + 1}} [/tex] = [tex] \displaystyle {\frac {1}{2}} [/tex]

Wow gauss202, that's amazing! Thanks for the help.


er...
Should the coefficient of the third term be $ 192$ instead of $ 92$?

I have just made a confirmation of the original, but I can't find typo.

[hide="answer"]$ (y+2x-2)(y-4x+4)(3y-2x-6)(3y+4x+12)=0$[/hide][hide="solution"]
It isn't helpful in general case, but here it works... ;)
We assume that it can be written as four lines equations.
We put $ x=0$ and we'll get $ y=2\vee y=-4$ and we put $ y=0$ and we'll get $ x=1\vee x=-3$, and if we put $ x=3$ we'll get $ y=\{\pm 4,\pm 8\}$, so we have 8 points:
$ (0,2)\ (1,0)\ (-3,0)\ (0,-4)\ (3,\pm 4)\ (3,\pm 8)$,  and now we look at these 8 points and we remember what we assumed before and we notice that points $ (0,2)\ (1,0)$ determine one line, $ (0,2)\ (-3,0)$ the second one, $ (-3,0)\ (0,-4)$ the third and $ (0,-4)\ (1,0)$ the last one, so:
$ (y+2x-2)(3y-2x-6)(3y+4x+12)(y-4x+4)=0$ and we should only now check is that correct... ;D and after ugly calculations we'll get what we want... ;P

I am askin' is there any more likeable solution?[/hide]


projective geometry
Gauss
geometry
circumcircle
geometry solved

Let $ABC$ be a triangle and $P$ a point in its interior, not lying on any of the medians of $ABC$. Let $A_1$, $B_1$, $C_1$ be the intersections of $PA$ with $BC$, of $PB$ with $CA$, of $PC$ with $AB$ respectively, and let $A_2$, $B_2$, $C_2$ be the intersections of $B_1C_1$ with $BC$, of $C_1A_1$ with $CA$, of $A_1B_1$ with $AB$, respectively. Prove that the midpoints of $A_1A_2$, $B_1B_2$, $C_1C_2$ are collinear.

One idea would be fixing $AP$ and moving $P$ around on this fixed line. The midpoint $A'$ of $A_1A_2$ would be fixed (because $A_1,A_2$ are fixed), and the mapping $B'=$midpoint of $B_1B_2\mapsto C'=$midpoint of $C_1C_2$ would be projective. In this case, in order to show that $B'C'$ passes through $A'$, it would be enough to show this for three particular positions of $P$: we can take $P\in BC$, and then $P$ being the intersection between the median from $B$ (and then $C$) with $AP$. We get some parallelisms which are easy to deal with.

Actually, that's why the problem forbids medians: we get parallel lines, and one of the points $B',C'$ becomes a point on the infinity line. We can still use that to our advantage though :).

Hey, Grobber, I thought you would recognize it...

Since the lines $AA_1$, $BB_1$, $CC_1$ concur at one point (namely, at the point P), the triangles ABC and $A_1B_1C_1$ are perspective, and thus, by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30323]the Desargues theorem[/url], the points $BC\cap B_1C_1 = A_2$, $CA\cap C_1A_1 = B_2$ and $AB\cap A_1B_1 = C_2$ lie on one line. This line, together with the three lines $B_1C_1A_2$, $C_1A_1B_2$ and $A_1B_1C_2$, forms a complete quadrilateral. The diagonals of this complete quadrilateral are the segments $A_1A_2$, $B_1B_2$ and $C_1C_2$. By [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=79751]the Gauss theorem[/url], the midpoints of the diagonals of a complete quadrilateral are always collinear. Hence, the midpoints of the segments $A_1A_2$, $B_1B_2$ and $C_1C_2$ are collinear. $\blacksquare$

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=827[/img]

  Darij

Here I briefly show my proof.

Let $O$ be the circumcenter of $ABC$ and $H$ be the orthocenter of $A_1B_1C_1$. Suppose $(a), (b), (c)$ are the circles with diameters $A_1A_2, B_1B_2, C_1C_2$ respectively (hence, their centers are the midpoints $M, N, P$ of $A_1A_2, B_1B_2, C_1C_2$ respectively). We will show that $OH$ is the common radical of $(a), (b), (c)$.

