Hello

Find the nth term of the sequence $ {a_{n}}$ such that $ a_{1}\equal{}1$, $ a_{2}\equal{}3$, $ a_{n\plus{}1}\minus{}3a_{n}\plus{}2a_{n\minus{}1}\equal{}2^{n}$.

I know we have to use the homogeneous solution somewhere.

[hide="Hint"] Let $ a_k \equal{} b_k \plus{} Ck \cdot 2^k$ where $ b_{k \plus{} 1} \equal{} 3b_k \minus{} 2b_{k \minus{} 1}$ and solve for $ C$. [/hide]

You can make it a normal one by noticing that $ 2^{n\plus{}1}\equal{}2\cdot 2^n$.

[quote="t0rajir0u"]The motivation for the important step (finding the nonhomogeneous solution) comes more from series considerations (that is, from generating functions) than anything else.[/quote]
What do you mean by the homogenous/nonhomogenous solution?  I know my motivation was pretty random - can you explain the generating functions approach? (not the actual generating functions solution, but how generating functions motivate the right substitution)

This is a term I've borrowed from the study of differential equations (the connection is very strong).  The [i]homogeneous[/i] solution is the solution $ h_k$ to the equation $ h_{k \plus{} 1} \minus{} 3h_k \plus{} 2h_{k \minus{} 1} \equal{} 0$.  We call it homogeneous because the terms are all of the same degree, so if $ h_k$ is a solution then so is $ c \cdot h_k$ for any $ c$.  Moreover, if $ a_k$ is a recurrence satisfying our original recurrence, then so is $ a_k \plus{} c \cdot h_k$.

The full solution to this recurrence then takes the form $ a_k \equal{} c \cdot h_k \plus{} n_k$, where $ n_k$ is called the [i]nonhomogeneous[/i] solution.  In particular, $ c \cdot n_k$ is not a solution unless $ c \equal{} 1$.

Motivation for this terminology:  multiply the recurrence by $ \frac {x^n}{n!}$ and sum across all $ n$.  Let $ A(x) \equal{} \sum_{k \equal{} 0}^{\infty} a_n \frac {x^n}{n!}$ be the [i]exponential generating function[/i] of the sequence $ a_n$.  The recurrence then becomes the differential equation

$ A'' \minus{} 3A' \plus{} 2A \equal{} 2e^{2x}$

and the methods for solving this equation are identical to the method above :)

So much for the exponential generating function.  It's actually easier to solve this problem using the [i]ordinary generating function[/i]

$ B(x) \equal{} \sum_{k \equal{} 0}^{\infty} a_n x^n$,

and in fact the nonhomogeneous solution is a natural consequence of this approach.  For more details, consult [url=http://www.math.upenn.edu/~wilf/DownldGF.html]generatingfunctionology[/url].

Is it not much easier to just do $ a_{n\plus{}1}\minus{}3a_n\plus{}2a_{n\minus{}1}\equal{}2(a_n\minus{}3a_{n\minus{}1}\plus{}2a_{n\minus{}2})$?

Well, yes, but we then have the cubic characteristic equation $ (x \minus{} 1)(x \minus{} 2)^2$, and the above methods I've given are (as far as I know) the standard way of handling a characteristic equation with roots of multiplicity greater than $ 1$.

Thanks t0rajir0u, I'll try to solve the problem with generating functions.

boxedexe, I think altheman was referring to the same thing.

My apologies for not taking a closer look at what people posted.  :blush:

[quote="matt276eagles"][hide="Solution"]Let $ b_{n} \equal{} a_{n} \plus{} n\cdot 2^{n}$.  Plugging this into the given equation, we get

$ b_{n \plus{} 1} \plus{} (n \plus{} 1)\cdot 2^{n \plus{} 1} \minus{} 3(b_{n} \plus{} n\cdot 2^{n}) \plus{} 2(b_{n \minus{} 1} \plus{} (n \minus{} 1)\cdot 2^{n \minus{} 1}) \equal{} 2^{n}$
$ b_{n \plus{} 1} \minus{} 3b_{n} \plus{} 2b_{n \minus{} 1} \plus{} (2n \plus{} 2)\cdot 2^{n} \minus{} 3n\cdot 2^{n} \plus{} (n \minus{} 1)\cdot 2^{n} \equal{} 2^{n}$
$ b_{n \plus{} 1} \minus{} 3b_{n} \plus{} 2b_{n \minus{} 1} \equal{} 0$

Now we can use characteristic equations.  We want the roots of $ x^{2} \minus{} 3x \plus{} 2 \equal{} 0$, which are 1 and 2.  Therefore, $ b_{n} \equal{} A\cdot 1^{n} \plus{} B\cdot 2^{n}$.  From our definition of $ b_{n}$, we have $ b_{1} \equal{} a_{1} \plus{} 2 \equal{} 3$ and $ b_{2} \equal{} a_{2} \plus{} 8 \equal{} 11$.  Therefore, $ A \plus{} 2B \equal{} 3$ and $ A \plus{} 4B \equal{} 11$.  Solving this, we get $ A \equal{} \minus{} 5$ and $ B \equal{} 4$.  Therefore, $ b_{n} \equal{} 2^{n \plus{} 2} \minus{} 5$, and $ a_{n} \equal{} 2^{n \plus{} 2} \minus{} n\cdot 2^{n} \minus{} 5$.

Motivation: We want to make a substitution to get rid of the $ 2^{n}$ in the given equation.  This is really a trial and error process, but one substitution that seems good to try is $ b_{n} \equal{} a_{n} \plus{} 2^{n}$.  All the powers of 2 cancel out, and we're left with the original recursion with $ b$'s instead of $ a$'s.  At this point we know we're pretty close.  Switching the $ n$ in $ 2^{n}$ with any $ n \plus{} k$ gives us the same result, but $ b_{n} \equal{} a_{n} \plus{} n\cdot 2^{n}$ happens to work.[/hide][/quote]

I think you might have made a mistake, for if you put $ n\equal{}3$, we get that $ a_{3}\equal{}3$, but it is really $ 8$.

The actual function if we do it your way, comes out to be $ a_{n}\equal{}2^{n}*n\minus{}2^{n\plus{}1}\plus{}3$

[quote="mathgeniuse^ln(x)"][color=darkred]Find the $ n^{th}$ term of the sequence $ {a_{n}}$ such that

$ a_{1} \equal{} 1$, $ a_{2} \equal{} 3$, $ a_{n \plus{} 1} \minus{} 3a_{n} \plus{} 2a_{n \minus{} 1} \equal{} 2^{n}$.[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Observe that for any $ n\in \mathcal N^*$ , $ a_{n \plus{} 1} \minus{} 3a_{n} \plus{} 2a_{n \minus{} 1} \equal{} 2^{n}$ $ \Longleftrightarrow$

$ \left(a_{n \plus{} 1} \minus{} a_n\right) \minus{} 2\left(a_n \minus{} a_{n \minus{} 1}\right) \equal{} 2^n$ $ \Longleftrightarrow$ $ \frac {a_{n \plus{} 1} \minus{} a_n}{2^n} \minus{} \frac {a_n \minus{} a_{n \minus{} 1}}{2^{n \minus{} 1}} \equal{} 1$ .

Denote for any $ n\in \mathcal N^*$ , $ b_n \equal{} \frac {a_{n \plus{} 1} \minus{} a_{n}}{2^n}$ , where $ b_1 \equal{} 1$ . Thus,

$ b_1 \equal{} 1$ and for any $ n\ge 2$ , $ b_n \minus{} b_{n \minus{} 1} \equal{} 1$ , i.e $ b_n \equal{} n$. Therefore,

for any $ n\in\mathcal N^*$ , $ a_{n \plus{} 1} \minus{} a_n \equal{} n\cdot 2^n$ , i.e. $ a_n \equal{} 1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}\left(k\cdot 2^k\right)$ .

You can ascertain easy that $ \sum_{k \equal{} 1}^{n }\left(k\cdot 2^k\right)\equal{}n\cdot 2^{n\plus{}2}\minus{}(n\plus{}1)\cdot 2^{n\plus{}1}\plus{}2$  .

In conclusion, for any $ n\in\mathcal N^*$ , $ \boxed {\ a_n\equal{}(n\minus{}1)\cdot 2^{n\plus{}1}\minus{}n\cdot 2^n\plus{}3\ }$ . [/color]


Jerry march tells that the base attacks the $ \alpha$ hydrogen in an unsymmetrical ketone at the side which is more in number(or atleast that's what i interpreted that to be)
Peter Sykes says the $ \ce{H}$ is lost from the site which forms the most stable enolate ion that is the alkene taht is most conjugated.
which of the following is ight
[hide="my opinion"]
i would beleive what peter sykes say...though probabilistically that's less but the stability factor clearly very gr8ly overweighs the other factors...what is ur opinion[/hide]

This issue was already discussed in another topic: enolate formation is a reaction that depends on base strenght, temperature, the counter ion of the base, solvent, and reaction time. So both autors can be correct, depending on reaction conditions.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73166]Look here[/url] (you can restart the discussion there).


