algebra
polynomial
real analysis
calculus
derivative
complex analysis
complex analysis unsolved

Suppose $ f$ is entire and every power series: $ f(z)\equal{}\sum_{n\equal{}0}^\infty c_n(z\minus{}a)^n$ at least one coefficient is $ 0$. Prove $ f$ is a polynomial.

Rudin gives the hint: $ n!c_n\equal{}f^{(n)}(a)$.

This hint isn't very useful since I would have tried this venue pretty quickly. So this obviously leads to the reformulation: If for every $ z$ there exists an $ n$ such that $ f^{(n)}(z)\equal{}0$, prove that $ f$ is a polynomial. I fiddled with series, limits, max-mod, all the tools of the chapter, yet nothing is connecting up.

The best I seem to be able to do is: Infinitely many of the coefficients must be zero. This still leaves lots of non-polynomials, though.

Any hints for the hint?

Try this:  show that some derivative must be bounded and apply Liouville's Theorem.

Some $ f^{n}$ must have uncountably many zeros and hence be identically zero

How incredibly simple! That agrees with my intuition which said to get a fact about the $ z$'s in terms of a specific $ n$ rather than the $ n$ dependent on the $ z$'s. Sometimes I wish I was better at stepping back and thinking of a good idea rather than just rushing into some symbol manipulation and details in hope of an idea emerging on its own.  :roll:


Let $ x:$ {$ x\ne1$} be the least positive integer that satisfies $ 2^{100}$ divides $ x^{2007} \minus{} 1$ where . What is the remainder if $ x$ is divided by $ 47$?

Try using lifting the exponent  and then it is just calculations

[quote="ZetaX"]Maybe you also want to exclude $ x \equal{} 1$  :wink:
And as $ x \mapsto x^{\text{odd}}$ permutates the odd residues $ \mod 2^{\text{anything}}$, the smallest such $ x$ is just $ 2^{100} \plus{} 1$.[/quote] I need clear solution.

Just prove what I said: If $ m,x,y$ are odd and $ x^m \equiv y^m \mod 2^k$, then $ x \equiv y \mod 2^k$.

Obviosly $ x\equal{}1\plus{}2^{99}\equal{}1\plus{}2^7\mod 47\equal{}35\mod 47$.

[quote="Rust"]Obviosly $ x \equal{} 1 \plus{} 2^{99} \equal{} 1 \plus{} 2^7\mod 47 \equal{} 35\mod 47$.[/quote] how do we know that $ x \equal{} 1 \plus{} 2^{99}$ is the least one?

[quote="ertanrock"]Let $ x:$ {$ x\ne1$} be the least positive integer that satisfies $ 2^{100}$ divides $ x^{2007} - 1$ where . What is the remainder if $ x$ is divided by $ 47$?[/quote]
We know that $ x^{2007} \equiv 1 \bmod 2^{100}$, but $ x^{2^{99}} \equiv 1 \bmod 2^{100}$ since $ \phi(2^{100}) = 2^{99}$. Then it is well known that the order of $ x$ $ \bmod 2^{100}$, let's call it $ o$, so $ o \mid \gcd(2^{99},2007)$. Since $ \gcd(2^{99},2007) = 1$ we know that $ o \mid 1$ so the order of $ x$ is $ 1$. Then $ x \equiv 1 \nmid 2^{100}$. The smallest is then $ x=1$, but i suppose that is not what you wanted? Else it is $ x = 2^{100} + 1$. Then $ x = 2^{100} + 1 \equiv 2^{100-46 \cdot 2}+1 = 2^8 + 1 \bmod 47$ since $ \phi(47) = 46$. Then $ 2^8 = 256$ So $ 2^8 + 1 = 257 \equiv 257 - 5 \cdot 47 = 22 \bmod 47$. So the answer is $ 22$ ! QED.

[quote="Mathias_DK"][quote="ertanrock"]Let $ x:$ {$ x\ne1$} be the least positive integer that satisfies $ 2^{100}$ divides $ x^{2007} - 1$ where . What is the remainder if $ x$ is divided by $ 47$?[/quote]
We know that $ x^{2007} \equiv 1 \bmod 2^{100}$, but $ x^{2^{99}} \equiv 1 \bmod 2^{100}$ since $ \phi(2^{100}) = 2^{99}$. Then it is well known that the order of $ x$ $ \bmod 2^{100}$, let's call it $ o$, so $ o \mid \gcd(2^{99},2007)$. Since $ \gcd(2^{99},2007) = 1$ we know that $ o \mid 1$ so the order of $ x$ is $ 1$. Then $ x \equiv 1 \nmid 2^{100}$. The smallest is then $ x = 1$, but i suppose that is not what you wanted? Else it is $ x = 2^{100} + 1$. Then $ x = 2^{100} + 1 \equiv 2^{100 - 46 \cdot 2} + 1 = 2^8 + 1 \bmod 47$ since $ \phi(47) = 46$. Then $ 2^8 = 256$ So $ 2^8 + 1 = 257 \equiv 257 - 5 \cdot 47 = 22 \bmod 47$. So the answer is $ 22$ ! QED.[/quote]
What about this solution. anything wrong?
If $ 2^{100}$ divides $ x^{2007} - 1$ then $ 2^{100}$ divides $ (x-1)(x^{2006}+.....+x+1)$
right part is odd so it is impossible. So the left part should be divisible by $ 2^{100}$  then the least $ x$ value is $ 2^{100}+1$

[quote="ertanrock"][quote="Mathias_DK"][quote="ertanrock"]Let $ x:$ {$ x\ne1$} be the least positive integer that satisfies $ 2^{100}$ divides $ x^{2007} - 1$ where . What is the remainder if $ x$ is divided by $ 47$?[/quote]
We know that $ x^{2007} \equiv 1 \bmod 2^{100}$, but $ x^{2^{99}} \equiv 1 \bmod 2^{100}$ since $ \phi(2^{100}) = 2^{99}$. Then it is well known that the order of $ x$ $ \bmod 2^{100}$, let's call it $ o$, so $ o \mid \gcd(2^{99},2007)$. Since $ \gcd(2^{99},2007) = 1$ we know that $ o \mid 1$ so the order of $ x$ is $ 1$. Then $ x \equiv 1 \nmid 2^{100}$. The smallest is then $ x = 1$, but i suppose that is not what you wanted? Else it is $ x = 2^{100} + 1$. Then $ x = 2^{100} + 1 \equiv 2^{100 - 46 \cdot 2} + 1 = 2^8 + 1 \bmod 47$ since $ \phi(47) = 46$. Then $ 2^8 = 256$ So $ 2^8 + 1 = 257 \equiv 257 - 5 \cdot 47 = 22 \bmod 47$. So the answer is $ 22$ ! QED.[/quote]
What about this solution. anything wrong?
If $ 2^{100}$ divides $ x^{2007} - 1$ then $ 2^{100}$ divides $ (x - 1)(x^{2006} + ..... + x + 1)$
right part is odd so it is impossible. So the left part should be divisible by $ 2^{100}$  then the least $ x$ value is $ 2^{100} + 1$[/quote]
Yes. That is a valid solution.


logarithms
integration
calculus
calculus computations

A not so elegant way to prove the divergence makes use of Stirling's formula $ n! \equal{} (n/e)^n\sqrt {2\pi n}(1 \plus{} o(1)).$ It is not difficult to show that the dominant term of $ 1/\ln(n!)$ is $ 1/(n\ln n)$ so determining the divergence.

