Let $ u_n$ satisfying $ u_0 > 0,u_{n + 1} = u_n - e^{\frac {- 1}{u_n^2}}$
Prove that $ \limits_{n\to \infty} (u_n^2 ln n) = 1$

[quote="Allnames"]Let $ u_n$ satisfying $ u_0 > 0,u_{n + 1} = u_n - e^{\frac { - 1}{u_n^2}}$
Prove that $ \limits_{n\to \infty} (u_n^2 ln n) = 1$[/quote]
No help for me  :o  :o  :o

I think It has some problems!
$ (U_n)$ is increase,don't exeist limit  -> $ (U_{n}^2)\minus{}>\infty$ and $ ln_n \minus{}>\infty$
->don't exist limit of your sequence! :wink:

[quote="Allnames"]No help for me  :o  :o  :o[/quote]
It would have had a better chance had you posted it where it belongs. You could try asking a moderator to move it to the calculus-analysis section.

[quote="Math pro"]I think It has some problems!
$ (U_n)$ is increase,don't exeist limit  -> $ (U_{n}^2) \minus{} > \infty$ and $ ln_n \minus{} > \infty$
->don't exist limit of your sequence! :wink:[/quote]
thank for your comment but my proof is true,and your idea is clearly wrong try it with the sequence $ x_n \equal{} \frac {1}{n}$
[quote="Kent Merryfield"]It would have had a better chance had you posted it where it belongs. You could try asking a moderator to move it to the calculus-analysis section.

[/quote]
thanks,but here is only a problem in a contest,and we must kill it by only all thing that we learned
Now .I proved it but  my proof is really and  used much hard knowledge ,,who can help me

It is clearly wrong?I don't think so. with $ U_n\equal{}\frac{1}{n} \minus{}> (U_n)$ have limit -> not satisfy your problem!
I can't find any mistake inmy ideal!I explained you in my class but why you don't understand?
I know that you r problem from a book!Can you detail the solution from the book? :maybe:

Uhm,ok,I have just check carefully your idea,and I knew where your mistake is
Can you detail why $ U_n$ doesn't have limit,I know it is increase,and you told me if exist $ c$ satisfies
$ U_n\ge c$ for all $ n$,we had the wrong thing
But you should note $ lim_{n\to \infty } u_n\equal{}0$ and it is a small  part  in my ugly proof(it is not obivious,and to proof it is quite hard)

[quote="Math pro"]but if $ (U_n)$ exist limit $ L \minus{} > L \equal{} L \minus{} e^{ \minus{} \frac {1}{L^2}} \minus{} > e^{ \minus{} \frac {1}{L^2}} \equal{} 0$ Is it true? :wink:[/quote]
wrong,note $ L\equal{}L\minus{}e^{lim \minus{}\frac{1}{u_n^2}}$ and if $ lim u_n\equal{}0$ it is until true.To prove it,we only need Lagrangre theorem,only a few lines


algebra
polynomial
number theory
Diophantine equation
relatively prime
number theory unsolved

Find x,y:
$x+x^{2}=y+y^{2}+y^{3}$
[b]that x is natural.[/b]


[color=blue][i]Κάπου την ξαναείδα τελευταία και τη θεωρώ καλή άσκηση.[/i][/color]

Δίνεται ισοσκελές τρίγωνο  $ ABC$ με $ AB \equal{}AC$  και το μέσο $ M$ της πλευράς $ AB$ .

Από το $ A$  φέρνουμε κάθετη προς την $ CM$ που συναντά την κάθετη της  $ AB$ ,στο  $ B$ , σε σημείο  $ D$.

Να αποδείξετε  ότι $ DC \equal{} DA$

 Μπάμπης

Λύνεται εύκολα με μιγαδικούς , αρκεί να δείτε ότι $ z_D\equal{}ki(z_C\minus{}z_M)\equal{}mi(z_A\minus{}z_B)\plus{}z_B$

Σε μερικά προβλήματα η λύση μπορεί να είναι χωρίς λόγια. Το σχήμα μόνο, είναι αρκετό.

Κώστας Βήττας.

[quote="vittasko"]Σε μερικά προβλήματα η λύση μπορεί να είναι χωρίς λόγια. Το σχήμα μόνο είναι αρκετό.

Κώστας Βήττας.[/quote]
Συμφωνώ απόλυτα Κώστα,μόνο που σε αυτή την περίπτωση θεωρείς το σημείο Ν και φέρνεις τον κόκκινο κύκλο (ζαβολιά?). Η λύση με αναλυτική είναι εξίσου εύκολη,αλλά πάντα πιο όμορφη είναι η Ευκλείδεια λύση. Πρότεινα τους μιγαδικούς σαν εξάσκηση της Γ λυκείου σ' αυτούς

Ζητώ συγνώμη Ροδόλφε, αλλά καμιά φορά νομίζουμε ότι κάτι φαίνεται εύκολα από το σχήμα, επειδή έτυχε να το δούμε εμείς.

Για το τετράπλευρο $ CDEN,$ προκύπτει ότι είναι εγγράψιμο, απο τη σχέση $ (AE)\cdot (AD) \equal{} (AM)\cdot (AB) \equal{} (AN)\cdot (AC),$ από το εγγράψιμο $ BDEM$ και το ισοσκελές $ \bigtriangleup ABC.$

Θα βάλω αργότερα περισσότερες λεπτομέρειες, αν χρειαστεί.

Κώστας Βήττας.


AB,CD are two chord of a circle .and P,Q are the midpoint of them.prove that;

if AB bisects the angle CPD,CD will bisect the angle AQB.

Let the chords AB, CD of the circle (O) intersect at a point X, obviously inside this circle. Denote a = XA, b = XB, c = XC, d = XD. The power of the point X to the circle (O) is equal to ab = cd. Let the lines QA, QB, PC, PD intersect the circle (O) at points B', A', D', C' different from B, A, D, C, respectively. Since P, Q are the midpoints of AB, CD, we have QA' = QA, QB' = QB, PC' = PC, PD' = PD and

$QA \cdot QB = QA \cdot QB' = QC \cdot QD = \left(\dfrac{CD}{2}\right)^2 = \dfrac{(c + d)^2}{4}$

$PC \cdot PD = PC \cdot PD' = PA \cdot PB = \left(\dfrac{AB}{2}\right)^2 = \dfrac{(a + b)^2}{4}$

WLOG, assume that the line AB bisects the angle $\angle CPD$. Then $\dfrac{PC}{PD} = \dfrac{XC}{XD} = \frac c d$. Knowing both the product and the ratio of the segments PC, PD, we can calculate their lenghts:

$PC = \dfrac{a + b}{2} \sqrt{\frac c d},\ \ PD = \dfrac{a + b}{2} \sqrt{\frac d c}$. Knowing the sides CD, PC, PD of the triangle $\triangle CPD$, we can calculate the length of its median m = PQ:

$4m^2 = 2(PC^2 + PD^2) - CD^2 = \dfrac{(a + b)^2}{2} \left(\frac c d + \frac d c \right) - (c + d)^2$

We have to show that $x = \dfrac{QA}{QB} = \dfrac{XA}{XB} =\frac a b$. Knowing $QA \cdot QB = \dfrac{(c + d)^2}{4}$, we have

$x = \dfrac{QA}{QB} = \dfrac{(c + d)^2}{4QB^2},\ \ \frac 1 x = \dfrac{QB}{QA} = \dfrac{(c + d)^2}{4QA^2}$

$QA^2 = \dfrac{(c + d)^2}{4} \cdot x,\ \ QB^2 = \dfrac{(c + d)^2}{4} \cdot \frac 1 x$

The segment m = PQ is also a median of the triangle $\triangle AQB$, hence,

$4m^2 = 2(QA^2 + QB^2) - AB^2 = \dfrac{(c + d)^2}{2} \left(x + \frac 1 x \right) - (a + b)^2$

Comparing the 2 expressions for the median m = PQ, we get a simple equation for $y = x + \frac 1 x$:

$\dfrac{(c + d)^2}{2} \left(x + \frac 1 x \right) - (a + b)^2 = \dfrac{(a + b)^2}{2} \left(\frac c d + \frac d c \right) - (c + d)^2$

Solving this equation for y,

$y = x + \frac 1 x = \left(\dfrac{a + b}{c + d}\right)^2 \left( \frac c d + \frac d c \right) - 2 + 2\left(\dfrac{a + b}{c + d}\right)^2 =$

$= \left(\dfrac{a + b}{c + d}\right)^2 \dfrac{c^2 + 2cd + d^2}{cd} - 2 =$

$= \dfrac{(a + b)^2}{ab} - 2 = \dfrac{a^2 + b^2}{ab} = \frac a b + \frac b a$

It is now obviuus that $x = \frac a b$ or $\frac b a$, but let's calculate it. Since $y = x + \frac 1 x$, it follows that $x^2 - xy + 1 = 0$ and

$x^2 - \left(\frac a b + \frac b a \right) x + 1 = 0$

The product of the 2 roots of this quadratic equation is 1 and their sum $\frac a b + \frac b a$ hence, they are equal to $\frac a b$ or $\frac b a$. Neglecting the trivial degenerated case, when a = b and both roots are equal to 1, we have one root greater than and the other less than 1. But it is clear that $x = \dfrac{QA}{QB} > 1\ \Longleftrightarrow\ \frac a b > 1$ and the other way around. Consequently, only one root $x = \frac a b$ is acceptable.

Dear Mathlinkers,
you can see
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=610405
iincerely
Jean-Louis


In a 150-year period, what is the maximum possible number of
leap years?

Isn't there some weird rule that if the year is divisible by 100, it doesn't count as a leap year,
however, if the year is divisible by 400, it is considered a leap year?

For example:

1900 is not a leap year while 2000 is a leap year.

Yeah so you just want to make sure 2000 is the only turn of the century.

If you start on a leap year, say 1920, you get 1920,1924....2068

Dividing all by 4:
480,481....517

Subtracting 479:

1,2,3...38

so 38 is the maximum.

Also, since a leap year occurs every four years, we have $ \left \lceil \frac{150}{4} \right \rceil$, or $ \boxed{38}$.

That doesn't always work. What if I say in a 200 year period?


algebra
polynomial
superior algebra
superior algebra unsolved

why is it true that f(x) is irreducible iff f(x+1) is irreducible.

Because $ s_c: A[x] \to A[x]$, $ f(x) \mapsto f(x\plus{}c)$ for $ c \in A$ is an automorphism of $ A[x]$ wih inverse $ s_{\minus{}c}$. And automorphisms preserve irreducibility.

[quote="ZetaX"]And automorphisms preserve irreducibility.[/quote]
Proof by dogmatic assertion. :)

If $ f(x) \equal{} p(x)q(x)$, then $ f(x \plus{} 1) \equal{} p(x \plus{} 1)q(x \plus{} 1)$.  And vice versa.  So factoring $ f(x)$ is the same as factoring $ f(x \plus{} 1)$. (note that it is important in this argument to observe that $ \rm{deg}\ f(x) \equal{} \rm{deg}\ f(x \plus{} 1)$)


inequalities
real analysis
real analysis theorems

What are the suffice and necessary conditions for a sequance like $\{a_i\}$ such that

1.$\sum_{i=1}^{\infty} x_i=$ convergent.

2.$\sum_{i=1}^{\infty}(-1)^i x_i=$  convergent

2. ,my teacher told me : these 2 conditions are enought but i didnt managed to proof.So would someone confirme the rightness of this?
1.$lim|a_i| \longrightarrow 0$ when $i \longrightarrow \infty$
2.$|a_{i+1} | \le |a_i|$

[quote="Diarmuid"]The answer you have for (1) is wrong - the classical counterexample being $a_i=\frac{1}{i}$, which is decreasing and limits to zero as in the conditions you mention, but the sum of whose terms diverges.[/quote]
I ment this for $\sum_{i=1}^{\infty} (-1)^i x_i$ ,i have edited the post .

By the way ,For $a_i=\frac1i$ what will be $\sum_{i=1} (-1)^i a_i$ :?:

Well it's not really right for the second one either - e.g. $a_i=\frac{(-1)^i}{i}$ gives $\sum_{i=1}^\infty(-1)^ia_i=\sum_{i=1}^\infty\frac{1}{i}$, which is the same thing.

If you additionally assume that all the terms are positive (or negative), then you'd have a sufficient condition.

1) The only reasonable sufficient condition you're going to get is something like the Cauchy criterion, unless you impose much tighter restrictions on $\{a_i\}$.  In particular, $a_i\to 0$ is not strong enough.

2)  If each $a_i$ is positive, then it is enough to have $a_i\geq a_j$ for $i\leq j$, and $a_i\to 0$.  This is the alternating series test; you may prove it using partial summation.

i have a related question: knowing that $\sum_{i=1}^{\infty} |a_i|$ converges, how do you prove that $\sum_{i=1}^{\infty} a_i$ converges?

[quote="Diarmuid"]Triangle inequality on the partial sums?

(i.e. $\left|\sum a_i\right|\le\sum|a_i|$).[/quote]

is this inequality sufficient to prove it? i didn't manage to prove it from here :(    :blush:

Yes, it is enough.  Try using the Cauchy criterion.


inequalities
search
inequalities proposed

Let $ a,b,c>0$ find the best $ t$ such that

\[ \frac{a\plus{}b}{c}\plus{}\frac{a\plus{}c}{b}\plus{}\frac{c\plus{}b}{a}\plus{}t(\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2})\ge 6\plus{}t\]

[quote="tranvanluan"]I think $ k \equal{} 2\sqrt {2}$[/quote]
No, $ k\equal{}4 \sqrt{2}$.
:)

Yes, you right, can_hang2007  :) Setting:
\[ f(a,b,c,t)\equal{}LHS\minus{}RHS\]
We can assume that $ a\plus{}b\plus{}c\equal{}1$. Hence we have:
\[ f(1\minus{}2x,x,x,t)\ge 0\to k\equal{}4\sqrt{2}\]
Easy to prove this one  :wink: Setting $ p\equal{}a\plus{}b\plus{}c\equal{}1,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$ so we have
\[ q\le 1/3,r\le \frac{q^2(1\minus{}q)}{2(2\minus{}3q)}\]
The inequality is equivalent to:
\[ \frac{q}{r}\plus{}\frac{4q\sqrt{2}}{1\minus{}2q}\minus{}9\minus{}4\sqrt{2}\ge 0\]
\[ \iff \frac{2(2\minus{}3q)}{q(1\minus{}q)}\plus{}\frac{4q\sqrt{2}}{1\minus{}2q}\minus{}9\minus{}4\sqrt{2}\ge 0\]
\[ \iff \frac{(1\minus{}3q)[(4\sqrt{2}\plus{}6)q^2\minus{}(11\plus{}4\sqrt{2})q\plus{}4]}{q(1\minus{}q)(1\minus{}2q)}\ge 0,\forall q\le 1/3\]

We are done  :wink:

[quote="zaizai-hoang"]Yes, you right, can_hang2007  :) Setting:
\[ f(a,b,c,t) \equal{} LHS \minus{} RHS
\]
We can assume that $ a \plus{} b \plus{} c \equal{} 1$. Hence we have:
\[ f(1 \minus{} 2x,x,x,t)\ge 0\to k \equal{} 4\sqrt {2}
\]
Easy to prove this one  :wink: Setting $ p \equal{} a \plus{} b \plus{} c \equal{} 1,q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$ so we have
\[ q\le 1/3,r\le \frac {q^2(1 \minus{} q)}{2(2 \minus{} 3q)}
\]
The inequality is equivalent to:
\[ \frac {q}{r} \plus{} \frac {4q\sqrt {2}}{1 \minus{} 2q} \minus{} 9 \minus{} 4\sqrt {2}\ge 0
\]

\[ \iff \frac {2(2 \minus{} 3q)}{q(1 \minus{} q)} \plus{} \frac {4q\sqrt {2}}{1 \minus{} 2q} \minus{} 9 \minus{} 4\sqrt {2}\ge 0
\]

\[ \iff \frac {(1 \minus{} 3q)[(4\sqrt {2} \plus{} 6)q^2 \minus{} (11 \plus{} 4\sqrt {2})q \plus{} 4]}{q(1 \minus{} q)(1 \minus{} 2q)}\ge 0,\forall q\le 1/3
\]
We are done  :wink:[/quote]
Dear Hoang, in your proof, I dont see when does the second equality hold?