When did we learn the agricolem's euclid or whatever it's called?

o.O

Did you mean the Euclidean Theorem?
(and would this go in the Introduction to Number Theory board, because you're talking about the class?)

The Euclidean Algorithm is taught in the Introduction to Number Theory text and class.

I thought the Intro to Number Theory forum was closed... What class was it, wasn't it number 7?


algebra
polynomial
partial fractions
algebra proposed

If a polynomial $ P\left(x\right)$ of degree $ n\geq 1$ has $ n$ distinct roots $ x_1$, $ x_2$, ..., $ x_n$, then prove that
$ \sum_{i\equal{}1}^n\frac{P^{\prime\prime}\left(x_i\right)}{\left(P^{\prime}\left(x_i\right)\right)^3}\equal{}0$.

I do not know if the following is correct , but I try:
   Decompose P''(x)/P^3(x) into simple fractions:
  P''(x)/P^3(x)=sum A_i/x-x_i+sum B_i/(x-x_i)^2+sum C_i/(x-x_i)^3. Clear denominators and you will get somethind like sum A_1*(x-x_1)^2(x-x_2)^3...(x-x_n)^3+bla bla bla. This polynomial must be P''(x). Take x=x_i and find that C_i=P''(x_i)/(P'(x_i))^3. Then I think that we must look at the dominant coeficient of P" and that in the expansion above. But I'm not sure. Anyway, I think this is the idea.

errr...harazi?... I think the denominator is a first derivative...

Maybe the error term in Newton-Raphson method helps...? like it works great for polynomials? Just a thought, this COULD BE WRONG. My bad...

I do not understand, in my solution all denominators are first derivatives.

[quote]Decompose P''(x)/P^3(x) into simple fractions:
P''(x)/P^3(x)=sum A_i/x-x_i+sum B_i/(x-x_i)^2+sum C_i/(x-x_i)^3. Clear denominators and you will get somethind like sum A_1*(x-x_1)^2(x-x_2)^3...(x-x_n)^3+bla bla bla. This polynomial must be P''(x). Take x=x_i and find that C_i=P''(x_i)/(P'(x_i))^3[/quote]

It was decomposed into simple fractions using P^3(x). Now, C_i=P''(x_i)/(P'(x_i))^3 for x=x_i by substituting it into your decomposition, but where is P^3(x_i)? How did P^3 change into (P')^3? Not only that the solutions of P are generally not the same as P', especially when they are distinct.

Maybe I'm blind, but I cannot see the mistake. Isn't P'(x_i) in the denumerators of the requested relation? Nevermind, the solution must be wrong.

I don't think it's wrong. I think the idea of the decomposition IS in fact leading to the solution. Let's try to handle the expression more carefully. (I'm working on your expression and hopefully a solution, thx harazi)

is this an algebra question? it's more like analysis to me

I had an exercice in Kholle = (oral training mathematics) in
mathmatiques suprieures HX2, long time ago in 1985 at
Louis Le Grand high school Paris.
The "kholleur" said to me "you must see this problem at least one in your life, you will need it in the futur", I did not believe him.
He was wright, today 18 years after I need it to solve Namdung problem
on Mathlinks.

---------------------------------------------------------------------------------
We can suppose P is monic
P(X)=(X-x_1)...(X-x_n), x_i distincts roots,  let's decompose f(X)=1/(P(X))^2 in partial fractions.
---------------------------------------------------------------------------------

f(X) = - \sum _(i=1 to n)P"(x_i)/(P'(x_i))^3.1/(X-x_i) +

\sum _(i=1 to n)1/(P'(x_i))^2.1/(X-x_i)^2

we multiply the equality by X and take the limit when X-->+oo we get

0 = - \sum _(i=1 to n)P"(x_i)/(P'(x_i))^3 + 0

Beautiful algebra partial fractions problem.

You are right, Moubinool. This nice idea I took from the Book "Algebre Oral Maths mines-points" of Christian Leboef.

BTW, in 1983, I was at Lycce Louis Le Grand in Paris, Moubi. Glad to meet you again here.

Namdung

[quote="Namdung"]
BTW
[/quote]

What is BTW ?


Show that the product of any two values $x$ and $y$ can be expressed as the square of the average of the two numbers minus the square of half the difference.
source-me

Yeah... this is probably too easy for even this forum.

$xy=\displaystyle\left(\frac{x+y}{2}\right)^2 - \left(\frac{x-y}{2}\right)^2$
$xy=\frac{x^2+2xy+y^2}{4} - \frac{x^2-2xy+y^2}{4}$
$xy=\frac{x^2}{4} - \frac{x^2}{4} + \frac{2xy}{4} - \frac{-2xy}{4} + \frac{y^2}{4} - \frac{y^2}{4} = \frac{4xy}{4}=xy$

The reason it's simple is because all you need to do is expand the terms and you have the answer. What's generally posted in this forum are problems that require more of a proof; check some of the other problem topics to get a good idea of what's more challenging.

or....

$\displaystyle xy = (\frac{x+y}{2})^2 + (\frac{x-y}{2})^2$
$\displaystyle =(\frac{x+y}{2} + \frac{x-y}{2})(\frac{x+y}{2}-\frac{x-y}{2})$
$\displaystyle =(x)(y) = xy$.


1) classical calculus problem :

x(2*n+1) is increasing
x(2*n) is decreasing

. Graphically, just draw y=sqrt (1-x) for x=0..1 and y=x and look how x(n) turns around (-1/2+1/2*sqrt(5) ,-1/2+1/2*sqrt(5))

. study f(x) = sqrt (1-x) and the sign of f(f(x)) -x

x(n) -> -1/2+1/2*sqrt(5)

2) if a < b :

. a < sqrt(ab) < b  (geometric mean)
. a < (a+b)/2 < b  (arithmetic mean)

So x(n) is increasing and v(n) is decreasing. also x(n) < y(n),  hence they are convergent (and it's easy to prove it's the same limit)

3)

The same :

. a < 2*a*b/(a+b) < b  (harmonic mean)


[quote="inom"]$cos^4(2x)+6cos^2(2x)=\frac{25}{16}$
Solve for x?[/quote]
[hide]this is pretty straight-forward..
let $\cos^2(2x)$ be $a$, then the given equation becomes:
$a^2+6a-\frac{25}{16}=0$.
Multiply $16$ on the equation, we get:
$16a^2+96a-25=0$.
This is factored into:
$(4a-1)(4a+25)=0$..
from here it 's easy to find solutions to the problem.[/hide]

[hide]Let $2x = \alpha$ and $\cos^2 \alpha = a$.
$16a^2 + 96a - 25 = 0 \Rightarrow \cos^2 \alpha = -\frac{25}{4} \vee \cos^2 \alpha = \frac{1}{4}$.
Hence $\cos \alpha = \pm \frac{1}{2} \Rightarrow x = \frac{\pi}{6} + k\pi \vee x = \frac{\pi}{3} + k\pi$.[/hide]

I believe the solutions for $x$ are

     $\frac{\pi}{6}+\frac{k\pi}{2}, \frac{\pi}{3}+\frac{k\pi}{2}$


In right triangle ABC, <A=90 degrees, <B=30 degrees, and R[size=59]ABC[/size]=square root of 3. Find the distance between vertex C and the point of tangency of leg AB and the incircle of the triangle.

[hide="hint"]Use the fact that the distance between the reqd. point and the vertex A is $ s\minus{}a$[/hide]


I was reviewing the 2007 state sprint and i was looking through the solution to some of the problems, one problem i think it was 25 really helped if u knew divisibilty by 11, i thought if the sum if the odd numbered digits 1's 100's ect. subtracted by the sum of the even digits is zero then the number is divisible by 11, but i found sum numbers like 946, 847, and 748 that are divisibble by 11 but dont follow this rule can anyone tell me why this occurs and what other method i can use to test if a number is divisible and not actually dividing?

Thanks in Advance

alright thanks for the help but i think u typoed

[quote="xpmath"]The difference just needs to be divisible by 11.
So, like in 946, (9+6)-4=11, which is divisible by 11.
EDIT:Fixed. I think I make too many typoes.[/quote]

just to clear up any doubts differences by 0 work too. for example 253 works cuz (2+3) - 5 = 0

also, remember: divisibility is defined for negative integers as well. so if you use the divisibility rule and get
-11, then the original number is still divisible by 11.