[quote="Hong Quy"]I wanna see  the other solution...Use  of Stirling's formula maybe the latter method[/quote]

$ {H_{n}=\sum_{k=1}^{n}\frac{1}{k}}$,it's a harmonic sequence.Draw the graph of $ {\frac{1}{t}}$ on $ [0,n]$,we can get

$ H_{n}-1=\sum_{k=2}^{n}\frac{1}{k}<\int_{0}^{n}{\frac{1}{t}} dt=\ln n$

$ H_{n}-\frac{1}{n}=\sum_{k=1}^{n-1}\frac{1}{k}>\int_{0}^{n}{\frac{1}{t}} dt=\ln n$

apply above method to $ \ln (n!)=\sum_{k=2}^{n}\ln k$,we can get

$ \ln ((n-1)!)<\int_{1}^{n}{\ln t dt<\ln (n!)}$.Using integration by parts,$ \int_{1}^{n}{\ln t dt=n\ln n-n+1}$

$ n!>e^{n\ln n-n+1}=e(\frac{n}{e})^n$

$ n!=n(n-1)!<ne^{n\ln n-n+1}=ne(\frac{n}{e})^n$

$ \ln n!\le \ln n^n\equal{}n\ln n$ but $ \sum_n\frac 1{n\ln n}\equal{}\plus{}\infty$ (by the Cauchy condensation trick or by the integral test).


trigonometry
quadratics
geometry
angle bisector
algebra
geometry solved

Characterize all triangles $ABC$ s.t.
\[ AI_a : BI_b : CI_c = BC: CA : AB  \] where $I_a$ etc. are the corresponding excentres to the vertices $A, B , C$

I think there is the following relation, isn't it??

$sin\frac{A}{2}=\frac{r_a}{AI_a}$ where $r_a$ is the radious of the excircle!!

Now if we substitute and use $r_a=p \cdot  tg\frac{A}{2}$
I think we find that it is equilateral!

 :)

I believe we get some more cases-- not sure!!!

$\cos \frac A2=\frac {s}{AI_a}\Longleftrightarrow AI_a=\frac {s}{\cos \frac A2}$. Thus, $\frac {AI_a}{a}=\frac {BI_b}{b}=\frac {CI_c}{c}\Longleftrightarrow a\cos \frac A2=b\cos \frac B2=c\cos \frac C2\Longleftrightarrow $
$a^3(s-a)=b^3(s-b)=c^3(s-c)\Longleftrightarrow a=b=c.$

There are indeed "some more cases", what is because $a^3\left(s-a\right)=b^3\left(s-b\right)=c^3\left(s-c\right)$ does [b]not[/b] imply a = b = c.

Anyway, let me give a complete solution of the problem.

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle, and let $I_a$, $I_b$ and $I_c$ be the centers of its A-excircle, B-excircle and C-excircle, respectively. Then, we have

$AI_a: BI_b: CI_c=a: b: c$

if and only if

- either a = b = c,
- or b = c and $a=\frac{1+\sqrt5}{2}b$,
- or c = a and $b=\frac{1+\sqrt5}{2}c$,
- or a = b and $c=\frac{1+\sqrt5}{2}a$.[/color]

[i]Proof of Theorem 1.[/i] First, let's consider an arbitrary triangle ABC.

Let the A-excircle of triangle ABC touch the line CA at a point T. Then, since $I_a$ is the center of this A-excircle, $I_aT\perp CA$, so the triangle $I_aTA$ is right-angled at T; also, $\measuredangle I_aAT=\frac{A}{2}$ (since the point $I_a$, being the center of the A-excircle of triangle ABC, lies on the angle bisector of the angle CAB), and AT = s, where $s= \frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Thus, in the right-angled triangle $I_aTA$, we have $AI_a=\frac{AT}{\cos\measuredangle I_aAT}$, so that $AI_a=\frac{s}{\cos\frac{A}{2}}$. But by the half-angle formulas, $\cos\frac{A}{2}=\sqrt{\frac{s\left(s-a\right)}{bc}}$. Thus,

$AI_a=\frac{s}{\sqrt{\frac{s\left(s-a\right)}{bc}}}=\sqrt{\frac{sabc}{a\left(s-a\right)}}$.

Similarly, $BI_b=\sqrt{\frac{sabc}{b\left(s-b\right)}}$ and $CI_c=\sqrt{\frac{sabc}{c\left(s-c\right)}}$. Hence, we can equivalently transform the relation $AI_a: BI_b: CI_c=a: b: c$ as follows:

$AI_a: BI_b: CI_c=a: b: c$
$\Longleftrightarrow\ \ \ \ \ \sqrt{\frac{sabc}{a\left(s-a\right)}}: \sqrt{\frac{sabc}{b\left(s-b\right)}}: \sqrt{\frac{sabc}{c\left(s-c\right)}}=a: b: c$
$\Longleftrightarrow\ \ \ \ \ \sqrt{\frac{1}{a\left(s-a\right)}}: \sqrt{\frac{1}{b\left(s-b\right)}}: \sqrt{\frac{1}{c\left(s-c\right)}}=a: b: c$
$\Longleftrightarrow\ \ \ \ \ \frac{1}{a\left(s-a\right)}: \frac{1}{b\left(s-b\right)}: \frac{1}{c\left(s-c\right)}=a^2: b^2: c^2$
$\Longleftrightarrow\ \ \ \ \ 1: 1: 1=\left(a^3\left(s-a\right)\right): \left(b^3\left(s-b\right)\right): \left(c^3\left(s-c\right)\right)$
$\Longleftrightarrow\ \ \ \ \ a^3\left(s-a\right)=b^3\left(s-b\right)=c^3\left(s-c\right)$.

Since $s-a=\frac{a+b+c}{2}-a=\frac{b+c-a}{2}$ and similarly $s-b=\frac{c+a-b}{2}$ and $s-c=\frac{a+b-c}{2}$, this becomes

$a^3\left(b+c-a\right)=b^3\left(c+a-b\right)=c^3\left(a+b-c\right)$.

Now, let ABC be a triangle satisfying the condition $AI_a: BI_b: CI_c=a: b: c$. Then, we must therefore have $a^3\left(b+c-a\right)=b^3\left(c+a-b\right)=c^3\left(a+b-c\right)$. Now, since our situation is symmetric, we can WLOG assume that $a\geq b\geq c$. Since $b^3\left(c+a-b\right)=c^3\left(a+b-c\right)$, we have

$0=b^3\left(c+a-b\right)-c^3\left(a+b-c\right)=a\left(b^3-c^3\right)-\left(b^4-c^4-b^3c+c^3b\right)$
$=a\left(b^2+bc+c^2\right)\left(b-c\right)-\left(b^3+c^3\right)\left(b-c\right)$
$=\left(a\left(b^2+bc+c^2\right)-\left(b^3+c^3\right)\right)\left(b-c\right)$.