Since $(A_1A_2BC) = -1$, according to Newton's formula, we obatin $MA_1^2 = MB.MC = MO^2-R^2$. Thus $OM^2 - [r(a)]^2 = R^2$, here $R$ and $r(a)$ are the radii of the $(O)$ and $(a)$. By similar argument, we obtain $OM^2 - [r(a)]^2 =OM^2 - [r(b)]^2 = OM^2 - [r(c)]^2 =R^2$. Hence $O$ has the same power with respect to $(a), (b), (c)$.

It is not hard to show that $H$ also has the same power with respect to $(a), (b), (c)$. Therefore, the problem is solved.


$ S \equal{} \{1,2,\dots,n\}$ is divided into two subsets. How the set is divided, if there exist two elements whose sum is a perfect square, then $ n$ is at least  ?

$\textbf{(A)}\ 13  \qquad\textbf{(B)}\ 14  \qquad\textbf{(C)}\ 15  \qquad\textbf{(D)}\ 16  \qquad\textbf{(E)}\ 17$


Find all triples $ (k,m,n)$ of nonnegative integers satisfying $ 5^m\plus{}7^n\equal{}k^3.$
Sorry if this is posted before.

Consider mod 2 we have $ k$ is an even number so $ 8|k^3$ ,it follows that $ 8|5^m+7^n$ .Therefore $ n$ is an odd number . Because ${ k^3(\mod 7)\in \{1,-1}$ so we must have $ 7|5^{2m}-1$ .It is not hard to show that $ ord_{5} mod 7=3$ ,hence $ 3|m$ .Let $ m=3a$ then the equation is equivalent to the following equation : $ k^3-5^{3a}=7^n$ ,and this equation is not very to solve .


For $x,y,z$ and $a,b,c$ positive reals such that $a+b+c=x+y+z=1$prove that

$1)$ $x^4z^2+z^4y^2+y^4x^2 \geq xyz(x^3+y^3+z^3)$ (already posted)

$2)$ $x^4c^2+z^4b^2+y^4a^2 \geq ax^4b +cz^4a +by^4c$ (I tried to generalize 1) )

[quote="manlio"]$2)$ $x^4c^2+z^4b^2+y^4a^2 \geq ax^4b +cz^4a +by^4c$[/quote]

That's definitely wrong. A random counterexample: x = 1, y = 3, z = 1, a = 1, b = 3, c = 4. Did you mix up some variables?

  Darij

Sorry, I will think of another kind of generalization, if possible. :blush:

Sorry, but I have a counterexample for (1) as well.
If we let z=1, y=2, x=10000, then all terms are negligble except for $x^4 z^2=10000$ and $x^4 yz=20000$. So the RHS is clearly larger.

I will check for 1) and 2). Probably I forgot to say that x+y+z=a+b+c=1 or I made a big mistake.
I must find them in my notes as when I wrote them I tried only to remember them.

Sorry, very much.

I don't se the point for $x+y+z=1$ or $a+b+c=1$.
If $S=m+n+p$ and $x=\displaystyle \frac mS,y=\frac nS, z=\frac pS$, $T=r+s+t$ and $\displaystyle a=\frac rT,b=\frac sT,c=\frac tT$ then the condition that $x+y+z=a+b+c=1$ is fullfiled but if we introduce these variables in the inequalities we get the same inequalities:
$m^4p^2+p^4n^2+n^4m^2\ge mnp(m^3+n^3+p^3)$ but with $m,n,p>0$ were darij's exemple would be a counterexample for the inequality.
Analoguous for the second one.
So manlio I think that these two inequalities are wrong :(

Can anyone delete this my delirious post?

Thank you very much.

manlio: you can correct your first inequality by mulplying left side by 2 - this can be checked easily with majorization


calculus
real analysis
real analysis theorems

well the topic i know seems stupid but sorry couldn't get a better one  :D
anyways does any one know of a book(Preferably an e-book) which has all the theorems and formuales on  limits like those frequently posted in calculus computations and tutorials which are really helpful to solving problems on limits as an example take [url=http://www.mathlinks.ro/viewtopic.php?t=166255]this[/url]

Problems in Real analysis I,real number sequence and series
It is a famous book of Kaczkor, Nowak.