I learned my 11's strangely.... My 5th grade teacher said that if the two ends of the number equals the number in the middle, then it is a multiple of 11.

For example: 121     the edges (1) added together form the middle making it divisible by 11....

That works for three digit numbers since then the difference is 0, which is a multiple of 11.

The method can be explained through modular arithmetic:
$ 1 \equiv 1 \, (mod\,11)$
$ 10 \equiv \minus{}1 \, (mod\,11)$
$ 100 \equiv 1 \, (mod\,11)$
$ 1000 \equiv \minus{}1 \, (mod\,11)$
...

So, if you consider a 5-digit number, $ ABCDE$:
$ ABCDE \equal{} A \cdot 10000 \plus{} B \cdot 1000 \plus{} C \cdot 100 \plus{} D \cdot 10 \plus{} E$
$ ABCDE \, (mod\,11) \equal{} (A \cdot 10000 \plus{} B \cdot 1000 \plus{} C \cdot 100 \plus{} D \cdot 10 \plus{} E) \, (mod\,11)$
$ ABCDE \, (mod\,11) \equal{} (A \cdot 1 \plus{} B \cdot \minus{}1 \plus{} C \cdot 1 \plus{} D \cdot \minus{}1 \plus{} E) \, (mod\,11)$
$ ABCDE \, (mod\,11) \equal{} (A\plus{}C\plus{}E) \minus{} (B\plus{}D) \, (mod\,11)$

The reason we choose $ (mod\,11)$ is due to the following:
$ ABCDE$ is a multiple of $ 11$ iff $ ABCDE \equiv 0 (mod 11)$

From above,
$ ABCDE \, (mod\,11) \equal{} (A\plus{}C\plus{}E) \minus{} (B\plus{}D) \, (mod\,11)$, so
$ ABCDE$ is a multiple of $ 11$ iff $ (A\plus{}C\plus{}E) \minus{} (B\plus{}D) \equiv 0 \, (mod\,11)$

Since the powers of $ 10$ alternate between $ 1$ and $ \minus{}1$ $ (mod\,11)$, we can take the difference between the sum of the "even" digits and the sum of the "odd" digits.  The same method can be used to generate other divisibility rules that involve summing digits together.  ($ 3$ and $ 9$ for example)

The divisiblility rule for 11 works not only when the digits result in 0, but also in any multiple of 11. This is the case in your examples.


Solve in $ M_2(\mathbb{C})$ the equation $ X^2\equal{}\left(
\begin{array}{cc}
1 & 2 \\
3 & 6 \end{array}
\right)$

$ \left[\begin{array}{cc} 1 & 2 \\ 3 & 6 \end{array}\right] \equal{} \left[\begin{array}{cc} 2 & 1 \\ \minus{}1 & 3 \end{array}\right] \left[\begin{array}{cc} 0 & 0 \\ 0 & 7 \end{array}\right] \left[\begin{array}{cc} \frac37 & \minus{} \frac17 \\ \frac17 & \frac27 \end{array}\right]$

from which it follows that

$ X \equal{} \left[\begin{array}{cc} 2 & 1 \\ \minus{}1 & 3 \end{array}\right] \left[\begin{array}{cc} 0 & 0 \\ 0 & \pm \sqrt{7} \end{array}\right] \left[\begin{array}{cc} \frac37 & \minus{} \frac17 \\ \frac17 & \frac27 \end{array}\right] \equal{} \left[\begin{array}{cc} \pm 1/\sqrt{7} & \pm 2/\sqrt{7} \\ \pm 3/\sqrt{7} & \pm 6/\sqrt{7} \end{array}\right]$


Let $n \geq 3$ be an integer and let $x,y,z \in {\mathbb Z}$ with $x \neq y \neq z$. Show that the equation $x^{3n}+y^{3n}-z^{3n}+3x^{3n}y^{3n}z^{3n}=0$ has no solutions in ${\mathbb Z}$.

 :ninja:

Oh, great one.I bet  Fermat would kill himself now.


If an analytic real-valued function has uncountably many zeros, must it be identically zero?

Yes. If it has a set of zeros with an accumulation point, it is identically zero.

Right, but why does uncountably many zeros imply that there is an "accumulation point"?

Sorry if I'm being slow :(

You can write the plane as a countable union of compact sets (take unit squares for example). One of these sets must contain infinitely many zeroes (in fact uncountably many, but you only need infinite). An infinite subset of a compact set must contain an accumulation point.

Oh, ok. Thanks, jmerry and CaptainCanada!

A quibble: this should be on the line. We expect curves of zeros for a real-analytic real-valued function on the plane.

Of course. :blush: When I see "analytic", I immediately think "complex analytic".


algebra
polynomial
superior algebra
superior algebra solved

Prove that Z/2Z [x] / (x^2 + x+ 1) is a field.

This amounts to proving that $x^{2}+x+1$ is irreducible over $\mathbb Z_{2}$. But there are only two polynomials of degree $1$ over $\mathbb Z_{2}$, and their products are never equal to $x^{2}+x+1$.


Let [tex]a,b,c,d[/tex] be nonnegative numbers satisfying [tex]a+b+c+d=1[/tex]. Prove that [tex]abc+bcd+cda+dab\leq \frac{1+176abcd}{27}[/tex].

This is one from Shortlist 93.
A solution may be found here :
http://www.kalva.demon.co.uk/short/soln/sh9326.html

And this probably already appeared on ML.

Pierre.


ΣΥΝΟΠΤΙΚΑ : Θέτω $ a_{n\plus{}1}\equal{}f(a_n),a_0\equal{}0$ τότε  όλα τα $ a_n$ είναι διαφορετικά μεταξύ τους επειδή η $ a_n$ είναι γνήσια αύξουσα και ρίζες του $ P(x)\minus{}x$ αρα το προηγούμενο πολυώνυμο έχει άπειρες ρίζες οπότε  $ P(x)\equal{}x$
Παλιό θέμα της ΕΜΕ

[quote="Papadopoulos"][quote="r_boris"]ΣΥΝΟΠΤΙΚΑ : Θέτω $ a_{n \plus{} 1} \equal{} f(a_n),a_0 \equal{} 0$ τότε  όλα τα $ a_n$ είναι διαφορετικά μεταξύ τους επειδή η $ a_n$ είναι γνήσια αύξουσα και ρίζες του $ P(x) \minus{} x$ αρα το προηγούμενο πολυώνυμο έχει άπειρες ρίζες οπότε  $ P(x) \equal{} x$
Παλιό θέμα της ΕΜΕ[/quote]

Ουπς, τα $ a_{n}$ διαφορετικά μεταξύ τους; Αν π.χ. f(x)=2x,  τότε όλα τα  $ a_{n}$ είναι 0.
Πες ότι αλλάζεις το $ a_{0}$. Και πάλι δεν βλέπω γιατί τα $ a_{n}$ είναι ρίζες του P(x)-x.[/quote]

Εκανα ενα τυπογραφικο λαθος στην εκφωνηση, η γνησια ανισοτητα $ f(x) > x$ ισχυει και για $ x\equal{}0$, οποτε η συναρτηση δεν μπορει να ειναι η $ f(x) \equal{} 2x$ και τα $ a_n$ ειναι παντα διαφορετικα μεταξυ τους. Τα $ a_n$ ειναι προφανος ριζες του πολυωνυμου $ P(x) \minus{} x$ οταν $ a_0 \equal{} 0$.

[quote="NickNafplio"]Για τη συναρτηση $ f: R \rightarrow R$ και το πολυωνυμο $ P$ ισχυουν:

$ P(0) \equal{} 0$
$ f(x) > x, \forall x \in [0, \plus{} \infty)$
$ P(f(x)) \equal{} f(P(x)) \forall x \in R$

Να βρεθει ο τυπος του πολυωνυμου $ P$[/quote]

Ευχαριστώ για την διευκρίνιση.
Πολύ ωραία η λύση του κ. Μπόρη.

Τάκης


Prove that among 39 sequential natural numbers there always is a number with the sum of its digits divisible by 11.


algebra
polynomial
inequalities
complex numbers
superior algebra
group theory

[b]Problem.  [/b]   Let $G$ be a finite group, and let the irreducible complex characters of $G$ have degrees $d_1,\ldots,d_r$ (so that $d_1^2 + \cdots + d_r^2 = |G|$). Show that
\[
\max_{g \in G} |g| \leq d_1 + \cdots + d_r.
\]

Vess: what do you mean here by irreducible character and degree of a character?