Now,

$a\left(b^2+bc+c^2\right)-\left(b^3+c^3\right)=\underbrace{\left(a-b\right)}_{\geq 0\text{, since }a\geq b}b^2+\underbrace{\left(a-c\right)}_{\geq 0\text{, since }a\geq c}c^2+\underbrace{abc}_{>0}>0$;

thus, we must have b - c = 0, so that b = c. Hence, the equation $a^3\left(b+c-a\right)=b^3\left(c+a-b\right)$ becomes $a^3\left(b+b-a\right)=b^3\left(b+a-b\right)$, i. e. $a^3\left(2b-a\right)=b^3a$. Division by a transforms this into $a^2\left(2b-a\right)=b^3$, and thus

$0=a^2\left(2b-a\right)-b^3=\left(a-b\right)\left(ab-a^2+b^2\right)$.

Hence, either a - b = 0, what leads to a = b and thus to a = b = c, or $ab-a^2+b^2=0$, what is a quadratic equation in a and has the solutions $a=\frac{1+\sqrt5}{2}b$ and $a=\frac{1-\sqrt5}{2}b$, of which the second one can be excluded since it is negative (and sidelengths of a triangle cannot be negative), so we get the solution b = c and $a=\frac{1+\sqrt5}{2}b$. The other solutions can be obtained similarly (we WLOG assumed that $a\geq b\geq c$ and thus didn't get the cyclic permutations). By tracing the above argumentation backwards, we see that these solutions are indeed solutions.

Theorem 1 is proven.

  Darij

Darij, you are true. I was negligently (it is regrettably !) I will prove that at least two between $a,b,c$ are equally, i.e. $(a-b)(b-c)(c-a)=0$. Thus,
$a\ne b\ne c\ne a\ \wedge \ a^3(b+c-a)=b^3(c+a-b)=c^3(a+b-c)\Longrightarrow$
$a^3+b^3=c(a^2+ab+b^2),\ b^3+c^3=a(b^2+bc+c^2),\ c^3+a^3=b(c^2+ca+a^2)\Longrightarrow$
$a^2+b^2+c^2=0$, what is absurd. Therefore, $a=b\ \vee\ b=c\ \vee\ c=a$ a.s.o.

Ya , darij! I got the same answers... Those were the other cases I was talking about above!


I am trying to prove that a topological space $ T$ with a countable base has a countable everywhere dense subset.  The proof is simple but for one (probably simple) point that I cannot figure out. I am a beginner when it comes to these arguments, so don't criticize too much if it is a very simple thing.

Let $ \mathbf{G}\equal{}{G_1, G_2, \ldots}$ be a countable base for $ T$, and construct $ M\equal{}{x_1, x_2, \ldots}$ by taking $ x_j \in G_j$.  Then $ M$ is countable, and it remains to show that it is everywhere dense in $ T$.

If $ [M] \neq T$ ([] denotes closure), then $ G\equal{}T\minus{}[M] \neq \emptyset$ must contain no points of $ M$.

On the other hand, [i]IF[/i] $ G$ is open, then $ G\equal{}\bigcup_a G_a$ for some $ {a}$ and hence $ x_n \in G$ for at least one $ n$, which from the previous statement implies that $ T\equal{}[M]$.

The question is: why is G open?

$ G$ is open because $ [M]$ is closed (by definition).

Thank you, t0rajir0u.  My definition of 'closed set' until now has been one whose compliment is open, but it is just the same to let a closed set be one which coincides with its closure (and indeed $ [[M]]\equal{}[M]$) because one can then show that the former definition is equivalent.  Of course, it is the same in metric space theory. I was just having trouble making the transition  :oops: . Thanks again.

You can show that a set is closed if and only if its closure is itself, so yes, they are equivalent definitions.  Two other equivalent definitions are $ B_{\varepsilon}(x)\cap F\neq\emptyset$ for every $ \varepsilon>0\Rightarrow x\in F$, or if a sequence $ (x_{n})\subseteq F$ converges to some point $ x\in M$ then $ x\in F$ (in these definitions, $ F$ is the closed set and $ M$ is the metric space).


Θεωρούμε διανυσματικό χώρο V  με dimV=n
δείξτε ότι αν $ f_1,f_2,f_3,...f_k$ γραμμικά ανεξάρτητα στοιχεία του V* ( όπου V ο δυικός χώρος του V)
τότε¨
$ dim({ \bigcap_{n=0}^{ k} \(kerf_i)= n-k$

Συγνώμη παιδιά μια μικρή διόρθωση,ξέχασα το κ πάνω από την τομή  και είναι από i=1

Θεωρούμε διανυσματικό χώρο V  με dimV=n
δείξτε ότι αν $ f_1,f_2,f_3,...f_k$ γραμμικά ανεξάρτητα στοιχεία του V* ( όπου V ο δυικός χώρος του V)
τότε¨
$ dim({ \bigcap_{i=1}^{k} \(kerf_i)= n-k$

Ummm, είναι άμεση εφαρμογή του θεωρήματος rank-nullity: dim(im f) + dim(ker f) = dimV όπου f ο πίνακας που προκύπτει από τα διανύσματα-γραμμές f_1 ... f_k. Αφού τα f_i είναι γραμμικά ανεξάρτητα, έπεται ότι dim(im f) = k κ' το ζητούμενο προκύπτει αβίαστα.

Cheerio,

Durandal 1707

Tώρα πάλι, θυμάμαι ότι μια φορά μου έιχε χρειαστεί να αποδείξω μία γενίκευση αυτής της πρότασης για απειροδιάστατους χώρους και μου είχε βγει κυριολεκτικά η πίστη για να το κάνω (μέχρι να βρω κιόλας τη σωστή διατύπωση για να αποδείξω... ουφφφφφ).

Hehehe, νοσταλγία ;) αν τη βρω πουθενά αυτήν τη γενίκευση θα την ανεβάσω γιατί είναι πολύ ωραίο πρόβλημα!

Cheerio,

Durandal 1707


If $ h(x) \equal{} \sqrt{\frac{x\plus{}3}{2}}$, what is the value of $ h(\minus{}1)$?


When  a ditch 2 meters wide and 1 meter deep was dug around a square building (adjacent to it) the volume of earth removed was 192 cubic meters. Find the demensions of the building.

[hide]For each meter around the perimeter of the building, 2 cubic meters of dirt are removed.  If 192 cubic meters are removed, the perimeter is $\frac{192}{2}=96$.  If the building is a square, then the side length is $\frac{96}{4}=24$, making the deminsions [hide="deminsions"]$\boxed{\boxed{\boxed{\boxed{\boxed{24\times24}}}}}$[/hide][/hide]

EDIT: wrong answer here. Trying to figure out. Ignore this "solution"- see next reply

[hide]The "area" of the earth removed per meter is $2m^3$. Then we find the perimeter of the building by doing $\frac{192}{2}=96$. We have $96m^3$ for the perimeter. Then, we assume the building is a square to make matters easier and simply do $\frac{96}{4}=24$ so the dimensions of the building are $24 \times 24$ when we disregard the height in three dimensions.[/hide]

EDIT: b-flat [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=382213#p382213]already answered[/url]... well if you didn't understand his you probably couldn't understand mine. Nevertheless...