Try to look at a local bookstore for books about sequences, or go to the library. You should be able to find in any book about sequences the theorems you are referring to. Alternatively, make a google search. For example, look for the Cesaro-Stolz theorem, a very important one, and from which most other sequence limits (for example, the aritmetic mean limit, the geometric mean limit) can be derived.

carcul do u have any e-material with all the theorems in consolidated form

No, I don't. As I suggested before, try a google search. There is probably plenty of material out there.


Determine (for a positive integer n)

$\sum_{k=0}^n (-1)^k \binom{2n}{2k}$


function
number theory unsolved
number theory

I wasn't sure where else to put this:
If a function $ f(x)$ is multiplicative, how would we put "multiplicative" as a noun? For instance, if I wanted to say, "I proved the _________ of $ f(x)$." Multiplicity? Multiplicativity?  :|

"multiplicity" refers to the repetition of, say, roots of an equation, so you can't use that. "multiplicativity" would be the correct form. but really, why not just say "I proved that $ f$ is multiplicative"?


calculus
integration
calculus computations

Find the value of the constant $ k$ such:

$ \int_{-3}^{1}12(x-k)(x+3)(x-1)\, dx = 256$

I think this routine problem has a nice result.

[hide="Answer"]$ k=1$[/hide]

---

By the way, are you affiliated with any college, university, or educational institution?  Have you been, are you, or will you be a student or teacher there?  If so, what have you done, what are you doing, or what will you be doing there?

I'm going to be a student at Worcester Polytechnic Institute in Worcester, MA, in the fall.

$ \int_{\alpha }^{\gamma}(x-\alpha )(x-\beta )(x-\gamma)\ dx=\frac{1}{12}(2\beta-\gamma-\alpha )(\gamma-\alpha )^{3}\ (\alpha <\beta <\gamma )$


Let X = {x_1,x_2,x_3,...} be a sequence of real numbers such that x_(n+1) = |x_n| - x_(n-1) for n = 2,3,.... Prove that X has period 9.

[i]12th Junior Tournament of the Towns 1990 Autumn problems[/i]
http://www.kalva.demon.co.uk/junior/j90a.html

Orlando, do you have sln of this one. I find it on "20.000 problems under the sea" but there has no sln.

Namdung

Hi Namdung,

I will look. But at first I have to discover where I found the problem so that I can give the source.

Let xk=a, xk+1=b.

Consider some cases.
1) a>0, a/2>=b>=0.
xk=a,xk+1=b,-a+b,a-2b,2a-3b,a-b,-a+2b,-b,a-b,xk+9=a,xk+10=b.

2) a>0, 0>=b>=-a.
xk=a,xk+1=b,-a+b,a,2a+b,a+b,-a,b,a-b,xk+9=a,xk+10=b.

3) a<0, b<=0.
xk=a,xk+1=b,-a-b,-a-2b,-b,a+b,-a,-2a-b,-a-b,xk+9=a,xk+10=b.

4) a<0, -a>=b>=0.
xk=a,xk+1=b,-a+b,-a,-b,a+b,-a,-2a+b,-a+b,xk+9=a,xk+10=b.

Thus we must prove that sequence contains one of 1), 2), 3) or 4).

Actually, there are negative number within x1, x2, x3 or x4 because (x1+x3)+(x2+x4)=|x2|+|x3|. So sequence {xn} contains negative numbers. Let a=xk<0 and b=xk+1.
If b<=0 we have 3). Suppose b>0. If b<=|a| we have 4). Suppose b>|a|, then xk+2=b-a>0, xk+3=-a>0 and xk+3<xk+2/2 ==> 1).

Since recurrence xk+2=|xk+1|-xk is invertible we conclude {xn} has period 9.

Yes, this one appeared in the french 'Concours Gnral', but it is probably not its first appearance.
I do not know other solution than brute-force, as done above.

Pierre.

I saw this problem 10 years ago (without sln), but it was "second hand" and the source was unknown.

As far as I remember, this problem was proposed in a Polish olympiad long time ago and also in Tournament of the Towns ( for JUNIORS!).