Here, if $\Delta_1,\ldots,\Delta_r$ are (up to equivalence) the irreducible representations of $G$ over the complex numbers $\mathbb{C}$ (so that $r$ is the number of equivalence classes of $G$ with respect to conjugation), the numbers $d_1,\ldots,d_r$ are the degrees of the representations $\Delta_1,\ldots,\Delta_r$. In other words, if $\chi_i = \mathrm{tr} \, \Delta_i$ is the character afforded by $\Delta_i$, then $d_i = \chi_i (1)$, and $d_1^2  + \cdots + d_r^2 = |G|$.

I can show that every element of $G$ has order at most $d_1 + \cdots + d_r$; my question is: is this trivial? Is there a better estimate for the maximal order of $G$ in terms of the $d_i$? Of course, the estimate is sharp for $G$ cyclic, but seems very rough for non-cyclic groups.

[quote="vess"]Let $G$ be a finite group, and let the irreducible complex characters of $G$ have degrees $d_{1},\ldots,d_{r}$. Show that
\[\max_{g \in G}|g| \leq d_{1}+\cdots+d_{r}. \]
[/quote]

Let $g$ be an arbitrary element of $G$, and let $m$ be its order. We can consider $g$ as being an element of the complex group algebra $\mathbb C[G]$ and as such, we can look at complex polynomials which annihilate $g$. On the one hand, $1,g,\ldots,g^{m-1}$ are linearly independent over $\mathbb C$, so any such polynomial has degree $\ge m$. On the other hand, $\mathbb C[G]$ is isomorphic to the direct sum $\bigoplus_{i=1}^{r}\mathcal M_{d_{i}}(\mathbb C)$, so for each $i$, the $i$’th component of $g$ satisfies a polynomial equation of degree $\le d_{i}$. Multiplying all these polynomials together, we get a polynomial of degree $\le\sum d_{i}$ which annihilates $g$. Coupled with the previous observation that the degree of such a polynomial must be $\ge m$, this proves the desired inequality: $m\le\sum d_{i}$.

Exactly -- it's what I did also :) By the way, can you characterize the groups for which equality holds?


invariant
linear algebra
matrix
vector
analytic geometry
algorithm
function

what's a straightedge? Gonna ask before I post the method :P  (it's not in my dictionary)

Can you find the midpoint of BC without consider AB and CD?

I certainly don't think so.. If you have two points in the plane, I don't think their midpoint is constructible with a straightedge alone. You need a compass.

[quote="grobber"]If you have two points in the plane, [b]I don't think[/b] their midpoint is constructible with a straightedge alone. You need a compass.[/quote]

Replace "I don't think" but "I am sure that not", since any construction with a straightedge alone has to be invariant under a projective transformation, while the midpoint of a segment is not.

  Darij

Yes, yes, I know that, I just don't like to sound too sure of myself :D. When you guys post on this topic you're torturing me, because each time I think someone might have solved it :).

it depends on the formulation... mathematically one cannot I guess, but in reality, if you have a rectangular ruler, you can use it in such a way you can create angles of the same size, and then the problem is easy of course :)

[quote="grobber"]On a line points $A,B,C,D$ are given in this order s.t. $AB=CD$. Can we find the midpoint of $BC$ using only a straightedge?

I've been struggling for quite some time. I sure hope someone cracks it fast.[/quote]

Grobber, do ypu think that is possible, and you are looking for a such construction?

Or do you think is not possible and you are looking for a disproof.

PS
From the book "what is mathematics" Courant & Robbins:

Given two parallel l1 and l2 bisect a segmente AB on r1.

I suspect a typo: should be l1 instead of (a generic) r1.  In this case the problem is easily solved.

If I am wrong, how to do such a construction?

If I am not wrong how to prove that is not possible realize such a construction?

To tell you the truth, I don't know what to think, but I suspect that the construction is possible.

And regarding your problem, bisecting a segment on $\ell_1$ if $\ell_2\|\ell_1$ is given and vice-versa (constructing a parallel if the midpoint of a segment is given) is easy, and you definitely can't do more (in other words I agree that it should be $\ell_1$ instead of $r_1$), but I don't think I have a very rigorous argument.

It seems that there are projective transformation which save equality of segments but it doesn't save midpoint.

Is it true:
If P is projective transformation of line L then there are projective transformation P' of plane which acts on the line L as P.

I'm not sure what that means: are you asking if there are more projective transformations which are the same on a certain line? That's not true, as all projective transformations are determined by the images of four points.

I am sure that you didn't answer my question.
Projective transformations are determined by the images of four points, no three of which are collinear.

In my question P acts only on line L, not on whole plane.

I see. Well, we can try to work it out. The projective transformation is represented by an invertible $3\times 3$ real matrix $A$. We have three relations, $A{\bf x}=\alpha {\bf t},\ A{\bf y}=\beta {\bf u},\ A{\bf z}=\gamma {\bf v}$, where ${\bf x,y,z}$ are linearly dependent column vectors with three components, not all $0$ (we work with homogenous coordinates), and $\alpha, \beta, \gamma$ are non-$0$ reals. The question is: is there such an invertible matrix $A$, given the vectors ${\bf x,y,z,u,v,t}$?

Of course there are such matrices: since the vectors ${\bf x,y,z}$ are linearly dependent, this gives us lots of room to make $A$ invertible. For example, if ${\bf x,y,z}$ are $(1,0,0),(0,1,0),(1,1,0)$, then the three equations only give us info about the first two columns of $A$, and the equations make the columns independent (because ${\bf x,y}$ are independent). This means that we can choose the third columns to be whatever we want.

Another way to look at it: take two lines in space, one with points $X,Y,Z$ on it and the other one with points $U,V,T$ on it. Clearly, we can place them in space s.t. $XU,YV,ZT$ are concurrent at $O$ (a point in space). Now we can take our transformation to be the perspectivity of vertex $O$ from a plane containing the first line to a plane containing the second line. Moreover, there are lots and lots of such perspectivities, according to the planes we choose.

I hope this is what you were asking.

It's not clear to me whether the original problem of the thread has been answered or not, but I tend to thinking that it hasn't. So why not answer it now...

[quote="grobber"]On a line points $A,B,C,D$ are given in this order s.t. $AB=CD$. Can we find the midpoint of $BC$ using only a straightedge?[/quote]

Finally I have done it. The problem is so difficult because one really doesn't want to believe that the answer is "NO". But it is "NO" - we cannot construct the midpoint of the segment BC using a straightedge only.

The [i]proof[/i] is really easy:

Assume, for the sake of contradiction, that we can construct the midpoint of the segment BC using a straightedge only. This means that there is an algorithm $\mathcal{G}$ consisting of nothing but just joining points and intersecting lines, which, applied to four collinear points A, B, C, D lying on a line in this order and satisfying AB = CD, yields the midpoint of the segment BC.

At first it is clear that the order of the points A, B, C, D on the line is not really important, provided that we interpret the equation AB = CD as an equation with directed segments. In fact, our algorithm $\mathcal{G}$ is a rational algorithm, i. e., working in Cartesian coordinates, its output value is a rational function of the input values (this is because the operations of joining points and intersecting lines are rational operations in Cartesian coordinates). Hence, the assertion that "the algorithm $\mathcal{G}$ yields the midpoint of the segment BC for any four collinear points A, B, C, D lying on a line in this order and satisfying AB = CD (with directed segments)" can be written as a polynomial identity, where the condition that the points A, B, C, D lie on their line [i]in the given order[/i] is just an inequality-type restraint to the variables of the polynomial. This inequality-type restraint puts the variables of the polynomial into a certain domain. But this domain is easily seen to be sufficient to conclude that, if the polynomial identity is valid in this domain, then it is also valid in the whole space. This means that the algorithm $\mathcal{G}$ yields the midpoint of the segment BC for any four collinear points A, B, C, D satisfying AB = CD (with directed segments). Thus, the order is not important anymore. Of course, we are using directed segments here and we will continue using them in the rest of our proof.

Now, if we denote the midpoint of the segment BC by M, then we have BM = MC, and since we also have AB = CD, we see that AM = AB + BM = CD + MC = MD, so that this point M is also the midpoint of the segment AD. So the point M is the midpoint of the segments BC and AD; in other words, the points C and D are the images of the points B and A in the reflection in the point M. However, reflection in the point M can be seen as an involution, and M is one fixed point of this involution (the other fixed point is the infinite point of the line ABCD). Hence, the points C and D are the images of the points B and A in this special involution which is actually a reflection in a point; in other words, the pairs of points (B; C) and (A; D) are conjugate point pairs in this involution.

Summarizing, we see that our algorithm $\mathcal{G}$ does the following: If A, B, C, D are four collinear points having the property that the pairs of points (B; C) and (A; D) are conjugate points in a special involution which is a reflection in a point, then our algorithm $\mathcal{G}$ yields a fixed point of this involution.