Hope this helped,

I don't think we can solve this like you did, because you only took account the part of the ditch along the building sides, but there's also part of the ditch in the corners. (A sketch might help: Draw a square and a larger square around it, then extend the sides of the smaller square until they meet the sides of the larger square. You'll get the ditch divided into 4 rectangles along the sides of the smaller square, but also 4 very small squares in the corners of the larger square. You took into account only the rectangles and not the smallest squares.)

[hide="Correct solution"]If the side of the building is $a$, then the ditch consists of four rectangles whose dimensions are $a$ and $2$ and four squares whose dimensions are $2$ and $2$. Adding them up we get $4\times(a\times 2)+4\times(2\times 2)=8a+16$, and this is the area of the ditch. We know this area equals $192$, because the ditch is $1$ meter deep. So, the equation is

$8a+16=192$
$8a=176$
$a=22$

The building is $22$ meters long and wide.[/hide]

I am confused with the proposition of Farenhajt, I did what you said. I don't understand where are 4 squares that you are talking about. Please take a look at the picture below. Correct me please, if I am wrong. Thanks.

[img]http://img75.imageshack.us/img75/1803/ditch9gj.jpg[/img]

No, you didn't do what I said - u connected the points, but I said "extend the SIDES of the smaller square until they intersect the sides of the larger square". The picture should look like 2 vertical and 2 horizontal lines inside the larger square, dividing it into 9 regions. Try again :)

[img]http://img479.imageshack.us/img479/4025/ditch5rz.jpg[/img]

I don't know I am still confused.

You don't need those oblique lines connecting the vertices. And after you remove them, you see you have 4 rectangles (along the sides of the smaller square) and 4 small squares (in the corners).

Great thanks finally I have understood the question.

The crosssection of the ditch was never specified. What if the ditch was in the form of an isosceles or right triangle instead of a rectangle?


[i]This is a problem for the first phase of the Greek HMS which was held today[/i]

[u]Problem:[/u]
[b]Find the angles of isosceles triangles that have the following property:
[/b][i]"There is a segment that connects a peak with the opposite side in such a way that inside the isosceles triangle, two isosceles triangles are created"[/i]
[b]( Examine all the possible cases)[/b]

Sorry if its kinda hard to understand what i mean, i'm not familiar with English math terminology, i tried to translate it the best i could.

You have to consider several cases
If I haven´t miscalculated the only solution are
$ 36$ $ 36$ $ 108$
$ 72$ $ 72$ $ 36$
and
$ \frac{540}{7}$ $ \frac{540}{7}$ $ \frac{180}{7}$

Daniel

Thanks for the number results  :)
But could you explain what method you used to find them?
I was banging my head today trying to solve it .... it was the only one out of the 4 probs i couldn't do :S
It would really help if i could see the whole solution :D

Well I will tell you how I did it
It is really a nasty solution to write down
The idea is to consider two major cases
1)Suppose that the segment you draw lies in the non equal side
2)You suppose that the segment lies in one of the equal sides
Then in each  of these cases you must consider 2 or 3 sub cases depending on which sides form the isosceles triangles
Then it´s all about drawing diagrams and making calculations

Hope it is useful..try to see if you can reach the solution...if not I will make an effort and explain the things with more details :)

Daniel

Thanks! i think the information yo gave me is enough to guide me through the solution  :D
Thank you for your time :)

Oh thank´s for pointing out my mistake... :)
indeed it was the most obvious solution, I don´t know what I may have forgotten

Daniel


$f \in C[0,+%Error. "infinit" is a bad command.
)$ such that $f'(x)=\frac{1}{x^2+f^2(x)}$
prove $f(+ %Error. "infinit" is a bad command.
)$ is exist,and $f(+ %Error. "infine" is a bad command.
) \leq 1+\frac{\pi}{4}$

It has been posted for many times before.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=26567.


The results from the first level Olympiad are online:
http://www.wocomal.org/statistics/olympiad/200607/200607o_level1.html

There were issues with the data (missing grades, misspelled schools, all 22 students from Belmont having the last name Baratta*) so I did the best I could to clean it up.  If you notice anything that's still wrong, let me know.

Rick Yanco
Worcester Academy

*For finalists and semifinalists the list contained Sergei Baratta, Max Baratta, Daniel Baratta, and Jonathan Baratta, not to mention Peter Baratta, because I think that one's right.  I guessed Bernstein, Chase, Walker, and Sailor, but I know very little about Belmont.

WHOA I got 47th!!!!!

I'm the highest 8th grader...guess who I am.  :P
Anyway I beat all the LHS people except 4, so I'm happy.  :)

Hoping for top 20 Level 2...

Those Belmont names look right, I've never heard of Jonathan Sailor though...

Yay 7th!

Robert Iu from Acton-Boxborough (98th place) is wrong, it should be "Robert Liu"

[quote]Robert Iu from Acton-Boxborough (98th place) is wrong, it should be "Robert Liu"[/quote] Fixed.  Let me know if you see anything else.

[quote="13375P34K43V312"]I just realized I beat every freshman...
maybe they just suck seeing how much I suck.[/quote]

Or maybe they're just insane at solving "real" (aka olympiad) problems...seeing how Vlad Firoiu is there-he scored a 15 on the USAMO last year as an 8th grader.

dude... 13375P34K43V312 is mad annoying...

No one cares that you're an 8th grader and placed 47th.

And for these math contests, it really doesn't matter how old you are... I didn't learn anything in 9-12th grade that really helped me on these problems at school. I just basically learned the stuff myself.

Damn, go brag to your parents.

P.S. Vlad is a beast and is awesome.

P.P.S. Boston Latin pwned this test! but why are we so bad at GBML...

Hmm, since you asked, just thought I'd point out a minor mistake. Number 18 is supposed to be Stephen Xu not Stephe Xu. :-)

[quote]Hmm, since you asked, just thought I'd point out a minor mistake. Number 18 is supposed to be Stephen Xu not Stephe Xu[/quote] Also fixed.  Most of the problems come from bubbling errors -- either the scantron didn't see the N or it wasn't filled in.
Keep looking -- once the errors are worked out I'll move the listing to the MML website (with MML schools highlighted) and the MAML website (no highlighting).

Do you know where the questions are for this contest?

[quote]So it looks like they gave everyone the question without the correct answer as a choice, right?[/quote] Yes. That was the precedent, so that's what was done.

[quote="Phelpedo"][quote="mathgeniuse^ln(x)"]Huh?

What does that have to do with a link?[/quote]
... :roll: ...

It doesn't.

[quote="13375P34K43V312"]WHOA I got 47th!!!!!

I'm the highest 8th grader...guess who I am.  :P
Anyway I beat all the LHS people except 4, so I'm happy.  :)

Hoping for top 20 Level 2...

[/quote]

[quote="13375P34K43V312"]
I was being sarcastic, but sure.
[/quote]

hm...[/quote]

no the sucking part

[quote="perfect628"][quote="jhredsox"]wasn't sergei also top written for mathcounts nats '05?
-jorian[/quote]

Yes, the fell to 5th after CD. He isn't good at CD. I almost beat him.[/quote]

almost only counts in horseshoes and hand grenades buddy


modular arithmetic
function
number theory
least common multiple
relatively prime
totient function

I'm not sure if this problem is posted already.  If it is, please lock it and if it isn't, could anyone solve this for me please?
Let $m$ be a whole number.
If $m^{1003}$ is divided by 77, the remainder is 19.
what is the remainder when $m$ is divided by 77?