The best $ t$ :$ 4\sqrt{2}$

Let $ a\plus{}b\plus{}c\equal{}1$ and $ ab\plus{}bc\plus{}ca\equal{}\frac{1\minus{}s^2}{3}$such that $ 0\le s\le 1$ and $ r\equal{}abc$
then we have $ r\le \frac{(1\minus{}s)^2(1\plus{}2s)}{27}$

The problem is equivalent to

\[ s^2((2\plus{}\sqrt{2})s\plus{}1\minus{}\sqrt{2})^2\ge 0\]

Which is true :wink:

To zaizai-hoang:
You wrote " We can assume that $ a\plus{}b\plus{}c\equal{}1$." But it refers to a special case, doesn't it?? :huh:

Can u post ur proof can_hang? :)

[quote="Sunkern_sunflora"]To zaizai-hoang:
You wrote " We can assume that $ a \plus{} b \plus{} c \equal{} 1$." But it refers to a special case, doesn't it?? :huh:

Can u post ur proof can_hang? :)[/quote]
Dear Sunkern_sunflora,
The reason we can assume $ a\plus{}b\plus{}c\equal{}1$ is due to the following property
\[ f(ka,kb,kc) \equal{}k^n f(a,b,c)\]
for all $ k$.
So, in here, if we let $ k\equal{}\frac{1}{a\plus{}b\plus{}c}$, we will have
\[ f\left( \frac{a}{a\plus{}b\plus{}c},\frac{b}{a\plus{}b\plus{}c},\frac{c}{a\plus{}b\plus{}c}\right) \equal{}\frac{1}{(a\plus{}b\plus{}c)^n} f(a,b,c)\]
So if $ a,b,c>0$ or $ n$ is even, and our question is to prove $ f(a,b,c) \ge 0$, it suffices for us to prove that
\[ f\left( \frac{a}{a\plus{}b\plus{}c},\frac{b}{a\plus{}b\plus{}c},\frac{c}{a\plus{}b\plus{}c}\right) \ge 0\]
which is equivalent to
\[ f(a_1,b_1,c_1) \ge 0\]
for $ a_1\plus{}b_1\plus{}c_1\equal{}1$.
Now, in our original problem, if we denote $ f(a,b,c) \equal{}LHS\minus{}RHS$, then you will have
\[ f(ka,kb,kc) \equal{}f(a,b,c)\]
So we can apply our above property. :)
This is called "homogeneous". :)
By the way, you can try to prove this inequality by SOS or vornicu-schur or mixing variables or more...
Regards
:)

[quote="Sunkern_sunflora"]Thank u can_hang2007! There's so much to learn from u! :)
Could u also tell me what is vornicu-schur, what is SOS or what is mixing variables, please! :([/quote]
For the Vornicu-Schur: http://www.mathlinks.ro/viewtopic.php?t=49427
For the SOS: http://www.mathlinks.ro/viewtopic.php?t=80127
For the mixing variables, I am sorry, I cant find the link for you.

[quote="Sunkern_sunflora"][quote="can_hang2007"]For the Vornicu-Schur: http://www.mathlinks.ro/viewtopic.php?t=49427
For the SOS: http://www.mathlinks.ro/viewtopic.php?t=80127
For the mixing variables, I am sorry, I cant find the link for you.[/quote]
First of all, THANK U VERY MUCH for the two links u provided. I could understand the SOS technique, but not Valentin-Schur for so sure. I'll have to study on that link further.
As for the mixing variables, could u please jist it a bit for me?? U explain subjects very good. Thank u very much. In our bengali, we call it "DHANYOBAD"! :)[/quote]
I have just found the following lecture on mixing variables of Garbriel Dospinescu and Tran Nam Dung. :)

[quote="Sunkern_sunflora"]Thank u! I have downloaded your source, but it shows that it is an RAR file. What do I need to open a RAR file?[/quote]
You just need to go to http://www.google.com. And search for the winrar program, then install it. :)


abstract algebra
linear algebra
matrix
absolute value
linear algebra unsolved

Let  A be an ortogonal  matrix  and  b   an  eigenvalue of A.  Prove that  module(b)=1.

The English word would be "modulus." The word "module" means something else. (I'd more likely say "absolute value," anyway.)

So suppose $ A$ is orthogonal, which means that $ A^TA\equal{}AA^T\equal{}I,$ and that $ \lambda$ is an eigenvalue of $ A.$

Then $ Ax\equal{}\lambda x$ for some $ x\ne0.$ However, it is possible that $ \lambda$ and $ x$ are complex valued.

In what lies below, let $ x^*$ and $ A^*$ be the conjugate transpose of the matrix in question, and for real $ A,$ we have $ A^*\equal{}A^T.$

$ |\lambda|^2\|x\|^2\equal{}\overline{\lambda}x^*x\lambda \equal{}(\lambda x)^*(\lambda x)\equal{}(Ax)^*(Ax)\equal{}x^*A^*Ax\equal{}x^*Ix\equal{}x^*x\equal{}\|x\|^2.$

If we divided by $ \|x\|^2,$ which is nonzero, we get $ |\lambda|^2\equal{}1$ which gives us that $ |\lambda|\equal{}1.$

Note that we didn't actually use that $ A$ was real; we only used that $ A^*A\equal{}I.$ Hence we may change the hypothesis of "orthogonal" to "unitary."


Let $ h(t)$ and $ V(t)$ represent the height and volume of water in a tank at time $ t$. If water leaks through a hole with area a at the bottom of the tank, then Torricelli’s Law says that  $ \frac{dV}{dt}\equal{}\minus{}a\sqrt{2gh}$, where $ g$ is the acceleration due to gravity. So, the rate at which water flows from the tank is proportional to the square root of the water height.

Suppose the tank is cylindrical with height $ 6$ feet and radius $ 2$ feet. The hole is circular with radius $ 1$ inch. If $ g \equal{} 32$ feet per second squared, show that $ \frac{dh}{dt}\equal{}\minus{}\frac{1}{72}\sqrt{h}$.

Assuming that the tank is full at time t = 0, show that $ h(t)\equal{}\left(\minus{}\frac{1}{144}t\plus{}\sqrt{6}\right)^2$ is a solution to the differential equation in (a).

How long will it take for the water in the tank to drain completely?

I'm not deriving the solution but I'll do the second part.  Set $ h(t)\equal{}0$ so we get
$ \frac{1}{144}t\equal{}\sqrt{6}\Rightarrow t\equal{}144\sqrt{6}$
So at that time, the water is drained.


geometry
parallelogram
analytic geometry
complex numbers
geometric transformation

Prove, using [u]only geometry[/u] that the center of the squares constructed on the sides of a parallelogram form a square as well.

Do you mean Euclidean geometry? There's a relatively simple proof [hide="using"]complex numbers.[/hide]

[hide="The Relevant Diagram"]Consider the image below [may take a few moments to load]. We hope to show that the four red segments form a square. To aid the process, I've created four light green triangles out of various segments. Can you see their relevance?

[img]http://img.photobucket.com/albums/v194/mukmasterxps/pww_para_square.png[/img][/hide]

We could alternatively use the fact that given a triangle ABC with squares constructed on AB and AC, if M is the midpoint of AC and X and Y are the centers of the squares on AB and AC, respectively, then XMY is an isosceles right triangle.

But honestly, complex numbers would be my choice. =P

Does the Mean Geometry Theorem count? Since that can be proven using spiral similarity...

Got it from TZF's picture.

All you need to use is the basic congruence principle: If two triangles have the same side , angle in between, and the other side equal, then those triangles are congruent.

The whole point is to use most basic form of geometry, defeats the purpose using complex numbers.

Thanks for the attention guys, appreciate it

What is the proof of using complex numbers? I tried it (first assigning the coordinates) but it gets very messy...
Also, what is the Mean Geometry Theorem? I couldn't find it on the internet... :?:

I'm not sure why...the MGT just says that the weighted average of 2 directly similar figures is similar to the original figure...its a result of spiral similarity that makes the proof of Napoleon's theorem, Van Aubel's Theorem and a ton of other results easier I guess...its not really anything new in my opinion


If $ a\plus{}b\equal{}8$, $ b\plus{}c\equal{}\minus{}3$, and $ a\plus{}c\equal{} \minus{}5$, what is the value of the product $ abc$?

We are given that
\begin{align} a + b & = 8 \\
b + c & = - 3 \\
a + c & = - 5 \end{align}
Subtracting $ (2)$ from $ (1)$ and then adding $ (3)$, we get \[ 2a = 6\Rightarrow a = 3\]  Substituting, we find that \[ b = 5 \\ c = - 8\]
The product is $ (3)(5)( - 8) = \boxed{ - 120}$

hello, solving the equation system
$ a\plus{}b\equal{}8$
$ b\plus{}c\equal{}\minus{}3$
$ a\plus{}c\equal{}\minus{}5$
we get $ a\equal{}3,b\equal{}5,c\equal{}\minus{}8$, so our product has the value $ abc\equal{}\minus{}120$.
Sonnhard.


advanced fields
advanced fields unsolved

Let I = [0,1] and A = {0,1,1/2,1/3,...,}. Show that there is no continuous retraction
$ I \times I \to I \times {0} \cup {A \times I}$.

I am guessing we want to use continuous on a compact set implies uniform continuity and then consider points near the y axis. I can't seem to make it rigorous though.

$ I \times I$ is locally connected, but $ I \times 0 \cup A \times I$ not, and the retraction is closed

[quote="myan"]$ I \times I$ is locally connected, but $ I \times 0 \cup A \times I$ not, [/quote]

I agree.

[quote="myan"]and the retraction is closed[/quote]

Can you elaborate? Why is the retraction closed? Do closed retractions preserve local connectedness? I can't find that theorem in Munkres...


calculus
integration
logarithms
parameterization
derivative
calculus computations

Compute $\int_{0}^{1}\left( x-\frac12 \right) \cdot \ln \left(-\ln x \right) \, dx$.

Your integral equals: $-\frac{\ln 2}{2}$.

Recall that $\int_{0}^{\infty}e^{-x}\ln x dx=-\gamma$.

Let $I$ be the value of your integral. Make the substitution $x=e^{-t}$ and you get that

$I=\int_{0}^{\infty}\left(e^{-2t}-\frac{1}{2}e^{-t}\right)\ln t dt=\int_{0}^{\infty}e^{-2t}\ln t dt-\frac{\gamma}{2}$.

The substitution $2t=y$ shows that the integral equals $-\frac{\ln 2}{2}$.

Nice solution, didilica. A few comments:

1. We don't need to know what the "exact" value of $\int_{0}^\infty e^{-x}\ln x \, dx$ is. The fact that it converges is enough.

2. How do you prove that $\int_{0}^\infty e^{-x}\ln x \, dx =-\gamma$ anyway?

3. By the way, I was computing $\int_{0}^{1}\frac{x-1}{\ln x}\, dx$ (I found it in a post of Jimmy, together with a [color=white]"parameter differentiation"[/color] solution) and I reached this integral after an integration by parts.

Here is a method of attack.
$\Gamma(a)=\int_{0}^{\infty}x^{a-1}e^{-x}dx$, valid for $Re(a)>0$.

$\Gamma'(a)=\int_{0}^{\infty}x^{a-1}\ln x e^{-x}dx$.

it is known that $\Gamma'(1)=-\gamma$

$\Gamma'(1)=\int_{0}^{\infty}e^{-x}\ln(x)dx=-\gamma$.

$\psi(z)=-\frac{1}{z}-\gamma-\sum_{n=1}^{\infty}\left(\frac{1}{z+n}-\frac{1}{n}\right)$, $z\neq 0,-1,-2,\cdots$.

$\psi(1)=-\Gamma'(1)=-\gamma$.


[quote="Centy"]1. Let $a, b, c \geq 0$. Show that $a^5 + b^5 + c^5 \geq 5abc(b^2 - ac)$.
[/quote]
Take$\frac{a}{b}=m,\frac{c}{n}$(if $b=0$ the ineq. is done)
$m,n>0$
$a^5 + b^5 + c^5 \geq 5abc(b^2 - ac)$<=>$m^5+n^5+5m^2n^2+1\geq 5mn$
$m^5+n^5\geq 2\sqrt{m^5n^5}=2\sqrt{t^2}$=> we must only prove that
$f(t)=2t^5+5t^4-5t^2+1\geq 0$ when $t\geq 1$=>$5t^4\geq 5t^2$=>$f(t)\geq 0$.
Now let's find the minimum valu of $f$ in $[0;1]$.
$f'(t)=10t(t+1)(t-\frac{\sqrt{5}-1}{2})(t+\frac{\sqrt{5}+1}{2})$
$f(0)=1>0$,so we only need to see that$f(\frac{\sqrt{5}-1}{2})\geq 0$,I think that isn't hard to do :lol:  ;) .

[quote="arqady"][quote="Tiks"] we only need to see that$f(\frac{\sqrt{5}-1}{2})\geq 0$ [/quote]
Tiks, I checked it:
$f(t)=2t^5+5t^4-5t^2+1=(t^2+t-1)^2(2t+1).$  :?  :D  :)[/quote]
That is nicer way :lol: ,I couldn't find any root of $f$ :) .

[quote="Centy"]1. Let $a, b, c \geq 0$. Show that $a^5 + b^5 + c^5 \geq 5abc(b^2 - ac)$.[/quote]

I remember posting and solving this, a very nice and unusual inequality. My solution is [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=318045#p318045]here.[/url]

Why is $b = 1$ without loss of generality?

The inequality is homogenous, so we can divide or multiply $a$, $b$ and $c$ by any factor we want - it doesn't change anything. Ok?


Find all odd positive integers (a, b, c) such that
ab(a2+b2) = 2c4.

I thought of this an hour ago. I am not sure if anyone else has ever made this exact problem before, so if they have, my apologies.