Indeed I looked up the source. I found it in the ToT competition.

pas seulement concours gnral   :D
American Mathematical Monthly question 6439 1983 page 569
solutionne en monthly mars 1985 p 218

Je n'ai pas compris encore le mode d'utilisation de ce forum roumain
par exemple pour
retrouver une question,
savoir  partir d'un document imprim dans quel forum on tait
comment retrouver l'endroit de la question : par exemple
si par google on demande mathlinls Mindspa on voit apparaitre l'enonc
det(I+A/1!...)=1 et impossible de le retrouver dans ce forum.... :?  :D

Come on Vidiani, the common language here is english... :D

Pierre.


LaTeX
number theory
least common multiple
AMC

Three planets revolve about a star in coplanar circular orbits with the star at the center.  All planets revolve in the same direction, each at a constant speed, and the periods of their orbits are 60, 84, and 140 years.  The positions of the star and all three planets are currently collinear.  They will next be collinear after $n$ years.  Find $n$.

again, i have no latex skills...
[hide]
to do this, u first have to find what number is divisable by all three numbers. That would give u the place where they all line back up. To do that, u take 60=2^2*3*5, 84=2^2*7*3, 140=2^2*7*5. Thus, the answer is 2^2*3*5*7=420. However, that gives when the orbits return to the original place. In order to get when they are colinear, u have to imagine that they are all in the same line, and that happens at half of 420 and one-fourth of 420. The smallest of those is 420/4, which is 105. [/hide]

Hopefully someone else can explain that last part...

[hide]
Let the length of the orbit be 420 "units". The speeds of the planets are 3, 5, and 7
If $t$ is the number of years that passed, we have the equation
$7t\equiv 5t\equiv 3t\mod 210$
It's mod 210 instead of 420, because the planets would be collinear if they are on opposite sides of the star.
$4t\equiv 2t\equiv 0\mod 210$
$2t\equiv t\equiv 0\mod 105$
So t is a multiple of 105, and the smallest multiple of that is just $\boxed{105}$
[/hide]

[quote="Aryth"]Basically, take the LCM of the three orbit periods and divide by the GCF.[/quote]

Stop spreading [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=783607#783607]lies.[/url]

[hide="Nice way (in my opinion)"]
Let's call the number of years $n$. Taking the decimal representations of $\frac{n}{60}$, $\frac{n}{84}$, and $\frac{n}{140}$. Notice that after the decimal point, all of them differ by $\frac{1}{2}$, since they can lie on opposite sides.

Taking pairs and subtracting, we have $\frac{n}{60}-\frac{n}{84}= \frac{k_{1}}{2}$, $\frac{n}{84}-\frac{n}{140}= \frac{k_{2}}{2}$, and $\frac{n}{60}-\frac{n}{140}= \frac{k_{3}}{2}$, for some integers $k_{1}, k_{2}, k_{3}$. Simplifying, we have $\frac{n}{105}= k_{1}$, $\frac{n}{105}= k_{2}$, and $\frac{2n}{105}= k_{3}$. Therefore, $n$ must be divisible by the least common multiple of $105$ and $\frac{105}{2}$, so the smallest value that $n$ can be is $\boxed{105}$.[/hide]


algebra
polynomial
complex numbers
absolute value
algebra unsolved

Let $a_0, ... a_n, b_0, ..., b_n$ be complex numbers such that, for $0 \le k \le 2n$, $\displaystyle\sum_{i + j = k} a_i b_j$ is equal to $0$ for $k$ odd, $a_\frac{k}{2}$ for k even.  If $a_0 a_n \not= 0$, show that $|a_0| = |a_n|$.