But the algorithm $\mathcal{G}$ is a pure incidence algorithm; hence, it is invariant under projective transformations. We can easily find a projective transformation mapping a generic hyperbolic involution (i. e. an involution which has two fixed points but is not a reflection in a point) to an involution which is a reflection in a point (it is just necessary to map one fixed point to an infinite point). Hence, our algorithm $\mathcal{G}$ should also, if applied to four collinear points A, B, C, D having the property that the pairs of points (B; C) and (A; D) are conjugate point pairs in a generic hyperbolic involution, yield a fixed point of this involution.

But it is very easy to find a hyperbolic involution where we have two conjugate pairs of points (B; C) and (A; D) with [i]rational[/i] coordinates, while both of the fixed points of the involution have [i]irrational[/i] coordinates (in fact, constructing the fixed points of an involution provided two conjugate pairs of points is equivalent to solving a generic quadratic equation - this is why the fixed points of elliptic involutions are imaginary). This shows that there cannot be any rational algorithm to find these fixed points, what clearly contradicts the assumed existence of our algorithm $\mathcal{G}$. Hence, our assumption was wrong, and there is no way to construct the midpoint of the segment BC using ruler only. $\blacksquare$

An almost philosophical proof, you see. I also hope that it's correct ;) .

As for the other problem:

[quote="sprmnt21"]PS
From the book "what is mathematics" Courant & Robbins:

Given two parallel l1 and l2 bisect a segmente AB on r1.

I suspect a typo: should be l1 instead of (a generic) r1.  In this case the problem is easily solved.[/quote]

Indeed, if $r_1$ is just a generic line, the problem cannot be solved by ruler only. This is pretty clear: You can find a projective transformation that leaves the common infinite point of the lines $l_1$ and $l_2$ invariant, but messes up the infinite point of the line $r_1$, so that it is not infinite anymore, and thus the midpoint of the segment AB is also not the midpoint anymore, so that a ruler-only construction is impossible.

  Darij

Yes, Darij, I have actually read it (:)) and find it quite convincing :). Thank you!

The answer is no.

Impose a coordinate on the line. Consider the case where \begin{align*} 	A &= -1 \\ 	B &= \sqrt2-1 \\ 	C &= 1 \\ 	D &= \sqrt2+1. \end{align*} Then $(AC;BD) = -1$.
Assume for contradiction our algorithm constructed $M$, the common midpoint of $BC$ and $AD$. Let $\infty$ denote the common harmonic conjugate of $M$. Take a projective transformation so that on the real line, \begin{align*} 	A' &= 0 \\ 	B' &= 3 \\ 	C' &= 4 \\ 	D' &= 6. \end{align*} Then $M'$ is the point with the property such that \[ (B'C';M'\infty') = (A'D';M'\infty') = -1. \] However, we can compute the coordinates of $M'$ explicitly: \[ M' = \frac{-1 \pm \sqrt5}{2}. \] Yet it's clear that in the transformed problem, it is not possible to obtain a irrational $x$ or $y$ coordinate for any point at any time in the process. Contradiction.

[b][color=red]Remark:[/color][/b]  	The numbers are chosen concretely here only 	for comedic value. 	In truth, the ``right'' way to think about it is 	by solving for $m$ as a function of $a$, $b$, $c$, $d$. 	One finds that the resulting equations are quadratic, 	and hence contain a radical, making the task 	obviously impossible for almost all starting values. 	However, solving the system is really annoying, 	so we just picked some really convenient values 	to avoid having to do heavy calculations.


The discriminant of the equation $ x^2\plus{}2x\sqrt{3}\plus{}3\equal{}0$ is zero.  Hence, its roots are:

$ \textbf{(A)}\ \text{real and equal} \qquad
\textbf{(B)}\ \text{rational and equal} \qquad
\textbf{(C)}\ \text{rational and unequal} \\
\textbf{(D)}\ \text{irrational and unequal} \qquad
\textbf{(E)}\ \text{imaginary}$

From the quadratic formula, since the discriminant is zero. The root is
$ \frac {\minus{}2\sqrt{3}}{2}$
So the roots are real and equal $ \implies \boxed {A}$.

$ x_1\equal{}x_2\equal{}\frac {\minus{}b}{2a}\equal{}\frac {\minus{}2\sqrt {3}}{2}\equal{}\minus{}\sqrt {3}$ or $ \boxed A$


algorithm
number theory
greatest common divisor
number theory unsolved

Some months ago I solved this problem:
"Let $(x_{n})$ be a sequence of nonnegatives integers, such that $x_{0}$ and $x_{1}$ are positive and $x_{n+1}=|x_{n}-x_{n-1}|$ for all $n \geq 1$. Prove that this sequence contains inifinitely many terms equal to 0"

This problem is interesting to me because I found that  $(x_{n})$ has other properties, like: "If $x_{k}=0$ then $gcd(x_{0}, x_{1})=x_{k-1}$, i.e. this sequence give us an algorithm to get the gcd of two numbers.
[b]I'm looking for other solutions for the original problem, and other  problems  related to it. [/b] Like this problem (of KOMAL):
"Let $x_{0}$ and $x_{1}$ be positive integers less than 1000. Starting with them form the sequence $x_{n+1}=|x_{n}-x_{n-1}|$ for all $n \geq 1$. Show that exists and index $i<1500$ such that $x_{i}=0$"
The way I solved the original problem not helped me in the second one, I hope the you'll find a solution.

$Tipe$

[quote="tipe"]If $x_{k}=0$ then $gcd(x_{0}, x_{1})=x_{k-1}$[/quote]

This basically [b]is[/b] Euclid's Algorithm.

Note that when $1$ first appears in the sequence, thereafter every third term is $1$ and the other terms decrease by $2$ every $3$ terms, which is where the $1500$ value comes from in the second problem.  Then you just have to show that until a $1$ appears in the sequence it decreases asymptotically faster than this.

thanks, but, could you be more specific  :blush: ?

$Tipe$


trigonometry
absolute value
real analysis
real analysis unsolved

Hello,

Find out whether the following series converges (absolutely) and why does it so?

\[\sum_{n=1}^{\infty}\left(-1\right)^{n}\frac{\sin^{2}n}{n}\].

Mash'Allah

The series does not converge absolutely. It does converge conditionally. What I would do there is to write $\sin^{2}n=\frac12(1-\cos2n)$ and split it into two series, handling each one with a separate argument.

But I still have to show, why $\sum_{n=1}^{\infty}(-1)^{n}\frac{\cos 2n}{n}$ converges conditionally.

Mash'Allah

$A_{n}=\sum_{k=1}^{n}(-1)^{k}\cos(2k)=\text{Re }\left(\sum_{k=1}^{n}(-1)^{k}e^{2ik}\right)=\text{Re }\left(\frac{-e^{2i}+(-1)^{n}e^{2i(n+1)}}{1+e^{2i}}\right).$

As $n\to\infty,\ |A_{n}|$ is bounded.

By summation by parts,

$\sum_{k=1}^{n}(-1)^{k}\frac{\cos(2k)}{k}=\frac{A_{n}}{n}+\sum_{k=1}^{n-1}A_{k}\left(\frac1k-\frac1{k+1}\right).$

Take the limit as $n\to\infty$ to get

$\sum_{k=1}^{\infty}(-1)^{k}\frac{\cos(2k)}{k}= \sum_{k=1}^{\infty}A_{k}\left(\frac1k-\frac1{k+1}\right).$

Let $M$ be an upper bound for $|A_{k}|.$ The series on the right is then bounded in absolute value by

$\sum_{k=1}^{\infty}M\left(\frac1k-\frac1{k+1}\right)=M.$

That shows that it converges. It doesn't converge absolutely because $\cos(2k)$ is greater than some fixed positive number often enough to set up an adverse comparison with the harmonic series.

But how exactly do I prove that $1-\cos2k\geq\alpha>0$ for sufficiently enough $k$? I thought of fixing an $\varepsilon$ such that the numerator of any rational in a $\varepsilon$-neighbourhood of $2k\pi$ has a lower boundary. But I don't know how to go on. And maybe (definitely) there is a method easier than the one I proposed.

Mash'Allah


inequalities
function
algebra unsolved
algebra

Suppose that a,b,c,d are >0 and are in R.
   and a^3+b^3+3ab=c+d=1.
Prove:
        (a+1/a)^3+(b+1/b)^3+(c+1/c)^3  =>40

From the condidition we have that
a+b+(-1)=0 or a=b=-1.The second is impossible so a+b=1.
Now let the function $f(x)=(x+\frac{1}{x})^3=x^3+3x+\frac{3}{x}+\frac{1}{x^3}$
We find after algebra that $f''(x)=6x+\frac{6}{x^3}+\frac{12}{x^5}$ So $f''(x)>0$
So from Jensen ineq we have that LHS$\geq 4f(\frac{1}{2})=4(\frac{5}{2})^3=\frac{125}{2}>40$ ;)

Take $x = a + \frac{1}{a} \geq 2$ and $y = b + \frac{1}{b} \geq 2$, then

\[x^3 + y^3 \geq xy(x + y)  \geq 4(x + y) = 4\left(a + b + \frac{1}{a} + \frac{1}{b}\right) \geq 4\left(a + b + \frac{4}{a + b}\right) = 20.\]

Similarly for $c$ and $d$, add, we're done.

dear Arne, when our teacher gave us this problem there was no expression with
(d+1\d)^3.indeed it makes the problem difficult..