I came to this while solving this problem: $(e^4-f^4)^2+(e^4+f^4)^2=(2g^2)^2$  where $e,f$ are both odd and prime to each other. Is it easier from this point?

sorry for my ignorance, but why does the last fact imply that V must be 0?

he use the fact that : there are no distinct positive integers $x,y$ for which $x^2+y^2$ and $x^2-y^2$ are both squares

Yes,And the general problem is:
$xy(x+y)=2z^2$
(Instead of $2z^4$)

thank you pluricomplex, and actually upon looking at it again it is not very hard to deduce from what we have, since:

$(u^2-v^2)(2^2+v^2) = u^4 - v^4 = (2xy)^4$ so by Fermat-Wiles, we deduce v = 0  since 2xy > 0 :P


geometry
rectangle
incenter
geometry unsolved

Let $ ABC$ be a rectangle triangle in A. Where $ r$ is the radius of the circle inscribed in ABC, prove that:

$ r\equal{}\frac{AB\plus{}AC\minus{}BC}{2}$

[img]http://img43.imageshack.us/img43/7697/aosdjas.jpg[/img]

Let $ O$ be the incenter of $ \triangle ABC$, and points $ D,E,F$ on $ BC,AC,AB$ respectively such that $ OD,OE,OF$ are the radius of the incircle.

We have $ AE\equal{}r$, then $ DE\equal{}AC\minus{}r$, and it is obvious that $ DE\equal{}CD$ so $ CD\equal{}AC\minus{}r$
Analogously , $ FB\equal{}BD\equal{}AB\minus{}r$
$ BD\plus{}CD\equal{}BC$
$ AC\minus{}r\plus{}AB\minus{}r\equal{}BC$
$ r\equal{}\frac{AC\plus{}AB\minus{}BC}{2}$

QED

Thanks ;)
It's a nice method. Thanks very much.


geometry
geometric transformation
reflection
geometry proposed

[b]Easy modified enunciation.[/b]

Consider an $ A$-isosceles triangle $ ABC$ and the points : $ \boxed M$ for which  $ AM\perp BC$ ; $ \boxed N\in AB$ for which

$ N\not\equiv B$ and $ MN \equal{} MC$ ; $ \boxed P\in AM\cap CN$ ; $ \boxed R\in CM$ for which $ BR\parallel AC$ . Prove that $ PR \equal{} PC$ .

[b]Notations[/b] : $ AB$ - [u]line[/u] and $ [AB]$ - [u]segment[/u] !

Here is my solution. :)

[img]http://i524.photobucket.com/albums/cc322/khanhto-photo/untitled-109.jpg[/img]

Denote by $ T \equal{} (M,MC) \cap AC (T\not\equiv C)$
We see that :$ BT,CN,AM$ concur at $ P$.
We have :$ \angle {PNM} \equal{} \angle{PCM} \equal{} \angle {PBM}$ ,i.e $ N,P,M,B$ lie on a circle . (1)
But: $ \angle {NMC} \equal{} 2 \angle {NBC} \equal{} 2 ( 90^0 \minus{} \frac {\angle {A}}{2}) \equal{} 180^0 \minus{} \angle {A} \equal{} \angle {NBR}$,so $ N,B,R,M$ lie on a circle.. (2)

From (1) and (2) ,we have $ N,P,M,R$ lie on a circle  ,so $ \angle {PRM} \equal{} \angle {MNP} \equal{} \angle {PCM}$,i.e  $ PR \equal{} PC$

[b]Here is another approach:[/b]
Thanks to $ MN\equal{}MC\Longrightarrow \angle MNC\equal{}\angle MCN$. In the other hand, $ B$ is the reflection of $ C$ respect to $ AM$, thus $ \angle NCM\equiv \angle PCM\equal{}\angle PBM$, which implies that $ \angle PNM\equal{}\angle PBM\Longrightarrow \{N,P,M,B\}$ is the set of concyclic point.
$ BR//AC\Longrightarrow \angle BRC\equiv \angle BRM\equal{}\angle ACR\equiv \angle ACM\equal{}\angle ABM$.
But due to the fact that $ MN\equal{}MC\equal{}MB\Longrightarrow \angle MNB\equal{}\angle MBN\equiv \angle ABM$, which implies that $ \angle BRM\equal{}\angle BNM\Longrightarrow R\in (NBM)\Longrightarrow \{R,N,P,M,B\}$ is the set of concyclic point. Thus, $ \angle PRM\equal{}\angle PNM\equal{}\angle PCM$, which implies that $ \triangle PRC$ is $ P$-isosceles $ \Longrightarrow PR\equal{}PC$.
Our proof is completed then.

Dear Mathlinkers,
1. Note (0) the circle centered at M and passing through C, T, N, B (se message from ma 29)
1. the isoceles triangles MTB and MCN are similar
2. M, P, T, C lie on circle (1)
   M, P, B, C lie on circle (2) ; note that these two circles intersect at M and P ; we have working wrt M
3. By a converse of Reim's theorem applied to (1) and (2), R is on (2)
4. Now consider again (1) and (2) and work wrt P ; then PCR is isoceles at P...
Sincerely
Jean-Louis

I liked very much this problem. Thanks to all for interest and nice proofs.
Here is and my proof which can be similarly with one from these.

[b][u]Proof[/u].[/b] $ m(\angle MCN)=90^{\circ}-B=m(\angle MAC)$ $ \implies$ $ \widehat {MCP}\equiv\widehat {MAC}$ $ \Longleftrightarrow$ ${ \triangle MCP\sim \triangle MAC}$ $ \implies$ $ \widehat {MPC}\equiv \widehat {MCA}$ $ \Longleftrightarrow$

$ \widehat {MPB}\equiv\widehat {MRB}$ $ \Longleftrightarrow\  MPRB$ is cyclically $ \Longleftrightarrow$ $ \widehat {PRM}\equiv\widehat {PBM}$ $ \Longleftrightarrow$ $ \widehat {PRC}\equiv\widehat {PCM}$ $ \Longleftrightarrow$ $ \widehat {PRC}\equiv\widehat {PCR}$ $ \Longleftrightarrow$ $ PR=PC$ .


How many digits are in the number $ 25^{21}\times2^{48}$?

First, let's see how many 0's it ends in. We have plenty of 2's, so let's see how many fives we have. $ 25^{21} \equal{} 5^{2 \cdot 21} \equal{} 5^{42}$(coincidence? I think not.) Therefore, there will be 42 0's. Remaining is $ 2^6 \equal{} 64$, which has two digits, so our final answer is $ 42 \plus{} 2 \equal{} \boxed{44}$.


logarithms
ratio
arithmetic sequence
geometric sequence

$\boxed{1}$

(1) Find the $n$ th term of the sequence $\{a_{n}\}$ such that $a_{1}=2,\ a_{n+1}=2a_{n}+1\ (n=1,\ 2,\ 3,\ \cdots).$

(2) Find the $n$ th term of the sequence $\{b_{n}\}$ such that $b_{1}=2,\ b_{n+1}=2b_{n}+n\ (n=1,\ 2,\ 3,\ \cdots).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108627[/url]

$\boxed{2}$

Given the sequence $\{a_{n}\}$ which is defined by $a_{1}=1,\ a_{n+1}=2a_{n}+2^{n}\ (n=1,\ 2,\ 3,\ \cdots).$ Find the $n$ th term $a_{n}$ and sum $\sum_{k=1}^{n}a_{k}.$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108644[/url]

$\boxed{3}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that $a_{1}=1,\ a_{n+1}=na_{n}+n-1\ (n=1,\ 2,\ 3,\ \cdots).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108680[/url]

$\boxed{4}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that  $a_{1}=1,\ a_{n+1}=\frac{a_{n}}{2a_{n}+3}\ (n\geq 1).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108782[/url]

$\boxed{5}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that  $a_{1}=\frac{1}{2},\ (n-1)a_{n-1}=(n+1){a_{n}}\ (n\geq 2).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108784[/url]

$\boxed{6}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that  $a_{1}=1,\ a_{n+1}=2a_{n}^{2}\ (n=1,\ 2,\ 3,\ \cdots).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108785[/url]

$\boxed{7}$

Find the $n$ th term of the sequence $\{x_{n}\}$ such that  $x_{1}=2,\ x_{n+1}=(n+1)^{2}\left(\frac{2x_{n}}{n^{2}}-1\right)\ (n=1,\ 2,\ 3,\ \cdots).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108786[/url]

$\boxed{8}$

The positive sequence $\{a_{n}\}$ satisfies the following conditions $[A],\ [B].$

$[A]\ \ \ a_{1}=1.$

$[B]\ \ \ \log a_{n}-\log a_{n-1}=\log (n-1)-\log (n+1)\ \ (n\geq 2).$

Find $\sum_{k=1}^{n}a_{k}.$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108787[/url]

$\boxed{9}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that  $a_{1}=1,\ a_{2}=3,\ a_{n+1}-3a_{n}+2a_{n-1}=2^{n}\ \ (n\geq 2).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108911[/url]

$\boxed{10}$

The operation $*$ which makes two non zero integers $m,\ n$ correspond to the integers $m*n$ satisfies the following three conditions.

[A] $0*n=n+1.$

[B] $m*0=m+1.$

[C] $m*n=(m-1)*(m*(n-1)) \ \ (m\geq 1,\ n\geq 1).$

(1) Find $1*n.$
(2) Find $2*n.$
(3) Find $3*n.$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108919[/url]

$\boxed{11}$

Find the $n$ th term of the sequence $\{a_{n}\}$ which is defined by $a_{1}=0,\ a_{n}=\left(1-\frac{1}{n}\right)^{3}a_{n-1}+\frac{n-1}{n^{2}}\ (n=2,\ 3,\ \cdots).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109050[/url]

$\boxed{12}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that  $a_{1}=1,\ a_{n+1}^{2}=-\frac{1}{4}a_{n}^{2}+4\ \ (a_{n}>0,\ n\geq 1).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109051[/url]

$\boxed{13}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that  $a_{1}=1,\ a_{n+1}=2a_{n}-n^{2}+2n\ (n=1,\ 2, \ 3,\ \cdots).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109055[/url]

$\boxed{15}$

Find the $n$ th term of the sequence $\{x_{n}\}$ such that $x_{n+1}=x_{n}(2-x_{n})\ (n=1,\ 2,\ 3,\ \cdots)$ in terms of $x_{1}.$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109309[/url]

$\boxed{16}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that  $\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}=n+\frac{1}{n}\ \ (n=1,\ 2,\ 3,\ \cdots).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109310[/url]

$\boxed{17}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that $\sum_{k=1}^{n}a_{k}=n^{3}+3n^{2}+2n.$ and Calculate $\sum_{k=1}^{n}\frac{1}{a_{k}}.$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109312[/url]

$\boxed{18}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that $\sum_{k=1}^{n}a_{k}=3n^{2}+4n+2\ (n=1,\ 2,\ 3,\ \cdots)$ and calculate $\sum_{k=1}^{n}a_{k}^{2}.$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109318[/url]

$\boxed{19}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that $a_{1}=0,\ a_{2}=1,\ (n-1)^{2}a_{n}=\sum_{k=1}^{n}a_{k}\ \ (n\geq 1).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109320[/url]

$\boxed{20}$

Let $a_{n}$ be the $n$ th term of the arithmetic sequence with $a_{1}=7,$ the common difference $2,$ and $b_{n}$ be

 the $n$ th term of the geometric sequence with $b_{1}=\frac{1}{3},$ the common ratio $\frac{1}{3}.$
For the sequence $\{c_{n}\},$ If $\sum_{k=1}^{n}a_{k}b_{k}c_{k}=\frac{1}{3}(n+1)(n+2)(n+3)$ holds, then find $c_{n}$ and evaluate $\sum_{n=1}^{\infty}\frac{1}{c_{n}}.$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109325[/url]

$\boxed{21}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that $a_{1}=a\ (\neq-1),\ a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)\ (n\geq 1).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109472[/url]

$\boxed{22}$

Find the $n$ th term of the positive sequence $\{a_{n}\}$ such that $a_{1}=1,\ a_{2}=10,\ a_{n}^{2}a_{n-2}=a_{n-1}^{3}\ (n=3,\ 4,\ \cdots).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109473[/url]

$\boxed{23}$

Find the $n$ th term of the sequence $\{a_{n}\}$ such that $a_{1}=1,\ na_{n}=(n-1)\sum_{k=1}^{n}a_{k}\ (n=2,\ 3,\ \cdots).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109474[/url]

$\boxed{24}$

Find the $n$ th term of the seuence $\{a_{n}\}$ such that $a_{1}=\frac{1}{2},\ a_{2}=\frac{1}{3},\ a_{n+2}=\frac{a_{n}a_{n+1}}{2a_{n}-a_{n+1}+2a_{n}a_{n+1}}.$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109475[/url]

$\boxed{25}$

Define the sequence $\{a_{n}\}$ such that $a_{1}=-4,\ a_{n+1}=2a_{n}+2^{n+3}n-13\cdot 2^{n+1}\ (n=1,\ 2,\ 3,\ \cdots).$
Find the value of $n$ for which $a_{n}$ is minimized.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109479[/url]

$ \boxed{26}$

Find the $ n$ th term of the sequence $ \{a_{n}\}$ such that $ a_{1} \equal{} \frac 32,\ \ a_{n \plus{} 1} \equal{} 2a_{n}(a_n \plus{} 1)\ (n \geq 1).$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=329978[/url]


Let $M,N,P$ be arbitrary points on the sides $BC,CA,AB$ respectively of an acute-angled triangle $ABC$. Prove that at least one of the following inequalities is satisfied: $NP\geq \frac 12 BC,PM\geq \frac 12 CA, MN\geq \frac 12 AB$.

[quote]bebakhshd ke saal ro engilisi ferestadam[/quote]

 :P

azizan lotfan ide bedahid , masaleye asooniye ha  ;)


binomial coefficients
Pascal\u0027s Triangle

Can someone put Pascal's identity in layman's terms instead of binomials and coefficients?

Well binomial coefficients aren't too complicated if you think of $ \dbinom nk$ as $ n$ "choose" $ k$ or the number of ways to choose $ k$ things from $ n$ things, where the order of the $ k$ things doesn't matter.

Then there are several ways to remember this.

One way is to use Pascal's Triangle, where the zeroth row is $ \left\{\dbinom00\right\} \equal{} \{1\}$, the first row is $ \left\{\dbinom10,\dbinom11\right\} \equal{} \{1,1\}$, and so on so that the $ n^{th}$ row is $ \left\{\dbinom n0,\dbinom n1,\dbinom n2,\ldots,\dbinom nn\right\}$. But each element in Pascal's Triangle is the sum of the two elements directly above it! This quickly gives us $ \dbinom{n \minus{} 1}{k \minus{} 1} \plus{} \dbinom{n \minus{} 1}k \equal{} \dbinom nk$, as desired.