Let us assume without loss of generality that $a_n=1$, so that it suffices to prove that $|a_0|=1$.
It follows from the given conditions that for any complex number $x$,
$\begin{eqnarray*} \sum_{i=0}^{n}a_ix^i\sum_{j=0}^{n}b_jx^j&=&\sum_{k=0}^{2n}x^k\sum_{i+j=k}a_ib_j\\ &=&\sum_{i=0}^na_ix^{2i} \end{eqnarray*}.$
Let $\sum_{i=0}^{n}a_ix^i=P(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)$ .  Then $P(x)|P(x^2)$; that is,
\[ (x-\alpha_1)\cdots(x-\alpha_n)|(x-\sqrt{\alpha_1})(x+\sqrt{\alpha_1})\cdots(x-\sqrt{\alpha_n})(x+\sqrt{\alpha_n}). \]
Suppose that there is some $|\alpha_i|<1$.  Then take an $|\alpha_i|<1$ such that $|\alpha_i|$ is minimal; then $|\sqrt{\alpha_j}|>|\alpha_i|$ for all $j$, so $(x-\alpha_i)$ cannot divide $(x-\sqrt{\alpha_1})(x+\sqrt{\alpha_1})\cdots(x-\sqrt{\alpha_n})(x+\sqrt{\alpha_n})$, a contradiction.  Similarly if there is some $|\alpha_i|>1$ there must be such an $\alpha_i$ with $|\alpha_i|$ maximal, which again leads to a contradiction.

Hence $|\alpha_i|=1$ for all $i$, and so $a_0$, the constant term of $P(x)$, has absolute value $|\alpha_1\alpha_2\cdots\alpha_n|=1$ as required.

Sorry, these may seem like dumb questions, but I'm new to this sort of problem solving.
[quote="Agrippina"]Then $P(x)|P(x^2)$[/quote]Why is this necessarily true?[quote="Agrippina"]then $|\sqrt{\alpha_j}|>|\alpha_i|$ for all $j$[/quote]Why does this...[quote="Agrippina"]so $(x-\alpha_i)$ cannot divide $(x-\sqrt{\alpha_1})(x+\sqrt{\alpha_1})\cdots(x-\sqrt{\alpha_n})(x+\sqrt{\alpha_n})$[/quote]Imply this?

Having $\sum_{i=0}^{n}a_ix^i\sum_{j=0}^{n}b_jx^j=\sum_{i=0}^na_ix^{2i}$ implies that $P(x)|P(x^2)$ because we have defined $\sum_{i=0}^{n}a_ix^i=P(x)$, so $\sum_{i=0}^{n}a_ix^{2i}=P(x^2)$.  $\sum_{j=0}^{n}b_jx^j$ is just another polynomial in $x$, say $Q(x)$, and so the condition boils down to saying that $P(x^2)=P(x)Q(x)$ for some $Q(x)$; that is, that $P(x)$ is a factor of the polynomial we get by substituting $x^2$ for $x$ in $P(x)$.

If $(x-\alpha_i)$ divides $P(x^2)=(x-\sqrt{\alpha_1})(x+\sqrt{\alpha_1})\cdots(x-\sqrt{\alpha_n})(x+\sqrt{\alpha_n})$, then when  $x=\alpha_i$ $P(\alpha_i^2)=0$.  But if $|\sqrt{\alpha_j}|>|\alpha_i|$, then $\alpha_i\neq\pm\sqrt{\alpha_j}$ for any $j$, so $\alpha_i-\sqrt{\alpha_1}\neq 0$,  $\alpha_i+\sqrt{\alpha_1}\neq 0$, ... $\alpha_i-\sqrt{\alpha_n}\neq 0$, $\alpha_i+\sqrt{\alpha_n}\neq 0$ and hence their product $P(\alpha_i^2)$ is nonzero, a contradiction.


Given that $ \overline{AB}$ is tangent to circle $ P$ at $ B$, what is the ratio $ \frac{x}{x\plus{}y}$? Express your answer as a common fraction.

[asy]pair A,B,P;
P=(0,0);
B=(0,9);
A=(12,9);
draw(Circle(P,9));
draw(P--(0,-9));
draw(B--A--P);
label("12",(6,9),N);
label("$x$",(9.6,7.2),SE);
label("$y$",(3.6,2.7),NW);
label("9",(0,-4.5),E);
label("$P$",P,W);
label("$B$",B,N);
label("$A$",A,E);[/asy]

$ \overline{BP}$ is just another radius so it is equal to nine. Since all tangent lines make a right angle at the point of tangent, we can use Pythagorean Theorem. $ 9^2 \plus{} 12^2 \equal{} 15^2$. Thus, $ x \plus{} y \equal{} 15$. Since $ y$ is also another radius, it equals nine. To get $ x$, we subtract nine from fifteen to get six. The answer is $ \boxed{\frac {2}{5}}$.