Thanks silouan.i will analyze your answer

Look, spider_boy, I have seen the problem in a competition before with $a + \frac{1}{a}$, $b + \frac{1}{b}$, $c + \frac{1}{c}$ AND $d +  \frac{1}{d}$.

Well, and in your case, it doesn't work - take $a = b = \frac{1}{2}$ and $c \rightarrow 1$, then \[\left(a + \frac{1}{a}\right)^3 + \left(b + \frac{1}{b}\right)^3 + \left(c + \frac{1}{c}\right)^3 \rightarrow \frac{157}{4} < 40.\]

By the way, Silouan solved exactly the same problem as me: he also assumed there was a term $\left(d +  \frac{1}{d}\right)^3$ on the left hand side.


abstract algebra
vector
linear algebra
linear algebra unsolved

Suppose we have a morfism of $K-algebras$ from $K^{n\times n}$ in $K^{m\times m}$. Prove that $n$ divide $m$.

I'm assuming that by $K^{n\times n}$ you meant $\mathcal M_{n}(K)$, i.e. the ring of $n\times n$ matrices over $K$, right? Because the notation you used is confusing; it can easily be interpreted as the product of $n^{2}$ copies of the ring $K$, for instance, and in this case the result is no longer true.

For any field $K$, the ring $A_{n}=\mathcal M_{n}(K)$ has only one simple left module up to isomorphism, which can be identified with the set of columns of size $n$ with elements in $K$ on which $A_{n}$ acts in the usual way (multiplication of matrices).

If $f: A_{n}\to A_{m}$ is a $K$-algebra homomorphism, then a simple $A_{m}$-module $M$ can be turned into an $A_{n}$ module via by defining the left action of $A_{n}$ on $M$ as follows: $ax=f(a)x,\ \forall a\in A_{n},\ x\in M$. Notice that by restricting our attention to $a\in K$ (which can be regarded as a common subfield of $A_{n},A_{m}$), we find that $M$ is an $m$-dimensional vector space over the copy of $K$ lying in $A_{n}$.

Since $A_{n}$ is semisimple, the $A_{n}$-module $M$ can be written as a direct sum of simple $A_{n}$ modules. The dimension of $M$ as a vector space over $K$ is, as we noticed above, finite (it's equal to $m$), so this direct sum must be finite as well. Moreover, it follows from the discussion in the second paragraph that the dimension of this direct sum over $K$ is a multiple of $n$. We know that it's equal to $m$ as well, so we're done.

I hope it's correct.


limit
logarithms
function
calculus
calculus computations

Evaluate this limit:

$\lim_{n\to\infty}(ln\frac{1}{n})$

Does anyone know of a proof that uses sequences?
If the value, which in this case $-\infty$, was obtained by some means, and you were asked to prove that the limit in question approaches $-\infty$. Is this a legitimate question to ask?

I guess it would be legitimate, however why would you insist on using sequences? I believe limits imply sequences, and in case of continuous functions, they are equivalent.

[quote="paladin8"][quote="akech"]Direct computations leave no rooms for explanation. The tricks that lead to a result may not be apparent to just any common person so such approach does not completely kill suspicion about the truth of a given statement.[/quote]

What's that supposed to mean? Is there any suspicion that $-\lim_{n \rightarrow \infty}\ln{(n)}=-\infty$ is not true?[/quote]

No, the limit in question is correct, but actually it is a theorem, which says if $\lim(f(x)) = k < 0$ while $\lim(g(x)) =+\infty$,
then $\lim(f(x)g(x)) =-\infty$. Taking $k =-1$ and $g(x) = ln(x)$, what we have is an application of this theorem, you see now it looks like we are assuming that lim $(g(x))$ goes to $+\infty$

So you're saying you want a proof that $\lim_{n \rightarrow \infty}\ln{(n)}= \infty$? I suppose we could take the sequence $e, e^{2}, e^{3}, \ldots$ to do this.

Well, we can prove that by assuming that the sequence a_{n} approaces a value A, and then apply exponential function to find some contradiction.

Sequence $\{\ln{n}\}$ is increasing and above unbounded because $\ln{e^{k}}=k\to+\infty$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose X is a non empty set and G $\leqslant$ Sym(X)
Let a $\in$X and $H =$ {$g \in G : g(a)=a$}

(i) Prove that H is a subgroup of G
(ii) Prove that $g_{1}H = g_{2}H$ $\Longleftrightarrow$ $g_{1}(a) = g_{2}(a)$  ( for all $g_{1}g_{2}\in G$

How would one apply a symmetric group to these questions?

And for the second part , the furthest I got is showing g1H=g2H ..  wat is g(a) . I dont get this question

Any help would be appreciated

[quote]
(ii) $g_{1}H=g_{2}H$ means $g_{1}^{-1}g_{2}\in H$,[/quote]
Im not trying to be offensive, I just want to explain better to the original poster why this is. Since it seems he is just learning group theory he might not so easily see why.

Say $g_{1}H = g_{2}H$ then it means $g_{1}h_{1}= g_{2}h_{2}$ so, $g_{2}^{-1}g_{1}=h_{2}h_{1}^{-1}$ but $h_{2}h_{1}^{-1}= h_{3}\in H$ because $H$ is a group so its closed. Therefore, $g_{2}^{-1}g_{1}= h_{3}\in H$


Let a[b] 6 digit [i][u]even[/u][/i][/b]number be such that it is made up of 1,2,3,4,5,6 digits only and no digit should be repeated.Find the greatest common divisor/Highest common factor for all such nos.

I hope its 6.

The sum of digits is 21, divisible by 3; the last digit is even ,so gcd is divisible by 2. Now if you have number 1=  ****02 and ****20 their difference, 18 is divisible by gcd, so gcd | 18. It is at least 6. It cannot be 18 cuz the sum of digits is not divisible by 9 so gcd is not divisible by 9. So, gcd =6.

This should be in another place.


This ones a pretty hard question.

If xy= 6, yz= 9 and zx= 24, find the value of xyz. I would show my work for this one, but honestly i do not knwo where to begin.

[hide]If you multiply xy, yz, and xz together, you will get $1296=x^2y^2z^2$.  So just take the square root of that and you will get $xyz=36$.[/hide]

[hide]
Find the square root of the product.
$9\cdot6\cdot24=1216$
$\sqrt{1216}=36$[/hide]

Can't xyz be negative too? (4, 1.5, 6) and (-4, -1.5, -6) work, and in the latter, xyz = -36

In these kinds of problems, they usually say "where $a,b,c$ are positive integers" or something.  Or you could put $\pm$...


calculus
integration
function
LaTeX
algebra
floor function
calculus computations

integrate $ f(x) \equal{} \int_{0}^{x}\ x \minus{} [ x ] \minus{} (1/2)\,dx$..please give the ans in-terms of { x }
where [b]{ x }[/b] and[b] [ x][/b] denote fractional parts and greatest integer function of x respectively.

$ \int_0^x(\text{whatever})\,dx$ is bad notation.

$ x$ cannot range from $ 0$ to $ x.$

Please restate this using proper notation.

[hide="My attempt, dunno if its right"]
$ f(x) = \int_0^x \ ( frac \ part(x) - \frac{1}{2}) \ dx \\ \\ = \int_0^{[x]} \ frac \ part(x) \ dx + \int_{[x]}^x \ frac \ part (x) \ dx - \frac{1}{2}\int_0^x dx$

[b]frac part[/b] denotes { x } ... can't type {x} in   $ \text{\LaTeX}$    :!:

x = [x] + {x}

{x} is periodic with period 1

So,

$ f(x) = [x] \ \int_0^1 \ x \ dx + \int_{[x]}^{[x]+ frac \ part(x)} \ frac \ part (x) \ dx  - \frac{1}{2} \ [x]_0^x$

$ = [x] \frac{1}{2} + \int_0^{frac \ part(x)} \ x \ dx - \frac{x}{2}$

$ = \frac{[x] - x}{2} + \frac{(frac \ part(x))^2}{2}$

$ = \frac{- \ frac \ part(x)}{2} + \frac{(frac \ part(x))^2}{2}$
[/hide]

thanks nora even i got the same answer but wasnt sure if it was correct.

thanks a lot :)  :)

[b]My pleasure !! [/b]  :roll:

the only thing i wasn't sure of was the middle integral ( from [x] to x ) ...