$m^{1003}\equiv 19 \mod{(77)}$

Then by fermats little theorem:

$m^{1003 \mod{(18)}}\equiv 19 \mod{(77)}$

$m^{(13)}\equiv 19 \mod{(77)}$

So you have to get the 13th root of 19 mod 77. Or do it mod 11 and nod 7 sepretely.

Am I mistaking something? I thought fermat's little theorem worked only on prime modulus. By using Euler's totient theorem,

$m^{1003}\equiv 19 (\mod 77)$

$m^{\varphi(1003)}\equiv m^{43}(\mod 77)$ since $\varphi(77)=60$...

Pardon me, yes i for some reason blocked out the fact that 77 was composite. FLT says that

$a^{\varphi(m)}\equiv 1 \mod{m}$

For a and m relatively prime

No FLT says p is prime (a,p)=1

$a^{p-1}\cong 1 (\mod p)$

[quote="seamusoboyle"]Sorry, i quoted the eular-fermat theorem. Set $m$ equal to $p$ and you get FLT.[/quote]

Err how do you know m is prime

in the theorem i quoted, m dosnt have to be a prime. but if you set it equal to a prime you get FLT

Yes, Fermat's Little Theorem is a special case of Euler's Totient Theorem.

[hide]
By Euler's Totient Theorem, $m^{\phi(77)}\equiv 1\pmod{77}$. Since $\phi(77) = 77(6/7)(10/11) = 60$, this gives $m^{60}\equiv 1\pmod{77}$. Therefore, $m^{1003}\equiv \left(m^{60}\right)^{16}\cdot m^{43}\equiv m^{43}\pmod{77}$. So we need to find $m$ such that $m^{43}\equiv 19\pmod{77}$. Multiplying both sides by $m^{17}$ gives $m^{60}\equiv 19m^{17}\iff 19m^{17}\equiv 1 \pmod{77}$, or $m^{17}\equiv 73\pmod{77}$.

Well modulo 7 we get $m^{6}\equiv 1$ so $m^{18}\equiv 73m\iff 3m\equiv 1\iff m\equiv 5$. Modulo 11 we get $m^{10}\equiv 1$, so $m^{20}\equiv 73m^{3}\iff 7m^{3}\equiv 1\iff m^{3}\equiv 8\equiv m\equiv 2$.

So $m\equiv 5\pmod{7}$ and $m\equiv 2\pmod{11}$. Therefore $m\equiv \boxed{68}\pmod{77}$.
[/hide]

Is that right?

ya if you check that answer on ur calculator in windows..it works..good job..like $68^{1003}\equiv19\mod77$

Good job, everyone.  Just one comment.  I notice many students use Euler's totient function $\phi$ to find a power that becomes 1.  A better function is Charmichael's function $\lambda$:
[url]http://mathworld.wolfram.com/CarmichaelFunction.html[/url].

The key difference between the two is that $\phi$ uses multiplication, whereas $\lambda$ uses least common multiple.  For example,
\[\lambda(77) = \lambda(7 \cdot 11) = \mathrm{lcm}(\lambda(7), \lambda(11)) = \mathrm{lcm}(6, 10) = 30 \, . \]
As a consequence, we know that
\[a^{30}\equiv 1 \pmod{77}\]
for every integer $a$ relatively prime to 77.  Euler's theorem gives us the weaker result
\[a^{60}\equiv 1 \pmod{77}\, . \]
For this particular problem, the improvement isn't much, but for other problems, Carmichael's bound can be quite useful.

[quote="chess64"]Yes, Fermat's Little Theorem is a special case of Euler's Totient Theorem.

[hide]

Well modulo 7 we get $m^{6}\equiv 1$ so $m^{18}\equiv 73m\iff 3m\equiv 1\iff m\equiv 5$. Modulo 11 we get $m^{10}\equiv 1$, so $m^{20}\equiv 73m^{3}\iff 7m^{3}\equiv 1\iff m^{3}\equiv 8\equiv m\equiv 2$.

So $m\equiv 5\pmod{7}$ and $m\equiv 2\pmod{11}$. Therefore $m\equiv \boxed{68}\pmod{77}$.
[/hide]

quote]I dont understand that part


An equasion is given: $ x^{3} \plus{} m^{3}x^{2} \minus{} m^{2}x \minus{} 1\equal{}0$

Find all values of real parameter $ m$, for which there is only one real solution of - number $ 1$.

[quote="Invariant"]An equasion is given: $ x^{3} \plus{} m^{3}x^{2} \minus{} m^{2}x \minus{} 1 \equal{} 0$

Find all values of real parameter $ m$, for which there is only one real solution of - number $ 1$.[/quote]

[hide]
If $ x\minus{}1$ is a factor the sum of the coefficients is $ 0$.

Hence $ 1\plus{}m^3\minus{}m^2\minus{}1\equal{}0 \Rightarrow m^2(m\minus{}1)\equal{}0$

Either $ m\equal{}0$ or $ m\equal{}1$.

Easy to check two cases
[/hide]


Stanford
college
geometry
3D geometry
MATHCOUNTS
MIT
summer program

This is at Stanford. I'm the second from the left, and Laughinghead is third from left.

third from left is a bit ambiguous - do you count the little lady poking her head in from the back?

I think so, because as far as I know, laughinghead isn't Asian. But I might be wrong. :ninja:

Yes, the one behind everyone else is laughinghead :) .

Group Photo :) .

The dorms are in the backround.

Yeah, it was cool :) .

Except when laughinghead or easyas3.14159 tried to jump off it :roll: .

Hey, that's a bad picture of me! :mad:

That looks quite funny, but I don't remember that picture... :(

My camera is still broken, otherwise I would have posted some awesomely awesome pictures from camp. There were some pretty cool ones. Laura, remember the one where you are sitting on the windowsill in the hall of the E dorm in that pretty black skirt? That one's cool. :lol:

I'm not little! I was crouching to fit in between them... :P

i_like_pie, how do you know I'm not asian? HOW DOES EVERYONE HERE NOW A LOT OF STUFF ABOUT ME?!?!?!?!!?! :huh:  :)

:rotfl:  :rotfl:  :rotfl:

*sigh* good times... Once a deer chased me through that bridge. :P

:o  :o  :o  You forgot about the name tags!!!!!! :o

at stanford? that's where my sister goes to school lol

though I haven't been there yet

Alina, fix your camera :roll: !! Aren't there places where you can just use the memory card to get the pictures? Because you just broke the camera, not the memory card...

I'm going to put some more pictures up sometime. I'll make sure to put up a good picture of you :) .

Another picture of the lake, this one I took earlier in May. The lake's called Swift Creek Resevoir, btw.

The reproduction of the Godspeed. I can't believe people sailed across the ocean in something that small....

Dude that's so awesome!!!! It's in your [i]neighborhood[/i]? Lucky, the only water body around here is...nothing. *sigh*

Supernova, Solafidefarms, and LadyKn1ght.

[quote="laughinghead505"]Duuuuuuuuuude... how come you actually look at those? And just so you know, there are some russian-asian people. :P[/quote]

Its 'cause i_like_pie is Russian.