Another way is a combinatorial argument. In the simplest manner, we can choose $ k$ items from $ n$ things in $ \dbinom nk$ ways, where order doesn't matter. Alternatively, we can mark one item as "bad," so that there are $ \dbinom{n \minus{} 1}k$ ways to choose $ k$ items from the $ n$ items without the "bad" item in the group, and $ \dbinom{n \minus{} 1}{k \minus{} 1}$ ways to choose $ k$ items from the $ n$ items with the "bad" item in the group. However, every group must have or not have that "bad" item! Thus, $ \dbinom nk \equal{} \dbinom{n \minus{} 1}{k \minus{} 1} \plus{} \dbinom{n \minus{} 1}k$.

Finally, we can just use the algebraic definition $ \dbinom nk \equal{} \frac {n!}{k!(n \minus{} k)!}$, but this is hardly as interesting as the other two arguments. It doesn't help to remember it either.

The first method is the main reason why it's called Pascal's Identity, and is (in my opinion) the easiest way to remember it.


[hide]

The number of handshakes with $ n$ people is $ \binom{n}{2}$. Hence, $ n(n\minus{}1)/2\equal{}28$ and $ n\equal{}8\Longrightarrow\mathrm{ (D) }$

[/hide]

Another way: think that if there are n people at a party, then the first person can shake hands with (n-1) people. Then, the next person can shake hands with all but the first person, or (n-2) people. Then the third person can shake hands with (n-3) and etc. until the second to last person shakes hands with the last person and all handshakes have been completed. Realizing this, we can simply add consecutive integers to get 28: 1+2+3+4+5+6+7 = 28 and so there are 8 people at the party.

Since each person shook some one's hand once, 1 handshake counts for 1 for each pair of people. The first would have x, the second x-1 (b/c we already counted the one he had w/ the first person), the 3rd x-2, etc. The sum ends up being 1+2+3+...+x-1+x. The equation x(x+1)/2=28 where x+1 is the number of people (x is the no. of handshakes the 1st person has, and since that person didn't shake hands w/ himself, the no. of people is n+1). We get x(x+1)=56, and x^2+x=56; x^2+x-56=0; (x-7)(x+8)=0. Since x can't be negative, x=7, and x=1=[u]8 people[/u].
I wish i knew more latex, but oh well...

This means that there were $1+2+...+(n-1)$ handshakes, which in turn is

$\frac{n(n-1)}{2}=28$

Simplifying gives,

$n^2-n-56=0$, which is $(n-8)(n+7)=0$.

The only positive solution is $n=\boxed{8}$, which is $\boxed{\text{D}}$

[quote="Brut3Forc3"]A total of $ 28$ handshakes was exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:

$ \textbf{(A)}\ 14 \qquad\textbf{(B)}\ 28 \qquad\textbf{(C)}\ 56 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 7$[/quote]

The number of people determines the number of handshakes.  There is an equation: n people shake hands with n-1 people.  Therefore, $\frac{n(n-1)}{2}=28$, which becomes $n=8$.


[b]Formulas:[/b]
   tn=t1+(n-1)d . . . to find the position number of a term, I believe
   Sn=(n/2)[(2(t1))+(n-1)d] . . . to find the sum of an arithmetic series when the last term is not given
   Sn=(n/2)(t1+tn) . . . to find the sum of an arithmetic series when given the first [i]and[/i] last terms

[b]The Problem:[/b]
"Find the sum of the multiples of 7 from 84 to 371, inclusive". So, I need to find the sum of all multiples of 7 from 84 up to 371. When I asked my teacher how to do this, she said I needed to find the postion number, because I don't know how many terms there are between 84 and 371 that are multiples of 7.

She showed me...

tn=t1+(n-1)d
371=84+(n-1)7

84 is the first term, or t1
I do not know n, because n is the number of terms, which I need to find
d is 7, since it is the common difference between each number, making multiples of seven

But why is tn equal to 371? What is tn? Any help would be greatly appreciated.

What you're trying to evaluate is $ \sum_{k\equal{}12}^{53}7k$, right?

You seem like you're not understanding how these formulas are derived; but to solve this problem, all you need is the simple $ \sum_{k\equal{}0}^{n}k\equal{}\frac{n(n\plus{}1)}{2}$.

Your sum is equivalent to $ 7\left(\sum_{k\equal{}0}^{53}k\minus{}\sum_{k\equal{}0}^{11}k\right)$. Make sure you see why; the $ 7$ is a constant multiplier, so summing $ 84\plus{}96\plus{}\cdots\plus{}371$, you're summing $ 7(12\plus{}13\plus{}\cdots\plus{}53)$, which is the same thing. Additionally, to simplify things, we add everything from $ 7\times 0$ to $ 7\times 53$, then get rid of the excess.

If you understand the way these formulas are derived, you shouldn't be so confused.

[b]EDIT:[/b]
Oh, just in case you're not familiar with sigma notation, an example:
\[ \sum_{k\equal{}a}^{b}f(k)\equal{}f(a)\plus{}f(a\plus{}1)\plus{}\cdots\plus{}f(b).\]

Um...I'm sorry, but I have no idea what you're talking aboiut. I'm trying to solve the problem I posted under the bolded text titled "The Problem" and I have never seen my teacher, or anyone else, work an arithmetic series the way you just did.

I apologize for my ignorance.

Here's how I would do it:

Take 84-371 and divide it by 7. You now have 12-53.

Pairing the numbers, you get 53+12=65, 52+13=65, etc. You have 42 numbers or 21 pairs, so 21*65=1365. Now multiply this answer by 7 to get what you originally had.

Ok, I'll explain it as simply as possible.

An arithmetic series looks like $ S\equal{}a\plus{}(a\plus{}r)\plus{}(a\plus{}2r)\plus{}\cdots\plus{}(a\plus{}nr)$. If we want to find a "closed form" for $ S$ (that is, an expression without "..." that gives the same answer), we'll use a couple of tricks.

First off, let's find a closed form for $ 1\plus{}2\plus{}\cdots\plus{}n$.
Let $ S\equal{}1\plus{}2\plus{}\cdots\plus{}n$. Now, look at this:
$ S\equal{}1\plus{}2\plus{}\cdots\plus{}n$
$ S\equal{}n\plus{}(n\minus{}1)\plus{}\cdots\plus{}1$.
Add them together:
$ 2S\equal{}\underbrace{(1\plus{}n)\plus{}(1\plus{}n)\plus{}\cdots\plus{}(1\plus{}n)}_{n\text{ times}}$.

Therefore, $ 2S\equal{}n(1\plus{}n)$, and $ S\equal{}\frac{n(n\plus{}1)}{2}$.

To summarize, $ \sum_{k\equal{}0}^{n}k\equal{}\frac{n(n\plus{}1)}{2}$. That fancy sigma notation is just saying $ 0\plus{}1\plus{}\cdots\plus{}n$.
This is an extremely important result, so know it well.

Now what you want is $ \sum_{k\equal{}12}^{53}7k$. That's summing a bunch of multiples of 7 together, so we can just pull the 7 out (as I showed in the other post), so it's equivalent to $ 7\sum_{k\equal{}12}^{53}k$.

Now you should be feeling a bit better, because $ \sum k$ is familiar. However, those bounds are weird; we're used to sums from 0 to $ n$, but when the lower bound is not $ 0$, the sum becomes more complicated.

But we can be clever and recognize that $ \sum_{k\equal{}a}^{b}f(k)\equal{}\sum_{k\equal{}0}^{b}f(k)\minus{}\sum_{k\equal{}0}^{a}f(k)$. Chew on that for a while, list it out as terms instead of using sigma notation if it helps, and make sure you see why it works.

Then you get the clean expression $ 7\left(\sum_{k\equal{}0}^{53}k\minus{}\sum_{k\equal{}0}^{11}k\right)$ for your arithmetic series. Now, substitute $ \sum_{k\equal{}0}^{n}k\equal{}\frac{n(n\plus{}1)}{2}$ and you're done.

You have 7*12, 7*13, 7*14 ... 7*53. There are (53-12) + 1 = 42 terms in this arithmetic sequence. Now you can apply the summation formula for an arithmetic sequence.

(84 + 371)/2 times 42 =  9555.

This formula, for clarity, is simply (first term + last term)/2 times the number of terms.

That's equivalent to what I'm saying.

$ 7\left(\sum_{k\equal{}0}^{53}k\minus{}\sum_{k\equal{}0}^{11}k\right)\equal{}7\left(\frac{53\cdot 54}{2}\minus{}\frac{11\cdot 12}{2}\right)\equal{}\boxed{9555}$.

I just went through the derivation of the formula.


For how many positive integers $n$ is \[\left( 1999+\frac{1}{2}\right)^{n}+\left(2000+\frac{1}{2}\right)^{n}\] an integer?

$ \Leftrightarrow 2^n|3999^n + 4001^n$
The first we has $ n$ is odd .
Consider $ n\equiv 1(\mod 2)$
Then we has :
$ 3999^n + 4001^n = 8000(\frac {3999^n + 4001^n}{8000})$
Lemma: $ \gcd(a + b,\frac {a^n + b^n}{a+b}) = \gcd(n,a + b)$ where $ \gcd(a,b) = 1$
So $ \gcd(8000,(\frac {3999^n + 4001^n}{8000}) = \gcd(800,n)$
So $ 2^n|8000$ so $ n\in = {1,3,5\}}$
[u]Result [/u]There are 3 number in this sequence is interger.

Your solution looks correct, but can you please add a proof of the lemma in your post? :)

Yes it from Newton express.
$ \gcd(a\plus{}b,\frac{a^n\plus{}b^n}{a\plus{}b})\equal{}\gcd(a\plus{}b,na^n)\equal{}\gcd(n,a\plus{}b)$

We can rephrase the problem as: Solve for $n$ such that $$v_2(3999^n+4001^n) \ge n.$$

Note that if $n$ is an odd integer, we have $$3999^n+4001^n=(3999+4001)*(3999^{n-1}4001+3999^{n-2}4001^2 \dots +4001^{n-1}).$$ Clearly the latter part of the expansion is not divisible by $2$, because modulo $2$ on gives $1+1 \dots +1$ iterated $n$ times, which is odd. Now computing gives $v_2(3999^n+4001^n)=6v_2(n)=6$. This gives the solutions $n=1,3,5$.

For evens, we can simply note that for even $n$, $2$ trivially divides $3999^n+4001^n$ but $4$ does not divide $3999^n+4001^n$, as modulo $4$ gives $(-1)^n+(1)^n \equiv 2 \pmod{4}$. Thus no even $n$ holds.


Prove that  there existed a positive number $C,$ irrelevant to $n$ and $a_1,\ a_2,\ \cdots a_n,$ satisfying the following condition.

Condition: For arbiterary positive numbers $n$ and arbiterary real numbers $a_1,\ \cdots a_n,$ the following inequality holds.

\[ \max_{0\leq x\leq 2}\prod_{j=1}^n |x-a_j|\leq C^n \max_{0\leq x\leq 1}\prod_{j=1}^n |x-a_j|. \]

This ineq is hard.
This one was posted two times in unsolved section, by hxtung and flip2004. Try to search.
In flip2004's topic, he prove it for $C=6+\sqrt{5}$.


We have that $\sqrt{\frac{x-m}{10-m}}\geq \sqrt{\frac{x-(10-m)}{m}}$ if and only if $2(x-10)(m-5) \geq 0$.
Now suppose we have that $x\geq 10$. Then by our 'lemma' the following inequalities hold: \[\begin{aligned}& \sqrt{\frac{x-6}{4}}\geq \sqrt{\frac{x-4}{6}}\\ & \sqrt{\frac{x-7}{3}}\geq \sqrt{\frac{x-3}{7}}\\ & \sqrt{\frac{x-8}{2}}\geq \sqrt{\frac{x-2}{8}}\end{aligned}\] with equality iff $x=10$. But by this inequalities $LHS \geq RHS$, so we have equality. Therefore, $x=10$ if $x\geq 10$.
If $x\leq 10$, we use the same argument to show that $LHS \leq RHS$ with equality iff $x=10$, so $x=10$.

Oh yeah, don't forget to use a more significant title than "easy equation" and "hyper-easy" next time :)

[quote="Kurt Gödel"]
Oh yeah, don't forget to use a more significant title than "easy equation" and "hyper-easy" next time :)[/quote]
Me thinks [b]arkhammedos[/b] is following in the footsteps of [b]boxedexe[/b] :)

Hey  $Kurt$ my solution is $extra$ easyr than urs , we rewrite the equoition in this form
$\sqrt{\frac{x-6}{4}}-\sqrt{\frac{x-4}{6}}+\sqrt{\frac{x-7}{3}}-\sqrt{\frac{x-3}{7}}+\sqrt{\frac{x-8}{2}}-\sqrt{\frac{x-2}{8}}=0$
 We have  $\sqrt{\frac{x-a}{b}}-\sqrt{\frac{x-b}{a}}=\frac{(a-b)[x-(a+b)]}{ab[\sqrt{\frac{x-a}{b}}+\sqrt{\frac{x-b}{a}}]}$,then by substituting  a and b by 2;3,4,6,7,8 we will find $(x-10)[S]$ wish $S\neq 0$ then $x=10$
 :rotfl:


Euler
ratio
geometry
perpendicular bisector
geometry unsolved

The line perpendicular to $ AE_a$ through $ M_a$ contains side $ BC$. Let the intersection be $ H_a$, the foot of the altitude from $ A$. Extend $ AE_a$ through $ E_a$ to $ H$ so that $ AE_a\equal{}E_aH$. $ H$ is the orthocenter. (The Euler point $ E_a$ is the midpoint of $ AH$.)
The circle with diameter $ E_aM_a$ is the nine-point circle. Notice that $ \triangle E_aE_bE_c$ is homothetic to $ \triangle ABC$, with the center at $ H$ and the ratio being $ \frac12$. Then $ HH_a\perp BC\Rightarrow HH_a\perp E_bE_c$ and $ E_bE_c$ bisects $ HH_a$. Since $ E_b$ and $ E_c$ also lie on the nine-point circle, they are the intersections of the perpendicular bisector of $ HH_a$ with the nine-point circle.
$ B$ and $ C$ are the intersections of $ HE_b$ and $ HE_c$ with $ BC$.


search
function
complex numbers
irrational number

Choose the answer that is both irrational and complex.

A. $\pi i$
B. $\sqrt{3136}$
C. $\frac{-e}{4}$
D. $\sqrt3-2i$
E. $NOTA$

EDITED.

[quote="surge"]Solve for all real solutions of $x$.

$2^x=x^2$

[hide="hint"]there are 3 solutions[/hide][/quote]

use search function, this problem was posted many times.....

I guess because $0+\pi i$ is not irrational... while $\frac{-e}{4}$ can be expressed as complex? I dunno, that's why I posted it (from Mu Alpha Theta competition).

[quote="surge"]I guess because $0+\pi i$ is not irrational...[/quote]

$\pi i$ not irrational?  But that would mean that it's rational...  And the answer is that $\frac{-e}4$ is irrational and [b]complex[/b]?  Isn't $\frac{-e}4$ real?

Mu Alpha Theta must've made a mistake...