Type [code]\{ [/code] and [code]\}[/code] for the brackets $ \{ \quad \}$.

[quote="Tyl"]Type [code]\{ [/code] and [code]\}[/code] for the brackets $ \{\quad \}$.[/quote]

$ \{ \ Thanks \ \}$ for that, Tyl !!

As prof kent pointed out here is thye restated problem.

$ f(x) \equal{}\int_{0}^{x}\ y\minus{}[ y ]\minus{}(1/2)\,dy$


search
function
arithmetic sequence
number theory
relatively prime

Given an arithmetic progression of terms $ an\plus{}b$, for $ n\equal{}1, 2, ...$ prove that the series contains an infinite number of primes  if $ a$ and $ b$ are relatively prime, i.e., $ (a,b)\equal{}1$.

It has been posted many times before.  See, for example, http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=231067471&t=152559 .

Dirichlet's Theorem is one of the most impressive results of analytic number theory and I'd be surprised if anyone here could actually discover the proof. Granted, there are probably a number who know how it works, but I doubt that any of them actually discovered it.

If nothing else, the proof of Dirichlet's Theorem is far beyond the High School forums.  The first, and simplest, step involves considering functions of the form $ L(s) \equal{}\sum_{n\equal{}1}^{\infty}\frac{\chi(n) }{ n^{s}}$, which are known as Dirichlet L-series.  And if you're wondering about the rest of the proof,

[quote="MathWorld"]in their classic text on number theory, the usually explicit Hardy and Wright (1979) report "this theorem is too difficult for insertion in this book."[/quote]

To add, I believe that Dirichlet's theorem is the only result in Hardy and Wright's books which "is beyond the scope of the book". Even the prime number theory which appears in later editions has a proof supplied to it.


$ a,b,c\,\succ\,0$    and $ n$ an integer $ \succ$ 0
prove that :

$ \frac {a^n}{b \plus{} c} \plus{} \frac {b^n}{a \plus{} c} \plus{} \frac {c^n}{b \plus{} a}\,\geq\,\frac {a^{n\minus{}1}\plus{}b^{n\minus{}1}\plus{}c^{n\minus{}1}}{2}$

$ (a^{n\minus{}1}\minus{}b^{n\minus{}1})(a\minus{}b)\ge 0\Leftrightarrow \frac{a^n\plus{}b^n}{a\plus{}b}\ge \frac{a^{n\minus{}1}\plus{}b^{n\minus{}1}}{2}$

By rearrangement,
\[ 2\sum \frac{a^n}{b\plus{}c}\ge \sum\frac{a^n}{a\plus{}b} \plus{}\sum\frac{a^n}{a\plus{}c}\equal{}\sum \frac{a^n\plus{}b^n}{a\plus{}b}\ge\sum a^{n\minus{}1}
\]


A point source of light radiates 25 W at a wavelength of 5000 angstroms. A plate of metal is placed 100 cm from the source. Atoms in the metal have a radius of 1 angstrom. Assume that the atom can continually absorb light. The work function of the metal is 4 eV. How long is it before an electron is emitted from the metal?

I thought I knew how to do this, but I am not using the radius of the atoms in the metal. Is this extraneous information?

isn't that a violation of what photoelectric effect says .
first electrons can't store enrgy they rlease it quickly either by leaving the atom or jumping to the neaxt higher shell or by flipping.also photoelectric effect says the electrons if given  light of sufficient frequency(energy) then and only then they come out as such they don't 'store' energy

Seems to be an exercise that shows the  limitation of Maxwell's classical model of light.

   As an eletromagnetic wave, being absorbed continually, it would take

\[ {{16.1,6.{10}^{\minus{}19}}\over{25.{10}^{\minus{}20}}}\; \equal{}\;2,56 s\]

to the emittion, while experience shows that it takes less than 3 ns (due to photoelectric effect).


geometry
incenter
analytic geometry
geometric transformation
reflection
geometry proposed

The incircle  of $\triangle{ABC}$ touches the side $BC$ at $F$. Let $M$ and $N$ be the midpoints of $BC$ and $AF$. Show that the incenter $I$ of $\triangle{ABC}$ lies on $MN$.

Use baricentric coordinates:
$F=(0, \frac{p-c}{a},\frac{p-b}{a})$, $A=(1,0,0)$,
hence $N=(\frac{1}{2},\frac{p-c}{2a},\frac{p-b}{2a})$
$M=(0,\frac{1}{2},\frac{1}{2})$, $I = (a,b,c)$, thus

$det\displaystyle \left( \begin{array}{ccc} \frac{1}{2}, & \frac{p-c}{2a}, & \frac{p-b}{2a}) \\ a & b & c \\ 0 & \frac{1}{2}, & \frac{1}{2}\\ \end {array}\right)=0$ ,which is true.

Hence $I, M, N$ are collinear.

Let F' be the reflection of F wrt M. Then the result follows from the well known fact AF'//MI.


geometry
3D geometry
invariant
combinatorics unsolved
combinatorics

Let the 3-cross be the solid made up of one central unit square with six other unit cubes attached to its faces.Prove or disprove that the space can be tiled with 3-crosses in such a way that no two of them share any interior points.

You can be sure that it can tile the $3$-space because it has null [url=http://www.google.ro/url?sa=t&ct=res&cd=7&url=http%3A//dimacs.rutgers.edu/%7Esam/dehninv/&ei=FtsWQ7rZJrL4wQHf_dWsCw]Dehn invariant[/url], and I believe there's a theorem stating that a solid can tile the Euclidean space iff it has this property. So, now that you know it's possible, just play around with it until you manage to find the tiling :).

The theorem about Dehn invariants probably refers to signed tilings, not tilings in the usual sense where the interiors are disjoint.  Clearly you can cut out small sub-cubes in some complicated way to make an ordinary tiling impossible, and this does not change the Dehn invariant.   Problems about infinite tilings can be undecidable in two dimensions (certainly if you allow multiple types of pieces and coloring of the edges) so I would be astonished if some simple test exists for tiling by a polyhedron.


AMC AIME USA(J)MO USAMO logarithms geometry trigonometry

AIME A 2004 number 12 is one problem with logs. I don't know if there are other problems with logs...

[quote="bpms"]Are there any AIME (numbers 11-15) or USAMO problems with logarithms?
Thank you.[/quote]
Last year's (2006) AIME I #9 used logarithms (which is pretty obvious once you look at the problem):

[quote="2006 AIME I #9"]The sequence $a_{1}, a_{2}, \ldots$ is geometric with $a_{1}=a$ and common ratio $r$, where $a$ and $r$ are positive integers. Given that $\log_{8}a_{1}+\log_{8}a_{2}+\cdots+\log_{8}a_{12}= 2006$, find the number of possible ordered pairs $(a,r)$.[/quote]

I think that in general, logarithms are found pretty often on AMC problems, occasionally on AIME problems, and rarely on USAMO.

EDIT: Oh wait, I just realized he was asking about AIME #11-15.  :oops: Oh well, #9 is close enough...  :roll:

Yes, with regards to geometry, #11 - 15 is almost guaranteed to have two geometry problems.

[quote="Phelpedo"]Seems like there is a dense population of complex numbers and geometry among the harder AIME problems.[/quote]

Definitely. But geometry has always been a very prominent subject on the AIMEs, especially on the "harder" questions. I suck at geometry so I know that I always miss like half the geometry questions and by that point am completely screwed over on my goal of getting 13 or so.

Last year as a pretty notable exception though. Very few trig. Very few complex numbers. Just not the typical fare.

Usually there's at most one complex numbers problem on an AIME.  I'm not sure about trig, but I'd say it's probably around 1-2 an AIME, on average.


number theory
prime numbers
number theory unsolved

[quote="St. Elena"]For n a natural number we consider the number A_n=2^n+3^n.
   [list][*]Find a necessary condition so that A_n is a prime number.
  [*]Is this condition also a sufficient one?[/list][/quote]
A necessary condition: $ n \neq 15$. :P Seriously: $ n$ must either be a prime number or a power of two.

Another necessary condition: $ n$ must be a power of $ 2$.

No, it is not sufficient. For example, taking $ n \equal{} 8$ we get

$ 2^8 \plus{} 3^8 \equal{} 256 \plus{} 6561 \equal{} 6817 \equal{} 17*401$.

Do you really think you can obtain a sufficient condition? For so many years mathematicians have failed to find the formula that can generate all prime numbers like the one you have.