The Puzzle Hunt :D (and the Banana of Doom)

My Puzzle Hunt team (all except for me and two other members)

So where are you in the picture? Or did you take it?

yeah, I took it. All of the photos I put up I took (except the one that does have me in it. I put the camera on a timer to take that picture, so no one took it)


A rectangular 7-inch by 12-inch picture is framed with a mat so that four inches of the mat are seen on the top and on each of the sides.  Five inches of the mat are seen on the bottom.  What fraction of the finished product's area (picture and mat) is the picture?  Express your answer as a common fraction.
[metapost=7303035872d7d921204ec8a00fca51db] save u;

 vardef mm_bracket(expr A,B) =
  save pict,v; pair v; picture pict;
  pict:=nullpicture;
  v:=(A-B)/abs(A-B) rotated 90;
  addto pict doublepath A--B;
  addto pict doublepath (A+4*v)--(A-4*v);
  addto pict doublepath (B+4*v)--(B-4*v);
  pict
 enddef;

 % Set the value of u to scale the entire image
 u:=.2cm;

 draw (0,0)--(15u,0)--(15u,21u)--(0,21u)--cycle;
 fill (4u,5u)--(11u,5u)--(11u,17u)--(4u,17u)--cycle withcolor (.8,.4,.9);
 draw (4u,5u)--(11u,5u)--(11u,17u)--(4u,17u)--cycle;

 draw mm_bracket((0,21u+8),(4u,21u+8));
 draw mm_bracket((11u,21u+8),(15u,21u+8));
 draw mm_bracket((15u+8,21u),(15u+8,17u));
 draw mm_bracket((15u+8,5u),(15u+8,0));

 label.top(btex $ 4$ etex,(2u,23u));
 label.top(btex $ 4$ etex,(13u,23u));
 label.rt(btex $ 4$ etex,(17u,19u));
 label.rt(btex $ 5$ etex,(17u,2.5u));
 label.top(btex $ 7$ etex,(7.5u,17u));
 label.rt(btex $ 12$ etex,(11u,11.5u));
 [/metapost]

uhh whatever 'metapost' is, but i don't think it works here.


A bag contains 4 pieces of paper, each labeled with one of the digits 1,2,3,or4, no repeats. Three of these pieces are drawnone at a time without replacement, to construct a 3-digit number. What is the probablitity that the 3-digit number is a multiple of 3?

A. 1/4  B. 1/3  C. 1/2  D. 2/3  E. 3/4

Apparently, the answer is C. I need to figure out how to understand these kinds of problems and master them! Please help! :ninja:

A number is a multiple of $ 3$ if the sum of its digits is divisible by three and this is true for half the possibilities, that is if the numbers chosen are $ 1, 2, 3$ or $ 2, 3, 4$ but not $ 1, 2, 4$ or $ 1, 3, 4$.

Exact same question [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311384]here[/url], as well as a few different solutions.


Given a polynomial $x^{2}+px-\frac{1}{2p^{2}}$ where $p$ is a real number not equal to $0$. $x_{1}, x_{2}$ are the two roots of the polynomial. Prove that:

$x_{1}^{4}+x_{2}^{4}\geq 2+\sqrt{2}$


logarithms
linear algebra
linear algebra unsolved

Prove that there exists a power of $2$ which starts with $111$ (i.e. $\exists n\in {\mathbb N}$ such that $2^n=111...$)

 :ninja:

i dunno a systematic proof but i found
[hide]$2^{143}=111503\ldots$[/hide]

If $a$ is not a power of $10$, then given an sequence of digits (not beginning with $0$, of course :)), there is an $n$ s.t. $a^n$ begins with that sequence of digits. This has been discussed before around here somewhere. The main idea is to take logarithms and use the fact that $\log_{10}a$ is irrational and if $\alpha$ is irrational, then $\{n\alpha\}_n$ is dense in $(0,1)$.


inequalities
combinatorics unsolved
combinatorics

Let $ k$ be a positive integer. Set $ A \subseteq \mathbb{Z}$ is called a $ \textbf{k \minus{} set}$ if there exists $ x_1, x_2, \cdots, x_k \in \mathbb{Z}$ such that for any $ i \neq j$, $ (x_i \plus{} A) \cap (x_j \plus{} A) \equal{} \emptyset$, where $ x \plus{} A \equal{} \{ x \plus{} a \mid a \in A \}$. Prove that if $ A_i$ is $ \textbf{k}_i\textbf{ \minus{} set}$($ i \equal{} 1,2, \cdots, t$), and $ A_1 \cup A_2 \cup \cdots \cup A_t \equal{} \mathbb{Z}$, then $ \displaystyle \frac {1}{k_1} \plus{} \frac {1}{k_2} \plus{} \cdots \plus{} \frac {1}{k_t} \geq 1$.

Is there anything to this problem other than invoking a notion of density?

This is a difficult problem, I hope everyone post solution.

Let $\overline{d}(A) = \limsup_{n \rightarrow \infty} \frac{|\{-n,\ldots, -1,0,1,2,\ldots,n\} \cap A|}{2n} $ denote the upper density of a set $A\subset\mathbb{Z}.$ One can note that $\overline{d}(B_1 \cup B_2 \cup \cdots \cup B_m)\leq \sum_{i=1}^m \overline{d}(B_i)$.

[b]Lemma[/b]: For any $ \textbf{k - set}$ $A$ we have $\overline{d}(A)\leq \frac 1 k.$
The sets $(x_i+A)\cap [-n, n]$ are disjoint, thus $\sum_{i=1}^k |(x_i+A)\cap [-n, n]|\leq 2n$ or equivalently $\sum_1^k |A\cap [-n-x_i, n-x_i]|\leq 2n$. Let $l =\max \{x_1,\ldots ,x_k, -x_1,-x_2,\ldots, -x_k\}$, then $[-n+l, n-l]\subset \cap_i [-n-x_i, n-x_i]$ and thus $k|A\cap[-n+l, n-l]|\leq 2n$ which when $n\to+\infty$ gives  $\overline{d}(A)\leq \frac 1 k$.

Now we have
$1= \overline{d}(\mathbb{Z})= \overline{d}(A_1 \cup A_2 \cup \cdots \cup A_t)\leq \overline{d}(A_1)+\overline{d}(A_2)+\ldots \overline{d}(A_t)\leq \frac {1}{k_1} + \frac {1}{k_2} + \cdots + \frac {1}{k_t}.$


Show that the some of several positive reciprocals of integers has numerator an odd number when reduced to lowest form.

This is how I did it. But I dont like it  :mad:  because its long.