But all real numbers are also complex numbers with imaginary part equal to zero...
From most definitions I have met before, an irrational number is a real number that is not rational. So irrationals are real numbers but I have never seen them called "complex irrationals"....
For $\pi i$, this is not a real number. Maybe you can call $i$ as a unit but that's not how it's used in most high school's definition of irrational???  :ninja:  So maybe because of this, A is ruled out. B is an integer. D is definitely a complex number.


logarithms
number theory
prime factorization
rational number
irrational number

Suppose not.  Then for some coprime positive integers x,y    we have $10^{x/y} = 8$.  It implies $10^x = 8^y$.  By ideas from prime factorization, x = 3y.  But this doesnt work.

edit:  explanation for gopher :: We suppose that it is rational.  It implies this: ($10^{x/y} = 8$) [x,y integer].  We then show this cant be satisfied in integers.  So our assumption is incorrect.

[quote="AntonioMainenti"][quote="gopherhole112"]That only proves that it is not algebraic.  Couldn't it be transcendental?  Also, your proof is a bit vague.  Could you go through it a bit slower?  Thanks[/quote]I'm not sure where $x=3y$ came from, but I think "suppose not" was a typo, since we really want to suppose it is rational, and then find the contradiction.[/quote]
$10^x=8^y$
$2^x5^x=2^{3y}$
$x=3y$ by comparing the power of $2$

[quote="gopherhole112"]That only proves that it is not algebraic.  Couldn't it be transcendental?  Also, your proof is a bit vague.  Could you go through it a bit slower?  Thanks[/quote]
Whether it is algebraic or transcendental does not matter here; the question only tells us to prove whether it is rational or not. (In this case, we are going to prove it is irrational.) And singular's proof is indeed quite clear at all.

[ I do not know the method of writing 10 a bit lower than log and writing exponent a bit higher than the base.So I have written '10'
 wherever I have wanted to mean 10 based log and used "q"
whenever I have wanted to mean q as a exponent :oops:  :?  :?  :) ]

Suppose log'10'8=x
10°=1 , 10¹=10 and 10"x" =8.
So 0<x<1
So x is not a whole number.
Now suppose x is a rational number. So we can write x=p/q
where p, q are whole numbers, q>0, q≠1 and p, q are relatively prime.

      log'10'8 = p/q

 Or, log'10'8+ log'10'(5/4)=p/q+ log'10'(5/4)

 Or, log'10'[(8)(5/4)]= p/q+ log'10'(5/4)

 Or, log'10'10= p/q+ log'10'(5/4)

 Or, 1= p/q+ log'10'(5/4)

 Or, [p+ qlog'10'(5/4)]/q=1

Or, p+ [log'10'(5/4)"q"]/q =1

Or, p+ [log'10'(5/4)"q"]= q

 As q is a whole number so (5/4) "q" is a fraction .So  [log'10'(5/4)"q"] is a fraction.
So p+ [log'10'(5/4)"q"] is a fraction because p is a whole number.
But q is obviously a whole number.
So p+ [log'10'(5/4)"q"]≠ q
So log'10'8 is not a rational number.
So log'10'8 is irrational.
                                   [Proved]
[please help me in writing exponent and 10 based log correctly :) ]

Of course, it is rational, for everyone who works in hexadecimal system.  It is indeed  $0.C$ which is nothing nut $3/4$. :)

So my proof that it was irrational was wrong?????

Please answer soon and clearly.

Sorry - Your proof was correct , I was saying it  in a little  jest In Hexadicmal system $10$ is  $16$  in Decimal system, and of ccourse,  $16^{3/4} = 8$ ....If one worked with computers (specially IBM architecture - etc) - Hexadecimal system is quite common, in fact many carry calculators which do all calcualtions in hexadecimal..

Again sorry for confusion, but hope that helps.

[quote=leepakhin][quote="AntonioMainenti"][quote="gopherhole112"]That only proves that it is not algebraic.  Couldn't it be transcendental?  Also, your proof is a bit vague.  Could you go through it a bit slower?  Thanks[/quote]I'm not sure where x=3y came from, but I think "suppose not" was a typo, since we really want to suppose it is rational, and then find the contradiction.[/quote]
10^x=8^y
2^x5^x=2^{3y}
x=3y by comparing the power of 2[/quote]
But this is incorrect since 2^x5^x=2^{3y} does not lead to the conclusion x=3y, in fact it leads to the conclusion 5^x=2^{3y-x} since we are not comparing only the powers of 2 on both sides of the equation, right? (Sorry, as a new user here, I am not allowed to post images so I have to make do without the Latex code.)

(My solution was incorrect)

[s]$\log_{10}{8}$ is not rational.

Let's assume it is rational. Then there must be integers x and y where $10^{x/y} = 8$.

Whenever we take 10 to the power of 1/y, we get an irrational number because 10 is not a perfect square/cube/etc. We multiply $10^{1/y}$ by $10^x$, which is rational. Since 8 is rational, it's impossible for it to be the product of one rational and one irrational number. Therefore, $\log_{10}{8}$ cannot be rational.

Let me know if you think my reasoning is incorrect.[/s]

[quote=basketballguy]\log_{10}{8} is not rational.

Let's assume it is rational. Then there must be integers x and y where 10^{x/y} = 8$.

Whenever we take 10 to the power of 1/y, we get an irrational number because 10 is not a perfect square/cube/etc. We multiply 10^{1/y} by 10^x, which is rational. Since 8 is rational, it's impossible for it to be the product of one rational and one irrational number. Therefore, \log_{10}{8} cannot be rational.

Let me know if you think my reasoning is incorrect.[/quote]
If I'm not wrong, though, I think 10^{1/y} multiplied by 10^x is not equal to 10^{x/y}, in fact it is only equal to 10^{1/y+x}, and likewise 10^{x/y} can only be simplified or expressed as (10^{1/y})^x.

@pear333 you are correct. Here is a new solution:

$\log_{10}{8}$ is not rational.

Let's assume it is rational. Then there must be integers x and y where $10^{x/y} = 8$. We rewrite $10^{x/y} = 8$ as $10^x = 8^y$. Since $10 = 2*5$ and $8 = 2^3$, we can write that as:

$2^x * 5^x = 2^{3y}$

The above expression becomes:

$2^{3y-x} = 5^x$.

However, there are no integers x and y that satisfy $2^{3y-x} = 5^x$ except for x = 0 and y = 0. That's because there's no way to get a power of 2 through multiplying 5 by itself over and over again (and vice versa). The only way that works is if you set both x and y equal to 0, but $\log_{10}{8}$ is clearly not $0/0$.

Since we had originally set x/y to be the solution to $\log_{10}{8}$, and x and y cannot be integers, $\log_{10}{8}$ is not rational.


Prove that $\left\{ \frac{x}{2}\right\}$ is either equal to $\frac{\{x\}}{2}$ or $\frac{\{x\}}{2}+\frac{1}{2}$.

let $x=2a+b$ where $a\in\mathbb{Z}$ and $0\le b<2$


calculus
integration
limit
trigonometry
calculus computations

Evaluate
\[ \lim_{x\rightarrow 0}\frac {1}{x}\int_{0}^{x}\left(1 \plus{} \sin{2t}\right)^{1/t}\,dt.\]
I know how to do this via l'Hoptal's rule, but is there another approach?

Use the Mean Value theorem for the integral and the problem reduces to calculate the following limit

$ \lim\limits_{\theta\rightarrow 0^{\plus{}}}(1\plus{}\sin 2\theta)^{1/\theta}$.

Ah okay, I know the integrand is continuous. So just to make sure, the mean value theorem for the integral guarantees a number c in [0,x] such that our limit is basically the limit as x approaches 0 of f(c), where f is basically the integrand except at 0, where it is the value that makes the integrand continuous (removable discontinuity). But as x goes to zero, so does c, and hence the same result as if we had applied l'Hopital in the first place. Is this right? Thanks.


vector
linear algebra
linear algebra unsolved

Let $ W$ and $ Y$ be subspaces of a finite dimensional vector space $ V$.

Prove that $ \dim\left(W\plus{}Y\right) \equal{} \dim W \plus{} \dim Y \minus{} \dim\left(W \cap Y\right)$.

[quote="Cloak"]Let W and Y be subspaces of a finite dimensional vector space V.

Prove that dim(W+Y) = dimW + dimY - dim(W intersect Y)[/quote]
Pick a basis $ B_0$ of $ Z \equal{} W \cap V$.

Add basis vectors until you get a basis $ B_1$ of $ W$ (Steinitz).

Add basis vectors to $ B_0$ until you get a basis $ B_2$ of $ V$ (Steinitz).

Check that $ B_1 \cup B_2$ is a basis of $ V\plus{}W$.

Count.

[quote="hpe"][quote="Cloak"]Let W and Y be subspaces of a finite dimensional vector space V.

Prove that dim(W+Y) = dimW + dimY - dim(W intersect Y)[/quote]
Pick a basis $ B_0$ of $ Z \equal{} W \cap V$.

Add basis vectors until you get a basis $ B_1$ of $ W$ (Steinitz).

Add basis vectors to $ B_0$ until you get a basis $ B_2$ of $ V$ (Steinitz).

Check that $ B_1 \cup B_2$ is a basis of $ V \plus{} W$.

Count.[/quote]

How do you establish linear independence to show it's a basis?

$ B_{0} \equal{} \{u_{1},\ldots,u_{i}\}$
$ B_{1} \equal{} \{u_{1},\ldots,u_{i}, w_{1},\ldots,w_{j}\}$
$ B_{2} \equal{} \{u_{1},\ldots,u_{i}, v_{1},\ldots,v_{k}\}$

$ \alpha_{1}w_{1} \plus{} \ldots \plus{} \alpha_{j}w_{j} \plus{} \beta_{1}u_{1} \plus{} \ldots \plus{} \beta_{i}u_{i} \plus{} \gamma_{1}v_{1} \plus{} \ldots \plus{} \gamma_{k}v_{k} \equal{} 0$

$ \Leftrightarrow \alpha_{1}w_{1} \plus{} \ldots \plus{} \alpha_{j}w_{j} \equal{} \minus{} \left(\beta_{1}u_{1} \plus{} \ldots \plus{} \beta_{i}u_{i} \plus{} \gamma_{1}v_{1} \plus{} \ldots \plus{} \gamma_{k}v_{k}\right)$

$ \Rightarrow \alpha_{1}w_{1} \plus{} \ldots \plus{} \alpha_{j}w_{j}\in W$ and $ \alpha_{1}w_{1} \plus{} \ldots \plus{} \alpha_{j}w_{j}\in V$

$ \Rightarrow \alpha_{1}w_{1} \plus{} \ldots \plus{} \alpha_{j}w_{j}\in W \cap V$

$ \Rightarrow \alpha_{1}w_{1} \plus{} \ldots \plus{} \alpha_{j}w_{j} \equal{} \beta^{\prime}_{1}u_{1} \plus{} \ldots \plus{} \beta^{\prime}_{i}u_{i}$

Since $ u_{1}, \ldots, u_{i}, w_{1}, \ldots, w_{j}$ are lineary independent, we get $ \alpha_{1} \equal{} \alpha_{2} \equal{} \ldots \equal{} \alpha_{i} \equal{} 0$. Now we can conclude also, that all $ \beta$-s and $ \gamma$-s are $ 0$.

[b]Lemma:[/b] The complement $ R$ of the subspace $ Q$ in the vector space $ P$ is a subspace, for which $ P$ is the direct sum of $ Q$ and $ R$; $ P\equal{}Q \oplus R$. The complement is not unique. In fact there are infinitely many of them for a non-trivial subspace $ Q$. But they are all isomorphic and $ \dim P\equal{}\dim Q \plus{} \dim R$.
[b]
Problem:[/b]
Let us denote some complements of $ W\cap Y$ in $ W, Y$ and $ W\plus{}Y$ by $ W^{\prime}, Y^{\prime}$ and $ (W\plus{}Y)^{\prime}$.

Since $ (W\cap Y) \cap W^{\prime}\equal{} \{0\}, (W\cap Y) \cap Y^{\prime}\equal{}\{0\}$ and $ W^{\prime} \cap Y^{\prime}\equal{}\{0\}$, we can take $ (W\plus{}Y)^{\prime}\equal{}W^{\prime}\oplus Y^{\prime}$.

Hence $ W \equal{} (W\cap Y) \oplus W^{\prime}, Y \equal{} (W\cap Y) \oplus Y^{\prime}$ and  $ W\plus{}Y \equal{} (W\cap Y) \oplus W^{\prime} \oplus Y^{\prime}$.

Therefore $ \dim W \equal{} \dim (W\cap Y) \plus{} \dim W^{\prime}, \dim Y \equal{} \dim (W\cap Y) \plus{} \dim Y^{\prime}$ and $ \dim W\plus{}Y \equal{} \dim (W\cap Y) \plus{} \dim W^{\prime} \plus{} \dim Y^{\prime}$.

It remains to express $ \dim W^{\prime}, \dim Y^{\prime}$ from the first two equalities and put them into the third one.
We get $ \dim(W\cap Y) \plus{} \dim (W\plus{}Y)\equal{} \dim W\plus{}\dim Y$, which was wanted.


We cannot clarify the problems while the round is in progress.  Please read the problems carefully.


Prove that there are $ a_0 ,a_1 ,...,a_n \in \mathbb{Z}$ not all of them equal to $ 0$ ,$ |a_i| \leq 9$ ,$ i\equal{}\overline{1,n}$ ,such that

$ |a_0\plus{}a_1\sqrt{2}\plus{}...\plus{}a_n \sqrt{n\plus{}1}|<\frac{1}{10^n\minus{}1}$

[i]author :Lavinia Savu[/i]

Consider all sums $ S(b)\equal{}S(b_0,b_1,...,b_n)\equal{}\sum_{k\equal{}0}^nb_k\sqrt{k\plus{}1}$ with $ 0\le b_i\le 9$.
We had $ 10^{n\plus{}1}$ means of sum and all of them $ 0\le 9A, \ A\equal{}\sum_{k\equal{}1}^{n\plus{}1}\sqrt k$
Therefore exist 2 means of nabors $ b^1\equal{}(b_0^1,b_1^1,...,b_n^1), b^2\equal{}(b_0^2,...,b_n^2)$ suth that $ 0<S(b^1)\minus{}S(b^2)<\frac{9A}{10^{n\plus{}1}\minus{}1}$. Take $ a_i\equal{}b_i^1\minus{}b_i^2$ and get, that $ 0<S(a_0,..,a_n)<\frac{9A}{10^{n\plus{}1}\minus{}1}$.

[quote="Rust"]$ 0 < S(a_0,..,a_n) < \frac {9A}{10^{n \plus{} 1} \minus{} 1}$.[/quote]

I don't see how from this one can come reach the conclusion .Can you be more specific ?


symmetry
inequalities
inequalities proposed

Let $ a,b,c$ be three distinct nonnegative real numbers. Prove that

$ (a) \ \ \ \  \sum \frac {(a\minus{}b)(a\minus{}c)}{(b\minus{}c)^2} \ge \frac {15}{4}$;

$ (b) \ \ \ \  \sum \frac {(a\minus{}2b)(a\minus{}2c)}{(b\minus{}c)^2} \ge 5$.