---EDIT---
Thanks for pointing that out :) !

Except $ 15\minus{}9\equal{}6$ :wink:

so $ \frac{6}{15}\equal{}\boxed{\frac{2}{5}}$

Another solution is to extend $ AP$ to a point $ C$ on the circle, then use power of a point and $ y\equal{}9$ to get \[ 12^2\equal{}x(x\plus{}18) \Rightarrow x\equal{}6\] So $ \frac{x}{x\plus{}y}\equal{}\frac{2}{5}$.


algebra
functional equation
algebra proposed

One more fine olympiad-like functional equation:  ;)

Find all $f: \mathbb{R} \to \mathbb{R}$ fulfilling
\[
f(x^2+y+f(y)) = f(x)^2 +2y.
\]

1)$ x=0  \to f(y+f(y)) = 2y+ f^2(0) $
    $ \to \exists c: f(c) =0 $
 2) $ y=0 \to f^2(x) = f^2(-x) \to f(-c) = 0
     Let $ y =c ; y =-c  \to c=0 $
 $ \to f(x)=0 \iff x=0 $
 $ \to f(x^2+y+f(y)) = f(x^2) + f(y+f(y)) $
 We'll prove $ y+f(y) $ gets all values in $ R $
  $ X =x^2 $
  Let $ y =-\frac{f(X)}{2} $
  $ \to y+f(y) $ gets all values in $ R $
  Let $ X =-(y+f(y)) $
  $ \to f(y+f(y)) $ is an odd func => Q.E.D
  PS : Let's try a more difficult problem : Find $ f:R--R $ satisfying : $ f(x^2+y+f(y)) = f^2(x)  $


geometry
geometric transformation
dilation

Each side of an equilateral triangle equals $12\sqrt{3}$.  After the 3 altitudes are drawn, a trisection point is located along each altitude nearer the vertex from which that altitude is drawn.  These 3 trisection points are vertices of a triangle.  If the area of this triangle is $K\sqrt{3}$, with $K$ rational, find $K$.

This is in fact a dilation and the new equilateral triangle formed by the trisection points has lengths of 1/3 the corresponding lengths of the original triangle.  The area of the new triangle has 1/9 the area of the original one, therefore the area is 12 :rt3: / 9 = 4/3 * :rt3:.   K = 4/3.

[quote]1/3 the corresponding lengths of the original triangle[/quote]
1/3 or 1/2 ?

What do you mean by trisections?
That would cut each angle into 3 equal pieces right, which means each new angle would be 20 degrees?
I'm not sure what exactly you mean.

[quote]The area of the new triangle has 1/9 the area of the original one, therefore the area is 12 :rt3: / 9 = 4/3 * :rt3: . K = 4/3.[/quote]
Isn't 12 :rt3:  the side of the original triangle? Then its area would be 108 :rt3: .
108 :rt3: /9= 12 :rt3:
so K=12

Eh whoops, I thought 12 :rt3: was the area.  My goof.

Again Be careful - Check it AGAIN - Is the smaller traingle's each  side is really      one third of the larger traingle  ?<spoiler starts>  [hide]or   is it only half  and K=27  [/hide]

Yeah it's a half.  My bad, drawing was off too :-(.


Does there exist a number $\alpha, 0 < \alpha < 1$ such that there is an infinite sequence $\{a_n\}$ of positive numbers satisfying \[ 1 + a_{n+1} \leq a_n + \frac{\alpha}{n} \cdot \alpha_n, n = 1,2, \ldots? \]

[quote="orl"]Does there exist a number $ \alpha, 0 < \alpha < 1$ such that there is an infinite sequence $ \{a_n\}$ of positive numbers satisfying
\[ 1 \plus{} a_{n \plus{} 1} \leq a_n \plus{} \frac {\alpha}{n} \cdot \alpha_n, n \equal{} 1,2, \ldots?
\]
[/quote]

I think he meant $ a_n$. Anyway:

Suppose that there exists such $ {\alpha}$. If one of the $ a_{i}$ is less than $ {i}$ then $ a_{i \plus{} 1}$ is less than $ a_{i}$. You can find here easy contradition.