[hide="Step 1"]
$\frac{1}{n}+\frac{1}{n+1}= \frac{2n+1}{n(n+1)}$
This is in reduced form cuz $\gcd(2n+1,n(n+1)) = d$ then $d|n(n+1)$. Then $d|n$ or $d|(n+1)$ because $\gcd(n,n+1)=1$. If $d|n$ then $d|(2n+1) \implies d|1$ so $d=1$. And if $d|(n+1)$ then $d|(2n+1) \implies d|[2(n+1)-1] \implies d|(-1) \implies d=1$. So $d=1$.
[/hide]

[hide="Step 2"]
This one is messy. If $odd/even+odd/even = odd/even$ in reduced form.
[/hide]

[hide="Step 3"]
We want to show $\frac{1}{a_{1}}+...+\frac{1}{a_{n}}$ has numerator odd when brought to reduced form. If $n$ is even we can pair there numbers as $\left( \frac{1}{a_{1}}+\frac{1}{a_{2}}\right)+...+\left( \frac{1}{a_{n-1}}+\frac{1}{a_{n}}\right)$. Each paired number is $odd/even$ by Step 1. And thier sum is $odd/even$ by step 2.
If $n$ is odd then pait them as,
$\frac{1}{a_{1}}+\left( \frac{1}{a_{2}}+\frac{1}{a_{3}}\right)+...+\left( \frac{1}{a_{n-1}}+\frac{1}{a_{n}}\right)$
And apply the same arugement.
[/hide]

Any body have a faster method?

[hide="hint"]Consider the element divisible by the largest power of two (why must this be unique?). Then when we write this over a common fraction there exists a unique element divisible by the smallest power of two. [/hide]

You didn't specify consecutive in the problem statement.  We can also make the stronger claim that the denominator is even, and thereby show that the sum of consecutive reciprocals of positive integers cannot be integer.

By the way,

[quote="sylow_theory"][hide="Step 2"]
This one is messy. If $odd/even+odd/even = odd/even$ in reduced form.
[/hide][/quote]

This is in fact trivially false if the denominator of the first two fractions are equal.


[color=darkblue]Let $ (u_n)$ is a number sequence and let $ \begin {cases} 0 < u_n < 1 \\
u_{n \plus{} 1}(1 \minus{} u_n) > \frac {1}{4} \end {cases}$
Prove that $ u_{n \plus{} 1} > u_n$, $ \forall n \in N^*$.[/color]

$ 0 < u_n < 1, 4u_{n \plus{} 1} (1 \minus{} u_n) > 1$

$ y_n \equal{} u_n \minus{} \frac {1}{2}$

$ \minus{} \frac {1}{2} < y_n < \frac {1}{2}$

$ (2y_{n \plus{} 1} \plus{} 1)(1 \minus{} 2y_n) > 1$

$ y_{n \plus{} 1} > \frac {y_n}{1 \minus{} 2y_n}$

$ f(x) \equal{} x\left[ \frac {1}{1 \minus{} 2x} \minus{} 1 \right] \to y_{n \plus{} 1} \minus{} y_n > f(y_n)$

$ \forall \alpha \in (0,\frac {1}{2} )$

$ f(0) \equal{} 0$,
$ f( \minus{} \alpha) \equal{} \alpha\left [\frac { \minus{} 1}{1 \plus{} 2\alpha} \plus{} 1\right] > 0$
$ f(\alpha) \equal{} \alpha \left [ \frac {1}{1 \minus{} 2\alpha} \plus{} 1 \right] > 0$

so $ y_{n \plus{} 1} \minus{} y_n > 0 \to u_{n \plus{} 1} \minus{} u_n > 0 \to u_n$ increasing

Another way to formulate the problem is:
if $ 0 < x < 1$ and $ 0 < y < 1$ and $ y(1 \minus{} x) > \frac14$ then $ y > x$

In other words: The graph of $ y \equal{} \frac {1}{4(1 \minus{} x)}$ in the square $ (0,1)\times(0,1)$ lays on or above the line $ y \equal{} x$.
This is easy to verify, because $ \frac {1}{4(1 \minus{} x)} \equal{} x$ has 2 equal solutions $ x \equal{} \frac12$.


limit
complex analysis
complex analysis unsolved

For each $ c\in\mathbb C$, let $ f_c(z,0)\equal{}z$, and $ f_c(z,n)\equal{}f_c(z,n\minus{}1)^2\plus{}c$ for $ n\geq1$.
a) Prove that if $ |c|\leq\frac14$ then there is a neighborhood $ U$ of origin such that for each $ z\in U$ the sequence $ f_c(z,n),n\in\mathbb N$ is bounded.
b) Prove that if $ c>\frac14$ is a real number there is a neighborhood $ U$ of origin such that for each $ z\in U$  the sequence $ f_c(z,n),n\in\mathbb N$ is unbounded.

a) First consider sequence of positive real numbers defined:

$ a_0 \equal{} A > 0$
$ a_{n \plus{} 1} \equal{} a_n^2 \plus{} |c|, n \geq 0$

If $ (a_n)$ is convergent then it is bounded. So let's assume that $ \lim_{n \to \infty} a_n \equal{} g$. Then from the reccurence relation we have:

$ g \equal{} g^2 \plus{} |c|$

$ |c| \leq \frac {1}{4}$ so the above equation has real roots. One of them is

$ g_1 \equal{} \frac {1 \plus{} \sqrt {1 \minus{} 4|c|}}{2}$

It's easy to check that if $ a_1 \leq \frac {1 \plus{} \sqrt {1 \minus{} 4|c|}}{2}$ then $ (a_n)$ is bounded.

Now let's take $ |z| \leq \frac {1 \plus{} \sqrt {1 \minus{} 4|c|}}{2}$ and define new sequence of positive reals:

$ a_0 \equal{} |z|$
$ a_{n \plus{} 1} \equal{} a_n^2 \plus{} |c|, n \geq 0$

We have $ |f_c(z,n)| \leq a_n \leq \frac {1 \plus{} \sqrt {1 \minus{} 4|c|}}{2}$ for $ n \in \mathbb{N}$.
Thus $ U \equal{} \{z: |z| \leq \frac {1 \plus{} \sqrt {1 \minus{} 4|c|}}{2} \}$ is desired neighbourhood.


The letters $ A, B$ and $ C$ are used to form every possible three letter ``word.'' When these ``words'' are arranged in alphabetical order and numbered so that $ AAA$ is Word 1 and $ CCC$ is Word 27, what number will correspond to the position of word $ BAB$ on the list?

Notice that for each first letter, there are $ 9$ "words" corresponding to the first letter. Thus, "BAA" would be 10th since it is the first after all $ 9$ of the "words starting with $ A$. The next word in alphabetical order would be "BAB", in position $ \fbox{11}$.


give an example of continuum pairwise disjoint subsets of real line, each with positive outer measure.

Thanks very much :)

another problem about subsets of real line:
Is there an infinite countable family $ \{A_n\}$ of pairwise disjoint, dense in $ \mathbb{R}$ and uncountable sets, such that $ \bigcup A_n\equal{}\mathbb{R}$ ?


Find all pairs of integers $(x,y)$ satisfying the equality
$y(x^{2}+36)+x(y^{2}-36)+y^{2}(y-12)=0$

We write it as a quadratic of $x, \,\, yx^{2}+(y^{2}-36)x+y(y-6)^{2}= 0$ .

The discriminant must be positive : $D= (y-6)^{2}(6-y)(6+3y) \geq 0$ and since $y$

is an integer the possible values of $y$ are ${-2,-1,0,1,2,3,4,5,6 }$ .

The discriminant must also be a square or zero so the values $y$ can take are ${-2,-1,0,4,6 }$ .

For each value of $y$ we solve the equation to get $x$ .