I just found that:
$ \minus{} 4\sum_{cyc} (a \minus{} b)^3(b \minus{} c)^3 \minus{} 15 (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)2 \equal{} (a \minus{} 2b \plus{} c)^2(b \minus{} 2c \plus{} a)^2(c \minus{} 2a \plus{} b)^2 \ge 0$ :)
(That is: $ \sum_{cyc} \frac {(a \minus{} b)(a \minus{} c)}{(b \minus{} c)^2} \minus{} \frac {15}{4} \equal{} \frac{(a \minus{} 2b \plus{} c)^2(b \minus{} 2c \plus{} a)^2(c \minus{} 2a \plus{} b)^2}{4(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}$)

The idea to finding this identity was to set $ x \equal{} a\minus{}b$, $ y \equal{} b\minus{}c$ and $ z \equal{} c\minus{}a$, and then solving the equation when $ x\plus{}y\plus{}z\equal{}0$. I had posted the solution including my ideas, but unfortunately i deleted it by accident :(

[quote="Vasc"]Let $ a,b,c$ be three distinct nonnegative real numbers. Prove that

$ (b) \ \ \ \ \sum \frac {(a \minus{} 2b)(a \minus{} 2c)}{(b \minus{} c)^2} \ge 5$.[/quote]
$ (b) \ \ \ \ \sum \frac {(a \minus{} 2b)(a \minus{} 2c)}{(b \minus{} c)^2} \minus{} 5 \equal{} \frac{(\ \sum a^3 \minus{} 2\sum bc(b\plus{}c) \plus{} 9abc\ )^2}{(b\minus{}c)^2(c\minus{}a)^2(a\minus{}b)^2}.$

[quote="manlio"][quote="Mathias_DK"]I just found that:
$ \minus{} 4\sum_{cyc} (a \minus{} b)^3(b \minus{} c)^3 \minus{} 15 (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)2 \equal{} (a \minus{} 2b \plus{} c)^2(b \minus{} 2c \plus{} a)^2(c \minus{} 2a \plus{} b)^2 \ge 0$ :)
(That is: $ \sum_{cyc} \frac {(a \minus{} b)(a \minus{} c)}{(b \minus{} c)^2} \minus{} \frac {15}{4} \equal{} \frac {(a \minus{} 2b \plus{} c)^2(b \minus{} 2c \plus{} a)^2(c \minus{} 2a \plus{} b)^2}{4(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}$)

The idea to finding this identity was to set $ x \equal{} a \minus{} b$, $ y \equal{} b \minus{} c$ and $ z \equal{} c \minus{} a$, and then solving the equation when $ x \plus{} y \plus{} z \equal{} 0$. I had posted the solution including my ideas, but unfortunately i deleted it by accident :([/quote]

Dear Mathias,

can you please explain better your method?

Thank you very much. :)[/quote]
Of course! :)

Let $ x \equal{} a\minus{}b, y \equal{} b\minus{}c, z \equal{} c\minus{}a$.

Then notice that $ \sum_{cyc} \frac {(a \minus{} b)(a \minus{} c)}{(b \minus{} c)^2} \equal{} \minus{} \sum_{cyc} \frac {(a \minus{} b)(b \minus{} c)}{(b \minus{} c)^2} \equal{} \minus{} \sum_{cyc} \frac{xy}{z^2}$. Multiplying with $ 4x^2y^2z^2$ we have to prove:

$ \minus{}4\sum_{cyc} x^3y^3 \minus{} 15x^2y^2z^2 \geqslant 0$.

Since $ \sum_{cyc} x^3y^3 \equal{} \left ( \sum_{cyc} xy \right )^3 \minus{} 3xyz \left ( \sum_{cyc} x \right ) \left ( \sum_{cyc} xy \right ) \plus{} 3x^2y^2z^2 \equal{}$ $ \left ( \sum_{cyc} xy \right )^3 \plus{} 3x^2y^2z^2$. (Since $ x\plus{}y\plus{}z\equal{}0$)

So we have to prove:

$ \minus{}4 \left ( xy\plus{}yz\plus{}zx \right )^3 \ge 27x^2y^2z^2$

The most natural way from here is to substitute $ z \equal{} \minus{}(x\plus{}y)$. Then $ \minus{}(xy\plus{}yz\plus{}zx) \equal{} \minus{}(xy\minus{}(x\plus{}y)^2) \equal{} x^2\plus{}xy\plus{}y^2$. Hence we have to prove:

$ 4 (x^2\plus{}xy\plus{}y^2)^3 \ge 27x^2y^2(x\plus{}y)^2$

By the pidgeon hole principle there are two of $ x,y,z$ which have the same sign. Because of symmetry we can wlog assume that these are $ x,y$. Now since the the inequality holds for $ x,y,z$ if and only if it holds for $ \minus{}x,\minus{}y,\minus{}z$ we can assume $ x,y \ge 0$. (The substitution $ x,y,z \to \minus{}x,\minus{}y,\minus{}z$ is equivalent to multiplying with $ (\minus{}1)^6 \equal{} 1$)

Now it is obvious that $ 4(x^2\plus{}xy\plus{}y^2) \geqslant (x\plus{}y)^2 \iff (x\minus{}y)^2 \geqslant 0$ and $ x^2\plus{}xy\plus{}y^2 \geqslant 3xy \iff (x\minus{}y)^2 \geqslant 0$. And hence the inequality is true, with equality if and only if $ x\equal{}y$ or $ y\equal{}z$ or $ z\equal{}x$. (Because we assumed that $ x,y$ were positive). From this knoweledge it is "easy" to guess that $ (x\minus{}y)^2 \mid \minus{}4(xy\plus{}yz\plus{}zx)^3\minus{}27x^2y^2z^2$. And after some work we deduce that $ \minus{}4(xy\plus{}yz\plus{}zx)^3\minus{}27x^2y^2z^2 \equal{} (x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2$. (Repeatedly using $ x\plus{}y\plus{}z\equal{}0$)

Noticing that $ x\minus{}y\equal{} a\minus{}2b\plus{}c$ and so on, we put it back in and obtain:

$ \minus{}4\sum_{cyc} (a\minus{}b)^3(b\minus{}c)^3 \minus{} 15a^2b^2c^2 \equal{} (a\minus{}2b\plus{}c)^2(b\minus{}2c\plus{}a)^2(c\minus{}2a\plus{}b)^2$ and further:

$ \sum_{cyc} \frac{(a\minus{}b)(a\minus{}c)}{(b\minus{}c)^2} \minus{} \frac{15}{4} \equal{} \frac{(a\minus{}2b\plus{}c)^2(b\minus{}2c\plus{}a)^2(c\minus{}2a\plus{}b)^2}{4(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}$

Which we will post on mathlinks without any explanation at all! :P

Nice, Mathias and Guest. :lol:

There is the following generalization for $ k \ge 0$:

$ \sum \frac {(a \minus{} kb)(a \minus{} kc)}{(b \minus{} c)^2} \ge \frac {8\plus{}8k\minus{}k^2}{4}$.

[quote="Vasc"]Nice, Mathias and Guest. :lol:

There is the following generalization for $ k \ge 0$:

$ \sum \frac {(a \minus{} kb)(a \minus{} kc)}{(b \minus{} c)^2} \ge \frac {8 \plus{} 8k \minus{} k^2}{4}$.[/quote]
$ \sum \frac {(a \minus{} kb)(a \minus{} kc)}{(b \minus{} c)^2} \minus{} \frac {8 \plus{} 8k \minus{} k^2}{4} \equal{}
\frac {(2\sum a^3 \minus{}(k\plus{}2)\sum bc(b\plus{}c) \plus{}6(k\plus{}1)abc)^2}{4(b\minus{}c)^2(c\minus{}a)^2(a\minus{}b)^2}.$

[quote="Guest"][quote="Vasc"]Nice, Mathias and Guest. :lol:

There is the following generalization for $ k \ge 0$:

$ \sum \frac {(a \minus{} kb)(a \minus{} kc)}{(b \minus{} c)^2} \ge \frac {8 \plus{} 8k \minus{} k^2}{4}$.[/quote]
$ \sum \frac {(a \minus{} kb)(a \minus{} kc)}{(b \minus{} c)^2} \minus{} \frac {8 \plus{} 8k \minus{} k^2}{4} \equal{} \frac {(2\sum a^3 \minus{} (k \plus{} 2)\sum bc(b \plus{} c) \plus{} 6(k \plus{} 1)abc)^2}{4(b \minus{} c)^2(c \minus{} a)^2(a \minus{} b)^2}.$[/quote]
That is!
The following are also nice.

If $ a,b,c$ are distinct nonnegative real numbers, then

$ (a) \ \ \ \ \ \  \sum \frac {(a\plus{}3b)(a\plus{}3c)}{(b\minus{}c)^2} \ge 25;$

$ (b) \ \ \ \ \ \  \sum \frac {(a\plus{}8b)(a\plus{}8c)}{(b\minus{}c)^2} \ge 78.$

[quote="Vasc"]Nice, Mathias and Guest. :lol:

There is the following generalization for $ k \ge 0$:

$ \sum \frac {(a \minus{} kb)(a \minus{} kc)}{(b \minus{} c)^2} \ge \frac {8 \plus{} 8k \minus{} k^2}{4}$.[/quote]
We have
$ \frac {(a \minus{} kb)(a \minus{} kc)}{(b \minus{} c)^2} \plus{} \frac {k^2}{4} \equal{} \frac {(2a \minus{} kb \minus{} kc)^2}{4(b \minus{} c)^2},$
So we have to prove that
$ \sum \frac {(2a \minus{} kb \minus{} kc)^2}{(b \minus{} c)^2} \ge 8(k \plus{} 1).$
It is easy to verify that
$ \sum \frac {2a \minus{} kb \minus{} kc}{b \minus{} c}\cdot \frac {2b \minus{} kc \minus{} ka}{c \minus{} a} \equal{} \minus{} 4(k \plus{} 1),$
hence the above inequality is equivalent to
$ \sum \frac {(2a \minus{} kb \minus{} kc)^2}{(b \minus{} c)^2} \ge \minus{} 2\sum\frac {2a \minus{} kb \minus{} kc}{b \minus{} c}\cdot \frac {2b \minus{} kc \minus{} ka}{c \minus{} a},$
or
$ \left( \sum \frac {2a \minus{} kb \minus{} kc}{b \minus{} c}\right)^2 \ge 0,$
which is obviously true. :)

Nice solution, Can_hang. :lol:
Your result and Guest's result are the same. What is your proof, Guest ?

[quote="Vasc"]
The following are also nice.

If $ a,b,c$ are distinct nonnegative real numbers, then

$ (a) \ \ \ \ \ \ \sum \frac {(a \plus{} 3b)(a \plus{} 3c)}{(b \minus{} c)^2} \ge 25;$

$ (b) \ \ \ \ \ \ \sum \frac {(a \plus{} 8b)(a \plus{} 8c)}{(b \minus{} c)^2} \ge 78.$[/quote]

In addition, the inequality holds

$ (c) \ \ \ \ \ \ \sum (a\minus{}b)(a\minus{}c)(a\minus{}2b)(a\minus{}2c)(2a\minus{}b)(2a\minus{}c)\plus{}15(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2 \ge 0.$


Let p be a prime, d a positive integer, and n any integer. Prove that one can find d or
fewer dth powers whose sum is congruent to n (mod p).

Corollary 3 in http://www.mathlinks.ro/Forum/viewtopic.php?t=19496 . Anyone bored enough to try out what happens for $ p$ non-prime? (Okay, $ d\equal{}2$, $ n\equal{}3$, $ p\equal{}4$ doesn't work - but what if $ d$ and $ p$ are coprime?)

  darij

There is some b.

let z^b_1=1,y^b_2=-1(modx),and one can find b, b_1b_2|b,b/b_2 is odd.

I see the points now. I'll declare that I did something wrong in my paper and show my new thought. It's referred to a famous problem by Artin and Chevaley.


Care to explain number 2? (I suck at combinatorics).

[hide="1"]We want to choose 2 males out of 5 males, and 2 females out of 3 females.
${5\choose2}*{3\choose2}=10*3=30$[/hide]

[hide="2"]$73\%^3*27\%^2* \frac{5!}{3!*4!}\approx 28\%$
The player makes 3 free throws, and doesn't make 2, and there are $\frac{5!}{3!*4!}$ ways for the player to make 3 free thows.[/hide]

Explanation of #2:
The player must miss two throws (.27) and must make three (.73). Multiplying the respective probabilities together gives $.27 \times .27 \times .73 \times .73 \times .73$ but it can happen in several different ways, say he makes the first one, misses the second and third, and makes the fourth and fifth etc. The number of ways it can happen is $\frac{5!}{2!3!}$ or $\binom53$, which is the coefficient of $.27^2 \times .73^3$ in the binomial expansion of $(.27+.73)^5$. Finally, multiplying them together gives $\binom53 \times .27^2 \times .73^3$
To Pianoforte: are you sure it's not $\binom53$?

[quote="lkryptonicl"][quote="Zenith Xan"][quote="lkryptonicl"]1. How many different 4 person committees can be formed from a group of 5 males and 3 females, if the committe must be equally divided between males and females?

2. A basketball player makes 73% of all free throws. Assuming independence, if the player shoots five free throws during practice, what is the probablitily that she will make [b]exactly[/b] three of the free throws? [b]Round to the nearest percent.[/b]

[hide="answers"]
1. 30
2. 28
[/hide][/quote]

[hide="1."]$\binom {5} {2}\binom {3} {2}=\boxed {30}$[/hide]
[hide="2."]3rd term of $(.73+.27)^5=\binom {5} {3}.73^3.27^2=.28359=\boxed{28}$[/hide][/quote] Can you explain why you took your approach?[/quote]

I was taught to. See Hokage's explanation.


quadratics
algebra
function
domain
Ring Theory
superior algebra
superior algebra unsolved

Let K=Q(sqrt(d) (d square-free) be a quadratic field whose ring of integer O is a UFD.

Let p be a prime Z so that p|d

Prove that the principal ideal generated by p in I is the square of some ideal.

The ideal you want is generated by $ p$ and $ \sqrt{d}$.

so how would I show that

<p>= <p, sqrt(d)>^2?

: /

You can express $ p$ as a linear combination of $ p^2$ and $ d$. In the other direction, it's easy to show that the squares are divisible by $ p$.

oh ok. I did that.

I guess I need to show that <p, sqrt(d)> is a prime.

if  O/<p, sqrt(d)> is a domain then the ideal is prime. this is the part I'm having trouble with.

There's no need to bring domains into this; just use the definition of a prime ideal directly.  Multiply two elements and ask under what conditions their product can lie in this ideal.

It may be useful to consider the norm $ N(a\plus{}b\sqrt{d})\equal{}a^2\minus{}b^2d$.

Ok. I think i got it.