Now let's say that $ a_{n} \equal{} kn \plus{} t$  and $ t$ is less than $ n$.
We obtain:
\[ a_{n \plus{} 1} \leq a_n \plus{} \frac {\alpha}{n}a_n \minus{} 1 \equal{} kn \plus{} t \plus{} {\alpha}k \plus{} {\alpha}\frac {t}{n} \minus{} 1 \leq kn \plus{} t \plus{} {\alpha}k \leq k(n \plus{} 1) \plus{} t
\]
Now it's easy to see that that it's impossible that $ a_{n \plus{} 1} \equal{} k(n \plus{} 1) \plus{} t$. We can conclude that $ a_{n \plus{} 1} \leq k(n \plus{} 1) \plus{} t \minus{} 1$. Doing same trick we easily obtain that finally $ a_i$ must be lower than $ i$.

Hope it's clear and correct.

Sorry for editing all the time but cannot get used to LateX


Let ABC be a triangle and M be the middle of BC.Draw the circle with diagonal AM intersects the AB and AC in E and F respectively.
let the tangents from E,F for the circle intersect eachother on S.
PROVE that SB=SC.

It has already been discussed several times.

http://www.mathlinks.ro/Forum/viewtopic.php?t=16779
http://www.mathlinks.ro/Forum/viewtopic.php?p=151313

[quote="amir2"][size=150]aghaye kvmedya soalhaye tekrari naferest[/size] :D[/quote]
Is it in English? What does it mean? I understand nothing!

Hi Amir2
soale "Remike" ro didy ? Hasti javabesho nadim ta too khomarish bemmoneh ?? ;)

This is an english forum. Please retain from making comments in other languages. If there are many of you, we can start a forum for Iran in the [url=http://www.mathlinks.ro/Forum/index.php?f=248]National and Local Communities Forums[/url]. Thanks.


I think this is easy but I can't do it.
A dot in mathematical text (between doble-dollars) becomes a comma in the output, why?
I need help.

What do you mean?  A period?  A "cdot"?  Could you show us?

$ .$

Seems fine for me.

For example, I write in latex "0.3+0.2=0.5" and the result in pdf is "0,3+0,2=0,5" and I don't know why.

This is the code:
\documentclass[11pt,letterpaper]{article}
\usepackage[spanish]{babel}
\usepackage[ansinew]{inputenc}

\begin{document}
Aquí el probema: $ 0.3\plus{}0.2\equal{}0.5$
\end{document}

I think undefined117 is correct.  Most of the symbols in the spanish that you need are symbols that you can enter yourself, probably with a package.  Check the comprehensive list on this site.

[quote="undefined117"]Seems like you're using a Spanish package?

I know in Spanish and some other languages, they switch the use of periods and commas when representing numbers.

For example, one thousand in English would be represented as 1,000

In Spanish it's 1.000

0.001 in English would be 0,001 in Spanish.

sorry, but I don't know how you would fix this :| why would you need the Spanish package though? The accent marks should work either way..[/quote]
I need the spanish package because we use a lot accents and marks. And with spanish package it becomes easier

[quote]I need the spanish package because we use a lot accents and marks. And with spanish package it becomes easier[/quote]

That may be, but it very well may ruin your math unless you want to learn all the extra syntax.   I would just stick to the regular and put in the extras yourself.

If that's the way it is in the Spanish package, perhaps your audience will be more accustomed to that way, and you shouldn't change it.

... or you could just type , for . and . for ,


1) Find the minimum point of F(x,y) = 0.25x^4 + (x^2)y + y^2

2) Find the saddle point of G(x,y) = x^3 + xy - x

I was able to find that (0, 1) is the saddle point of G(x,y), but I couldn't find the minimum point of function F.

F(x,y)=((2*y+x^2)^2)/4  >=  0 . So it is obvious that the points of minimum are (x , -0.5x^2) , for all real x.


Cheldelin Middle School has 12 doors to enter or leave the
building. In how many ways is it possible to enter the building
by one door and leave the building by a different door?

You can enter through $ 12$ doors, and exit through $ 11$ because you can't exit through the one that you entered through. The answer is $ 12\cdot11\equal{}\boxed{132}$.