Yes with all such problem we always think of method applying condition the quaratic equation has root


Putnam
linear algebra
matrix
linear algebra unsolved

For positive integers $ n$ , let $ M_n$ be the $ (2n \plus{} 1)$x$ (2n \plus{} 1)$ by $ 2n \plus{} 1$ skew-symmetric matrix for which entry in the fist $ n$ subdiagonals below the main diagonal $ 1$ and each of the remaining entries below the main diagonal $ \minus{} 1$ ,Find . with proof the rank of $ M_n$ , (According to one definition , the rank of a matrix is the largest $ k$ such that there is a $ k$x$ k$ submatrix with nonzero determine)
 one may note that  :
  $ M_1 \equal{} \begin{bmatrix}0 & \minus{} 1 & 1 \\
1 & 0 & \minus{} 1 \\
\minus{} 1 & 1 & 0\end{bmatrix}$
  $ M_2 \equal{} \begin{bmatrix}0 & \minus{} 1 & \minus{} 1 & 1 & 1 \\
1 & 0 & \minus{} 1 & \minus{} 1 & 1 \\
1 & 1 & 0 & \minus{} 1 & \minus{} 1 \\
\minus{} 1 & 1 & 1 & 0 & \minus{} 1 \\
\minus{} 1 & \minus{} 1 & 1 & 1 & 0\end{bmatrix}$
[hide="Answer"] $ 2n$ I hope that it correct , I do'nt see a solution in putnam,Thank for help [/hide]

we can prove  M_n  is songular ,  so its rank must be less than or equal to 2n.

the submatrix A = (aij) obtained by
deleting row 2n+1 and column 2n+1 of Mn

it has has nonzero determinant.

Then A2 is equivalent modulo 2 to the 2n × 2n identity matrix,
so detA is not equal to 0.

Thus rank(Mn) = 2n.

A property nice (remember) :maybe:
 $ X \equal{} X_{2n} \equal{} \begin{bmatrix}0 & 1 & 1 & 1 & ... & 1 \\
1 & 0 & 1 & 1 & ... & 1 \\
1 & 1 & 0 & 1 & ... & 1 \\
... & ... & ... & ... & ... & ... \\
1 & 1 & 1 & 1 & ... & 0\end{bmatrix}$
 then $ \det(X_{2n}) \equal{} ( \minus{} 1)^{2n \minus{} 1}(2n \minus{} 1) \equal{} \minus{} (2n \minus{} 1)$
 And $ \det(A_2) \equal{} \det(2B \plus{} X) \equal{} 2k \plus{} \det(X) \equal{} 2k \minus{} (2n \minus{} 1) \equal{} 2(k \minus{} n) \minus{} 1 \neq 0$ Here $ k$ caculate follow $ B \in M_n{(\mathbb{Z})}$
 A  diferent solution
 Put $ J \equal{} J_{2n \plus{} 1} \equal{} \begin{bmatrix}0 & 1 & 0 & 0 & ... & 0 \\
0 & 0 & 1 & 0 & ... & 0 \\
0 & 0 & 0 & 1 & ... & 0 \\
... & ... & ... & ... & ... & ... \\0 & 0 & 0 & 0 & ... & 1\\
1 & 0 & 0 & 0 & ... & 0\end{bmatrix}$
 then $ J^{2n \plus{} 1} \equal{} I$ and $ rank(J \minus{} I) \equal{} 2n$ (very implies)
 Then $ M_n \equal{} \minus{} (J \plus{} J^2 \plus{} ... \plus{} J^n) \plus{} (J^{n \plus{} 1} \plus{} J^{n \plus{} 2} \plus{} ... \plus{} J^{2n}) \equal{} J(J^n \minus{} I)(I \plus{} J \plus{} ... \plus{} J^{n \minus{} 1})$
We have  $ \det(M_n) \equal{} 0$
 $ rank(M_n) \geq rank(J(J^n \plus{} I)(J^n \plus{} I)(J^n \minus{} I)(I \plus{} J \plus{} ... \plus{} J^{n \minus{} 1})) \equal{} rank(J(J^n \plus{} I)(I \minus{} J)(I \plus{} J \plus{} .. \plus{} J^{n \minus{} 1})) \equal{} rank(J(J^n \plus{} I)(I \minus{} J^n)) \equal{} rank(I \minus{} J) \equal{} 2n$
 Hence $ rank(M_n) \equal{} 2n$


algebra
polynomial
linear algebra
matrix
complex numbers
superior algebra
superior algebra solved

Let A \in Mn(R), n \geq 2, and m \geq 2 integer.Proove that
1) If P \in R[X] monic polynomial (with dominant coefficient 1) without
reals roots and degP=m, then det P(A)  \geq 0
2)If there exist positive integers p<q such that A^p = A^q + I
then det A>0

Conc. Vranceanu-Procopiu 2003
-----------------------------------------------------------------------------
Here my solution for 1) if you have a different one i want it, thanks.

1) Since dominant coefficient of P is 1, P(x) ~x^m >0 when x->oo,
m is even (if m is odd not there will a real root)
P has no real root its sign is constant, strictely positive
Lemma : If P real polynomial such that P(x)  \geq 0 the there exist C, D in R[X] such that P(X)=C(X)+D(X)
Proof it well known result

P(A)= C(A)+D(A)
det P(A) = det (C(A)+D(A)) = |det C(A)+iD(A)|  \geq 0
where |z| = z.z* , z* complex conjugate of z

Mr.Omarjee

[quote="Moubinool"]Let A \in Mn(R), n \geq 2, and m \geq 2 integer.Proove that
2)If there exist positive integers p<q such that A^p = A^q + I
then det A>0
[/quote]
For 2) i forgot something p and q are odd
(numerele natural impare p,q cu p<q)

Mr.Omarjee

Well I have the oficial solution for this one, but I will not post it yet, i will let other people think about it, I will send it to you soon Moubinool! :)

cheers!

Moubi wants an other solution. Here you are :
1)
Let v(1),,v(n) be the eigenvalues of A, so A = U*T*(U^-1), where T is a triangular matrix with v(1),..,v(n) on its diagonal.
Its easy to  see that P(A) = U*P(T)*(U^-1), where P(T) is a triangular matrix with P(v(1)),..,P(v(n)) on its diagonal, so det(P(A)) = P(v(1))*...*P(v(n))
P has real coefficients and no real root, so its degree is even and P(x) >0 for all real x (because its dominant coefficient is positive) .
So if  v(j) is real then P(v(j) > 0  .
If v(k) is non real complex we can associate its complex-conjugate  (v(k)) because the matrix A has real coefficients.
But P(v(k))=(P(v(k)), because the coefficients of P are real, so P(v(k))*P(v(k)) = |P(v(k)|^2 >= 0
Finally det(P(A))  >= 0 (q.e.d)
2)  det(A) = Product of the eigenvalues of A
The coefficients of A are real , so :
A has only  non real eigenvalues  ===>  det(A) > 0
A has at least one real eigenvalue ===> sign(detA)) = sign(product of real eigenvalues)
p,q are odd , A^p = A^q + I
Let x a real  eigenvalue of A ===> x^p = x^q + 1
p > q ===> x^p = x^q + 1 has only one real root and it is positive  ===> det(A) > 0

Whats the official solution ?