So, if we have two elemens, r and s.

r= a+b* (sqrt(d)) and s=(c+e*sqrt(d))

then rs=  (ac+bed) + (ae+bc) (sqrt(d))

and for this to be in the ideal generated by p and sqrt(d), then p|(ac+bed) and p| (ae+bc)

since p|d, the p|a or p|c.

If p|a then on the left side p|b so that r is factor. And it works the same way for the other side.

correct?

I don't know what "r is factor" means.  Do you mean to say $ r\in \langle p, \sqrt{d}\rangle$?  These distinctions are super-duper important.


ratio algebra polynomial induction blogs vector function

Prove Binet's Formula. (i.e. Find the formula for the Fibonacci Numbers)

What do you mean the formula for the Fibonacci numbers?  The golden ratio between them?

Prove that $ F_n \equal{} \frac1{\sqrt5}\left(\left(\frac {1 \plus{} \sqrt5}2\right)^n \minus{} \left(\frac {1 \minus{} \sqrt5}2\right)^n\right)$, for all integers $ n\in\mathbb{Z}^ \plus{}$.

Is this supposed to be related to the [url=http://www.artofproblemsolving.com/Wiki/index.php/Binet%27s_formula]WoW?[/url] [i]Edit[/i]: an explanation (by me, lol) similar to nateharman1234's is at that link

Since $ F_{n \plus{} 1} \equal{} F_{n} \plus{} F_{n \minus{} 1}$, we have a linear recurrence with characteristic polynomial $ x^2 \minus{} x \minus{} 1 \equal{} 0 \equal{} \left(x \minus{} \frac {1 \plus{} \sqrt {5}}{2}\right)\left(x \minus{} \frac {1 \minus{} \sqrt {5}}{2}\right)$. We know that $ F_0 \equal{} 0$, $ F_1 \equal{} 1$, and by the characteristic we have $ F_n \equal{} c_1 \left(\frac {1 \plus{} \sqrt {5}}{2}\right)^n \plus{} c_2 \left(\frac {1 \minus{} \sqrt {5}}{2}\right)^n$. So now we have a system of two equations (if we let $ n \equal{} 0,1$), $ 0 \equal{} c_1 \plus{} c_2 \Longrightarrow c_2 \equal{} \minus{} c_1$ and $ 1 \equal{} c_1\left(\frac {1 \plus{} \sqrt {5}}{2}\right) \plus{} c_2\left(\frac {1 \minus{} \sqrt {5}}{2}\right)$ $ \Longrightarrow 1 \equal{} c_1\left(\frac {1 \plus{} \sqrt {5}}{2} \minus{} \frac {1 \minus{} \sqrt {5}}{2}\right)$ $ \Longrightarrow c_1 \equal{} \frac {1}{\sqrt {5}}$.

Thus, $ F_n \equal{} \frac {1}{\sqrt {5}}\left(\frac {1 \plus{} \sqrt {5}}{2}\right)^n \minus{} \frac {1}{\sqrt {5}}\left(\frac {1 \minus{} \sqrt {5}}{2}\right)^n$.

Many people do the induction proof of this at some point in their life, and while it may be a good induction exercise, it gives no insight into why this works. Heres a proof I know, and for another you can read one of t0rajir0u's recent blog posts.

[hide]

Consider the set S of sequences of the form An= A(n-1)+A(n-2), A0= x and A1 = y for real x,y (real works here, but for different sequences you may need complex)

Note that if a sequence An is in S and a sequence Bn is in S, then (An+Bn) is in S.

Also note that this addition of sequences is associative and commutative and all that nice stuff as real additon is

Oh and of course if k is a real number and An is in S than so is the sequence  k*An, and this multiplication distributes over the sequence addition I talked about before. and note that 1*An=An

Ok then note the sequence An=0 is in S and for all An in S -An is in S too.

Well if you haven't realized already I am trying to establish that S forms a vector space, I think thats just about everything if I forgot an axiom I appologize.

Anyway, now that we have a nice vector space it is of course nice to find a basis for it. Well i hope it is painfully obvious that S is two dimensional.

Ok now lets see if there are any nicely behaved sequences in S that have a closed form, perhaps any geometric sequences...

Solve X^n = X^(n-1)+X^(n-2), x =/= 0...
Ooh... two linearly independent geometric sequences, that would make a lovely basis, now to just express the fibbonacci numbers in terms of them...

I'll stop there, because I don't feel like doing any of the actual computations.
[/hide]

The proof nateharman is talking about can be found [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=175257]here[/url].

[hide="Another solution by generating functions"] Let $ G(x) \equal{} \sum_{k\equal{}0}^{\infty} F_k x^k$, where $ F_0 \equal{} 0, F_1 \equal{} 1$.  The Fibonacci relation implies (where $ \phi, \varphi$ are the positive and negative roots of $ x^2 \equal{} x \plus{} 1$):

$ x G(x) \plus{} x^2 G(x) \equal{} G(x) \minus{} x \Leftrightarrow$
$ G(x) \equal{} \frac{x}{1 \minus{} x \minus{} x^2} \equal{} \frac{x}{(1 \minus{} \phi x)(1 \minus{} \varphi x)} \equal{} \frac{1}{\phi \minus{} \varphi} \left( \frac{1}{1 \minus{} \phi x} \minus{} \frac{1}{1 \minus{} \varphi x} \right)$

(The last step is partial fraction decomposition.)  Using the expansion $ \frac{1}{1 \minus{} ax} \equal{} 1 \plus{} ax \plus{} a^2 x^2 \plus{} ...$, we conclude that

$ \sum_{k\equal{}0}^{\infty} F_k x^k \equal{} \frac{1}{\phi \minus{} \varphi} \left( \sum_{k\equal{}0}^{\infty} (\phi^k \minus{} \varphi^k) x^k \right) \Leftrightarrow$
$ F_k \equal{} \frac{\phi^k \minus{} \varphi^k}{\phi \minus{} \varphi}$

as desired (this is the nice form of Binet's formula). [/hide]

Let $ \tau \equal{} \frac {(1\plus{}\sqrt{5})}{2}$ and $ \sigma \equal{} \frac {1\minus{}\sqrt{5}}{2}$

Since $ \tau$ and $ \sigma$ are the roots of $ x^2\equal{}x\plus{}1$ then $ \tau^2\equal{}\tau\plus{}1$ and similarly with $ \sigma$.

Now multiplying both by their corresponding n-powers, $ \tau^{n\plus{}2} \equal{} \tau^{n\plus{}1}\plus{}\tau^n$, and $ \sigma^{n\plus{}2} \equal{} \sigma^{n\plus{}1}\plus{}\sigma^n$

Subtracting the first from the latter yeilds $ \tau^{n\plus{}2}\minus{}\sigma^{n\plus{}2}\equal{} \tau^{n\plus{}1}\plus{}\tau^n\minus{}\sigma^{n\plus{}1}\minus{}\sigma^n$

so $ (\tau^{n\plus{}2}\minus{}\sigma^{n\plus{}2})\equal{} (\tau^{n\plus{}1}\minus{}\sigma^{n\plus{}1})\plus{}(\tau^n\minus{}\sigma^n)$.

Since $ \tau\minus{}\sigma \equal{} \sqrt{5}$, and $ \tau^2\minus{}\sigma^2\equal{}\sqrt{5}$ we must divide by $ \tau\minus{}\sigma$

so $ \frac {\tau^{n\plus{}2}\minus{}\sigma^{n\plus{}2}}{\tau\minus{}\sigma}\equal{} \frac{\tau^{n\plus{}1}\minus{}\sigma^{n\plus{}1}}{\tau\minus{}\sigma}\plus{}\frac{\tau^n\minus{}\sigma^n}{\tau\minus{}\sigma}$.

If you would like it in your form with the denomiator being the radical,

so $ \frac {\tau^{n\plus{}2}\minus{}\sigma^{n\plus{}2}}{\sqrt{5}}\equal{} \frac{\tau^{n\plus{}1}\minus{}\sigma^{n\plus{}1}}{\sqrt{5}}\plus{}\frac{\tau^n\minus{}\sigma^n}{\sqrt{5}}$.

Let $ u_n \equal{} \frac{\tau^n\minus{}\sigma^n}{\tau\minus{}\sigma}$, then the sequence is found as $ u_{n\plus{}2}\equal{}u_{n\plus{}1}\plus{}u{n}$

and this completes the proof.


when $ \text{1,3,5,5\minus{}tetramethyl cyclohexa\minus{}1,3\minus{}diene}$ is disssolved in cold conc. $ \text{H}_2\text{SO}_4$ then a lowering of Freezing point is observed that correspond to 2 particles for each molecule of diene.When the solution is treated with water the diene is regenerated.Describe what happens with relevant equations

The part "that correspond to 2 particles for each molecule of diene" means that for each molecule of diene that reacts with H2SO4, two species are formed. Well, the diene has double bonds and H2SO4 is an acid: what do you think that might happen?

[hide="Answer"]A salt of the form cyclohexadiene+HSO4- is formed.[/hide]

[b]Question[/b]: what would happen if the 1,3,5,5-tetramethylcyclohexa-1,3-diene had a C=O bond in carbon 6 and were treated with sulphuric acid?

i got the salt but couldn't know how on adding water that get's converted to diene

In the salt formation reaction, the diene acts as a base, accepting an H+ ion from sulphuric acid. When the solution is treated with water, the diene acts as an acid with a molecule of water acting as a base, regenerating the diene. These are simply acid-base reactions.


algebra
polynomial
induction
algebra unsolved

Any polynomial $g(x)$ of degree d or less satisfies the recursion

$\sum_{i=0}^{d+1}(-1)^i\binom{d+1}{i}g(x+i)=0$

Can someone explain this to me?

It sufficies to prove the identity for $g(x)=x^k$, $k\leq d$.
We will use an induction.
1. For $d=1$ the assertion is true.
2. We have \[\sum_{i=0}^d(-1)^i\binom{d+1}{i}(x+i)^k=\sum_{i=0}^d(-1)^i\binom{d+1}{i}(x+i)^{k-1}(x+i)=\] \[=x\cdot \sum_{i=0}^d(-1)^i\binom{d+1}{i}(x+i)^{k-1}+\sum_{i=0}^d(-1)^ii\binom{d+1}{i}(x+i)^{k-1}.\]
We know that $\sum_{i=0}^d(-1)^i\binom{d+1}{i}(x+i)^{k-1}=0$ using induction proposition for the pair $(d,k-1)$. Moreover, \[\sum_{i=0}^d(-1)^ii\binom{d+1}{i}(x+i)^{k-1}=\sum_{i=0}^{d-1}(d+1)(-1)^i\binom{d}{i}((x+1)+i)^{k-1}=0\] using induction proposition for pair $(d-1,k-1)$.

Or like this... For polynomial $g(x)$ let $\Delta g(x)=g(x+1)-g(x)$. Observe that $\Delta g(x)$ is polynomial which degree is smaller than degree of $g(x)$. Also let $\Delta_2 g(x)=\Delta(\Delta g(x))$ and $\Delta_n g(x)=\Delta (\Delta_{n-1} g(x) )$. It is very easy to proove (by induction) that $\Delta_n g(x)=\sum_{i=1}^n (-1)^n {n \choose i}g(x+i)$. We need to proove that $\Delta_{d+1} g(x) =0$. But, this is now trivial, because $\deg \Delta g(x) \leq d-1$, $\deg \Delta_2 g(x) \leq d-2$, and so on, and $\deg \Delta_d g(x) =0$ so it is constant and that is why $\Delta_{d+1} g(x) =0$

How do you prove $\Delta_n g(x)=\sum_{i=1}^n (-1)^n {n \choose i}g(x+i)?$

For n=1 wouldn't you have $\Delta g(x)=-g(x+1)?$

[quote="pilot"]If you're inducting on d how can you assume the pair (d,k-1) is true?[/quote]
Sorry, I forgot to say that we use the double induction on pairs $(d,k)$ and for $k=0$ we get the well known equality $\sum_{i=0}^{d+1}(-1)^i\binom{d+1}{i}=0$.

I typed it wrong. It should be $\Delta_n g(x)=\sum_{i=0}^n (-1)^{n-i} {n \choose i}g(x+i)$. For $n=1$ it becomes $\Delta g(x) = g(x+1)-g(x)$, for $n=2$ it should be $\Delta_2 g(x)=(g(x+2)-g(x+1))-(g(x+1)-g(x))=g(x+2)-2g(x+1)+g(x)$ and so on. If it's true for $n$ then $\Delta_{n+1} g(x) = \Delta_n g(x+1) - \Delta_n g(x)=\sum_{i=0}^n (-1)^{n-i} {n \choose i}g(x+1+i) - \sum_{i=0}^n (-1)^{n-i} {n \choose i}g(x+i)=g(x+n+1)-g(x+n)({n \choose n-1}+{n \choose n})+g(x+n-1)({n \choose n-2}+{n \choose n-1})+\cdots =\sum_{i=0}^{n+1} (-1)^{n+1-i} {n+1 \choose i}g(x+i)$.

I hope I didn't make any mistake again...


Quadrilateral ABCD is inscribed in a circle with diameter AD=4.  If sides AB and BC each have length 1, then find CD.

[hide="Solution"]By Ptolemy's Theorem, $ AB \cdot CD \plus{} BC \cdot AD \equal{} CA \cdot BD$, we get
$ 4 \plus{} CD \equal{} CA \cdot BD$.
Because $ AD$ is the diameter, we know that triangles $ ABD$ and $ DCA$ are right. We then use the Pythagorean Theorem to find the values of CA and BD.
$ 4 \plus{} CD \equal{} \sqrt {15} \cdot \sqrt {16 \minus{} CD}$.
This is just a quadratic, so we solve and get $ CD \equal{} \frac {7}{2}$.[/hide]

[hide="solution"][img]http://i.imgur.com/063bPwZ.png[/img]
As $AB=BC=1$, $ \overarc{AB} = \overarc{BC}  \Rightarrow \angle{BDA}=\angle{BDC}=\frac{1}{2}\angle{CDA}$
$\frac{CD}{AD}=\cos(CDA)=\cos(2BDA)=\cos^{2}(BDA)-\sin^{2}(BDA)=1-2\sin^{2}(BDA)=1-2(\frac{AB}{AD})^2$
$\Rightarrow CD=\frac{7}{2}$[/hide]

[hide=Nice problem!] We use Pot Theorem to get AB*CD+BC*AD=CA*BD, therefore 4+CD=CA*BD. Because AD is diameter, we know triangles ABD and DCA are right. By Pythagorean Theorem, to find CA and BD. 4+CD=root(15)*root(16-CD). This is just a quadratic, so we solve and get CD=7/2. [/hide]

Oh no! Isn't this the same thing as sup3rcrash3r's solution? Oh well post count inflation is fun, so I won't delete.

It should be 4+CD=root(15)*root(16-CD^2).


can anyone give an exact number of how many AoPS users participated at nationals?
There seems like an awful lot.

[quote="236factorial"]can anyone give an exact number of how many AoPS users participated at nationals?[/quote]No, I don't think anyone can give you an [b]exact[/b] number :)

AoPS is getting really big. There were a lot of AoPS shirts (I saw about 7), plus you shouldn't forget the people like me who don't have AoPS shirts. I would say there were a good 20 there. Then again, I really have no sense for how many there are. Maybe more, maybe less.

too many. Two off my team mates are here on aops and so is (was) my coach. then one of them (Secret Asian) went around asking everyone for their usrn.

Hey!  Not everyone....er....ok yes everyone.  I just wanted to see who was who.  It was mathgeek who decided to go around and also tell people his username.

mathgeek? All he ever do is jump around in the halls and hit you. And through that you say he found time to ask people for their usrn??

I said Secret Asian cause I remebered at that moment you asking kingof21penguins.

Ok, I asked:  New Jersey(which one is treething/aopser), Entropywins(but he was wearing an aops shirt, hence me asking) the whole reason I started asking texas was because asdf4(something like that, he's from tenessee) asked [b]me[/b] if I was an aopser...I asked more people than that, but I can't remember all of them

The only people I saw were Naga, Xantos. C. Guin, and Solafidefarms. :(  :(  :(

You didn't see any of the TX or CA teams?

[quote="solafidefarms"]I saw jb05, xantos, xxreddevilxx, zepelinator, smiley, smartnerd666, treething, biffandoc, 09husbri, entropywins, Hexahedron, mathgeek2006, Secret Asian, Dnas(though I didn't know it was him), kchande, aznness, bob123, piggypi, cutiepi, nosoupforyou, falconwing64, shlomadapenguin, jaeminlee, natedawg, mdean09, topper(who was at Word Power nats too), rorrimimage, jasonho007, dengmi, MooGoesCow?, RC-7th, and seungsoo. Oh, yes, and rcv, naga, frost13, and rrusczyk for the adult AoPSersAt least those are the ones I know the usernames for. 35. I didn't meet asdf4 from Tennessee, or if I did, I didn't ask if he was a AoPSer. Would someone please PM me his real name?

Billy[/quote]

Holy Monkey!!

That's even more than I saw. Can you email me that list of everyone's emails and AoPS usernames?

It was nice meeting people who I knew only by username before.  Now I can see some faces while reading posts.

Ohh, yes, Monkey, and Kingof21 penguins. Did I forget anymore???

WOW 35 AoPSers including 3 adults
I know mdean09 he's at my school. Doesn't post much though. (sry dean)
Let's see...
57*5=285
285/38=7.5%!!!
plus the undoubtable 10 poeple we missed or so,
we have about 10%!!!!!!!!!
Woohoo!
~1 out of 10 national people was an AoPSer!!!! :D  :D  :D  :D

Billy - since you do have my e-mail, could I just request that you not give it out to anybody unless you know for a fact that they were at nationals? Thanks!

And everybody else, if you want my e-mail and you fit the above prerequisite, Billy has it!

[quote="solafidefarms"]I saw jb05, xantos, xxreddevilxx, zepelinator, smiley, smartnerd666, treething, biffandoc, 09husbri, entropywins, Hexahedron, mathgeek2006, Secret Asian, Dnas(though I didn't know it was him), kchande, aznness, bob123, piggypi, cutiepi, nosoupforyou, falconwing64, shlomadapenguin, jaeminlee, natedawg, mdean09, topper(who was at Word Power nats too), rorrimimage, jasonho007, dengmi, MooGoesCow?, RC-7th, and seungsoo. Oh, yes, and rcv, naga, frost13, and rrusczyk for the adult AoPSersAt least those are the ones I know the usernames for. 35. I didn't meet asdf4 from Tennessee, or if I did, I didn't ask if he was a AoPSer. Would someone please PM me his real name?

Billy[/quote]

You saw me too, remember at Joe DUmars

OK OK, I forgot some people. ... Does that make 36 now?

I think you saw me... although I was being swarmed by people who wanted that extremely rare and valuable CT pin.

[quote="Treething"]Yeah, a lot of people were running up to me asking "Are you Treething?"

I'M A CELEBRITY :P ![/quote]

...I just wanted to know.  Basically my rules were if you were wearing an Aops shirt you were fair game...or if I knew where you were from.  I think Asod and mathgeek told people who I was though.  I wanted to stay "Secret"

hmmm. I've been thinking, and I don't think that I saw xxreddevilzxx, since he wasn't from Washington. hmmmm. Sorry!


These question(s) are from [i]Gravity from the Ground up[/i] - I want to know if my answer(s) are in fact solutions.

The first question is:  Suppose that at the $ n^{th}$ time-step $ t_n$, the vertical speed is $ v_n$ and the vertical distance above the ground is $ h_n$.  Show that at the next time-step $ t_{n\plus{}1}$ $ \equal{}$ $ t_n$ $ \plus{}$ $ \Delta$$ t$, the vertical speed is $ v_{n\plus{}1}$ $ \equal{}$ $ v_n$ $ \minus{}$ $ g$ $ \Delta$ $ t$.

My answer:

$ v_{n\plus{}1}$ = $ v_n$ $ \plus{}$ $ v_1$

$ v_1 \equal{} at$

In this case $ a \equal{} g$ (gravity)

So $ v_n$ + $ v_1$ $ \equal{}$ $ v_n$ $ \minus{}g$ $ \Delta$ $ t$

I'm not sure why, but my answer doesn't seem complete to me.

I'll post the rest of the questions/my answers tomorrow :)


If the base $8$ representation of a perfect square is $ab3c$, where $a\ne 0$, then $c$ equals

$\textbf{(A) } 0\qquad \textbf{(B) }1 \qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } \text{not uniquely determined}$

[hide="NT Solution"]Recall that a perfect square must be  equivilant to $0$, $1$, or $4$ mod $8$. Thus, c=0,1, or 4.

$c=0$ is impossible because we're left with $8(8^2a+8b+3)$, which cannot be a square because $8$ isn't a square.

If $c=4$, we are left with $4(8N +7)$, where N is $4\cdot 8^2a+32b$. This is once again impossible, because numbers of the form $8N+7$ cannot be a perfect square. (8 isn't a quadratic residue mod 7)

Thus, the answer is $n=1$
[/hide]

[quote="4everwise"]If the base $8$ representation of a perfect square is $ab3c$, where $a\ngeq 0$, then $c$ equals

$\text{(A)} 0\qquad \text{(B)}1 \qquad \text{(C)} 3\qquad \text{(D)} 4\qquad \text{(E)} \text{not uniquely determined}$[/quote]

you might want to edit the

$a0$, $\$$a\neeq 0$\$$ to
$a\ne 0$, $\$$a\ne 0$\$$ i did not notice this until i looked at the code

anyway, your solution is good, i was thinking mod 4, but that is not as easy, but also 25 is 31 base 8 so 1, answers are never "not determined"...


There are five friends. Josh says that his birth date is 31st dec 1878. Rick says that his birth date is 14th nov 1902, Derik says his birth date is 12th july 1898, Usha says her birth date is 29th feb 1900 and Mike says his birth date is 15th May 1904.In which year will be the sum of their ages is equal to a prime number between 175 and 240.

Considering that this is probably homework, I'll just give you a hint.

[hide]Start from the beginning of 1905 and note that each year the sum of the ages increase by 5.

Btw, I think there might be more than one solution, but I'll have to check that. [/hide]

[quote="1=2"]Never; February 29 1900 did not happen because 1900 is not a leap year.[/quote]

Uhh...... yes it is. Any year divisible by 4 is a leap year...

[quote="hsiljak"]Except for those divisible with 100 and not divisible with 400.

[quote="Wikipedia"]However, some exceptions to this rule are required since the duration of a solar year is slightly less than 365.25 days. Years that are evenly divisible by 100 are not leap years, unless they are also evenly divisible by 400, in which case they are leap years.[1][2] For example, 1600 and 2000 were leap years, but 1700, 1800 and 1900 were not. Similarly, 2100, 2200, 2300, 2500, 2600, 2700, 2900, and 3000 will not be leap years, but 2400 and 2800 will be. By this rule, the average number of days per year will be 365 + 1/4 − 1/100 + 1/400 = 365.2425, which is 365 days, 5 hours, 49 minutes, and 12 seconds.[/quote][/quote]

Oh, yes. You are right. I've never heard that before.... I wonder why....

There are four solutions that I found, 1931, 1935, 1941, and 1943


trigonometry
complex analysis
function
logarithms
real analysis
real analysis unsolved

Let $ \bold{f}: \mathbb{R}^2\to\mathbb{R}^2$ be defined by $ \bold{f}(x,y)\equal{}(e^x\cos y, e^x\sin y)$
Show that $ \bold{f}$ is injective on the strip $ \{(x,y): \minus{}\pi<y<\pi\}.$

$ f$ is the holomorphic function $ \mathbb{C} \to \mathbb{C}$ given by $ f(z) \equal{} e^z$.  It defines a conformal map from the given strip to $ \mathbb{C} \minus{} i \mathbb{R}_{\ge 0}$, and the logarithm makes sense here.

Maybe more direct -- say $ f(x_1,y_1)\equal{}f(x_2,y_2)$.  Then $ e^{x_1}\equal{}\|f(x_1,y_1)\|\equal{}\|f(x_2,y_2)\|\equal{}e^{x_2}$, so $ x_1\equal{}x_2$.  Since $ y_1,y_2\in (\minus{}\pi,\pi)$, $ y_1\equal{}y_2$ as well.  Hence $ f$ is injective.


What level do you think these books are:

mathematical olympiad treasures
and mathematical olympiad challenges.

Olympiad...  I feel kinda redundant saying this since the word ''olympiad'' is in their titles...but that's what they are.


Er...I assume this is for some homework, but all it's saying is basically

$ I_x\equal{}3I_N$
$ \log(I_X)\equal{}\log(I_N)\plus{}\log(3)\equal{}5\plus{}\log(3)$

This should be moved to the appropriate forum.


inequalities
triangle inequality
inequalities proposed

But
$ \sqrt {r_aw_a}(\frac {1}{w_b} \plus{} \frac {1}{w_c}) \geq 2$
is easy.
BQ


This isn't math-related problem I supposed. Anyway, imagine you have tons of mails with your addresses, sin#, all sorts of stuffs that you don't want other people to know. Obviously, you don't want to just dump them in a recycling bin due to the possibility of identity theft. And buying a shredder is expensive.

I was thinking about soaking all those mails into water in order to make them easier to destroy. Maybe I could add some colourful ink to mess up all the writing in the mails. Unfortunately I haven't taken any chemistry, so I don't know which ink to use.

I hope somebody could give me a suggestion. Hopefully I am posting in the right forum.

You are posting them in the right forum, or that's what I think, and your idea of adding ink seemed good.

Eat them. But seriously, you could just burn it.

there was an article once, in the Atlantic Monthly I think on the topic of destroying information and failing to do so(not really related to what you want) that was amusing and may also have had information on how to properly dispose of stuff if anyone remembers the article maybe they could describe it better.

I would suggest tearing them to pieces with your teeth. Not only will this mimic the effect of a conventional shredder, your saliva will both partially disintegrate the paper and smudge the ink.

since people are describing the best ways to destroy the documents i recommend feeding them to a sibling or letting (significantly) younger siblings shred them and then feed them to (older more annoying) sibling.  This may be biased by the fact that i do not like the taste of paper maybe others do and want to eat themselves.  alternatively if you dont have siblings you could cover the paper in birdseed you can let a flock of pigeons or crows do the dirty work.

is a regular shreder not the best option?

[quote="dogseatcheese"]Umm....why spend money to buy a shredder when SCISSORS work just as well....or you can flush it down the toilet.[/quote]
are you crazy?!?!
1) do you know how long it would take to cut up a bunch of sheets of paper?
2) you can get a shreder for like 20 dollars
3) flush it? do you want to back up your pipes? or maybe that was after you have cut it up into small peices w/ your scissors.

Matches are cheap.  I got 1,500 of them at Wal-Mart for $\$$3.  They lasted me almost two years!

Burning them would have to happen outside, and that can get nasty if it's a lot of paper

In Belgium you can't burn stuff less than hundred meters (about thirty feet) from houses.

I would suggest buying a good shredder.  A shredder is an INVESTMENT.  You'll be sorry after your identity gets stolen.

make information soup. ya know, make soup with it. >_>

hamsters are extremely handy if you spilled food, just let them walk over it and it's gone!
but i think it's really bad to feed paper to a hamster :(

i was thinking of lining the cage, not feeding paper to it.

tear it apart with your hands.

Just fold it up and tear, repeat step until disposes of properly. If you are that worried, burning it is the best way to go.

[quote="Gyan"]Only sure way is to hurl it into a black hole. All information will  be lost.[/quote]

Didn't Stephen Hawking just lose a bet (a set of encyclopaedias on baseball, I believe) because he decided that information is actually not lost?

[quote]Didn't Stephen Hawking just lose a bet (a set of encyclopaedias on baseball, I believe) because he decided that information is actually not lost?[/quote]
Yes, that is true. (to be honest, I was almost sure that some would point this out, when I posted the message .. I was just curious, how long willl it take :)

Also in 1978, before US  embassy was  taken over , embassy employees  were able to  shred  most of the  secret documents (with professional shredders)  , but Iranians were able to put   most of the documents back.. the information leaked thus, according to reports,  was very significant.

What you do, is you get a professional shredder. Then you take the shredded paper and use it to line the cage of your rabbit/hamster/guinea pig. See? Then it's professionally nastied!

Well...I it isn't a lot of items....the SCISSORS. if a lot....SHREDDER. if some....HAMSTER

Their shredders cut the documents into 1/4 inch strips.  For greater security use the cross cut type or the type that cuts the paper into strips that are about as wide as they are long.

Does anyone else remember that one of the soldiers helped the Iranians to piece together the documents, as well as telling them which documents were most important then sued when he didn't get a medal, like the other soldiers.

You can shred the papers split them up into 4 piles and travel to the four corners of the earth and hide them that should be good enough. :)

Fast - run over important information with a permanent marker or whiteboard marker. If you really want to be secure, run over the corresponding area on the other side of the paper too. Noone will see.

Just cremate all the papers in your fireplace.  :D

[quote="cheerful coffin"]first of all, thank you all for the suggestions.

It looks like for paper, shredder would be the best option since it is fast (I'm not gonna spend a whole day tearing paper), cheap (hamster and whiteboard pen costs a fortune), and clean (burning is dirty + toxic too).

Though I admit, I am quite phobic towards shredder after reading about some disturbing accidents in the past, which I would rather not talk about it.

Even though I am really annoyed by the birds' regular fly-by, I don't want to feed those poor things with paper... yet.[/quote]
Eh?  how would a whiteboard marker cost more than a shredder?  Personally, I think burning is the best idea.  Do it in the sink. And have a pail of water on standby.

[quote="nr1337"]Thermite ;)

But that would count as burning, wouldn't it....[/quote]

I made thermite thats stuff is pretty scary (but really cool) but it would be effective.

Hmmm...If it's not too many pages, burning it seems best. Use a bunsen burner...advancement of science...

[quote="dogseatcheese"]Umm....why spend money to buy a shredder when SCISSORS work just as well....or you can flush it down the toilet.[/quote]

Because a shredder is so much fun!