If $ n \equal{} (2k \plus{} 1)2^e$ where $ e\ge0$ and $ k\ge1$ then $ A_{2^e}|A_{n}$.

So I am checking the case　of $ n \equal{} 2^e$.


Increasing a number by 10$ \%$ is the same as multiplying it by
$ y$. What is the value of $ y$? Express your answer as a decimal to
the nearest tenth.

So increasing a number $ x$ is basically $ x+x(10%)
$ $ x+\frac{x}{10}$ or $ \frac{11x}{10}$ or $ 1.1x$
so yaeh :) $ 1.1x=y*x$ so $ y=1.1$

When u increase by 10 percent, u times 1.1. Why? Because in decimals, 10 percent is 0.1. But if u multiplied by 0.1, the price would decrease. If a number times another number, what number does it have to times to stay the same? 1. Therefore, you have to do the number times 1.1. So y would equal 1.1 because it is the same as a number being increased by 10 percent. Yay.

Let x be the number you are increasing by 10%. It would equal 100%x+10%x which equals x(1+0.1) which therefore equals x(1.1). That means y is 1.1

Everyone, be sure to hide your solutions!


calculus
integration
calculus computations

Thank you very much, [b]vidyamanohar[/b]. I will continue to post problems.

For $n>2,$ prove the following inequality.

\[ \frac{1}{2}<\int_0^{\frac{1}{2}} \frac{1}{\sqrt{1-x^n}} dx<\frac{\pi}{6}. \]

it is sufficient to show that
[hide]

$1<\frac{1}{\sqrt{1-x^n}}<\frac{1}{\sqrt{1-x^2}}$ for $0\leq x\leq \frac{1}{2}$ and $n>2$  :)
[/hide]


algebra
polynomial
inequalities
triangle inequality
complex numbers
algebra unsolved

Let f be a polynomial all of whose coefficients are 0 or 1 and such that $f(2)$ is prime. Prove that f is irreducible in $Z[X]$. I know several similar problems (Balkan and some other contests), but this one seems much tougher. Or have I completely lost my mind?

[quote="harazi"]Or have I completely lost my mind?[/quote] I am not a psychiatrist, but...

 :lol:  :lol:  :lol:

Myth, anyway I do not trust in your words in these cases :D  :D . But if you have a nice short solution, I will consider you a doctor!  ;)

Problem.
Let $f$ be a polynomial all of whose coefficients are $0$ or $1$ and such that $f(2)$ is prime. Prove that $f$ is irreducible in $\mathbb{Z}[X]$.

Solution.
Let $f= X^n + a_1 X^{n-1}+ \dots + a_n$, with $a_i \in \{0,1\}$. We first prove:

Claim 1. For any root $\alpha$ of $f$, we have $|2-\alpha| > |1-\alpha|$, or equivalently, $\text{Re}(\alpha) < 3/2$.

By absurd, let us suppose that $\text{Re}(\alpha) \geq 3/2$.  Let $\beta = \alpha^{-1}$. Then $|\beta| < 1$ and we have \[0= 1+ \sum_{i=1}^n a_i \beta^{i} = \left(1 + \frac{1}{2}\sum_{i=1}^N \beta^i\right)+ \left(\frac{1}{2}\sum_{i=1}^N \pm \beta^i \right), \quad \text{for all} \; N \geq n.\]
The sum in the left parenthesis equals $(2-\beta-\beta^{N+1})/(2(1-\beta))$ and both parentheses have the same modulus,
thus (after multiplying by $2$ and using the triangle inequality):
\[\left| \frac{2-\beta}{1-\beta} \right|  - \frac{|\beta|^{N+1}}{|1-\beta|}\leq \left| \frac{2-\beta-\beta^{N+1}}{1-\beta} \right| = \left| \sum_{i=1}^N \pm \beta^i \right| \leq \sum_{i=1}^N |\beta|^i \leq \frac{|\beta|}{1-|\beta|}.\] Letting $N \longrightarrow \infty$ and using $|\beta| < 1$, we deduce that: ${\left| \frac{2-\beta}{1-\beta} \right|  \leq
  \frac{|\beta|}{1-|\beta|}}.$ In terms of $\alpha$, this inequality reads: $\left| \frac{2\alpha - 1}{\alpha-1}  \right| \leq \frac{1}{|\alpha|-1}.  \quad \quad (1)$

On the other hand, $\text{Re}(\alpha) \geq 3/2> 3/4$, so $|\alpha -1/2| > |\alpha-1|$, which in turn gives: $\left| \frac{2\alpha - 1}{\alpha-1}  \right| >2.   \quad \quad (2)$

The inequalities $(1)$ and $(2)$ together imply that $|\alpha|<3/2$, which contradicts our assumption that $\text{Re}(\alpha) \geq 3/2$. Thus, Claim 1 is proved.

Let us now suppose that $f$ can be written as $f(X) = g(X) h(X)$, with $g,h$ (monic) polynomials in $\mathbb{Z}[X]$ of degree $>1$.

Claim 2. We have $|g(2)| \geq 2$ and $|h(2)| \geq 2$.

Now, clearly, Claim 2 establishes the initial result, since $f(2) = g(2) h(2)$ was supposed to be a prime number. It remains to prove Claim 2.

All the roots of $g$ are also roots of $f$, thus, by Claim 1, $g$ factorizes as $g(X) = \prod_{\alpha \in \mathcal{T}} (X-\alpha)$, where the product is taken over a finite set $\mathcal{T}$
of complex numbers $\alpha$ such that $|2-\alpha| > |1-\alpha|$. It follows that $|g(2)| = \prod_{\alpha \in \mathcal{T}} |2-\alpha| > \prod_{\alpha \in \mathcal{T}} |1-\alpha|=|g(1)| \geq 1$, so $|g(2)| \geq 2$.
Similarly, $|h(2)| \geq 2$. This concludes the proof.

[quote="alinbostan"]
$ \prod_{\alpha \in \mathcal{T}} |1-\alpha|=|g(1)| \geq 1$,[/quote]

How do you justify that $|g(1)|\geq 1$

Moubinool, if $ g(1)=0$, then 1 is a root of f, which has positive coefficients!  :?

There is a general rezult of this named by theorem of Cohn:
If $p$ reprezents in some baze $q$ as $a_0a_1..a_n $ then
$f(x) = a_0x^n +.. +a_{n-1}x+a_n$ is irreducible in $Z[X]$

Solution is same as [b]alinbostan[/b] did:

Let $f(n) =g(n)h(n)$
1) $Re(z) < n-1/2$
 from ($|g(n-1/2-t)| <|g(n-1/2+t)|$ for all $t>0$)

2) $f(n-1) <> 0$
3) $f(n) = g(n)h(n) > g(n-1)h(n-1)$ -contradiction.

[quote="prowler"]There is a general rezult of this named by theorem of Cohn:
If $p$ reprezents in some baze $q$ as $a_0a_1..a_n $ then
$f(x) = a_0x^n +.. +a_{n-1}x+a_n$ is irreducible in $Z[X]$.[/quote]
Indeed, the result and its proof seem to be classical, but Polya write in "Aufgaben und lehrsatze aus der analysis", that the following results is a generalization of the statement belonging to O. Gmelin:
[i]Suppose $P(x)\in\mathbb Z$, $n\in\mathbb Z$, and
1) all roots of $P$ lies in halfplane $\textrm{Re }z<n-\frac{1}{2}$;
2) $P(n-1)\neq 0$;
3) $P(n)$ is a prime number.
Then polynomial $P(x)$ is irreducible.[/i]

I probably made an extremely ugly error, but anyway I post my "solution" here:
We have to prove that [tex]f \in \{ 0, 1 \} [ X ] \wedge f ( 2 ) \in P \Rightarrow
\neg ( \exists x_0 \in Z : f ( x_0 ) = 0 )[/tex]

Suppose the contrary [tex]\exists x_0 \in Z : f ( x_0 ) = 0[/tex] it does mean that
[tex]\exists Q ( x ) \in Z [ X ] : f ( x ) = ( x - x_0 ) \cdot Q ( x )[/tex] (since Z[X] is a ring)

But because [tex]f ( 2 )[/tex] is prime we must have [tex]|2 - x_0 | = 1[/tex] (in the other
case we would get an immediate contradiction because [tex]f ( 2 )[/tex] wouldn't be
prime) it does mean that [tex]x_0 = 1[/tex] or [tex]x_0 = 3[/tex]

(Let's write [tex]f(x) = \sum_{j = 0}^n a_j \cdot x^j[/tex] where [tex]a_j[/tex] belongs to [tex]{0,1}[/tex] for [tex]j=0,1,2,...,n[/tex])

If [tex]x_0 = 1[/tex] then [tex]f ( 1 ) = \sum_{j = 0}^n a_j^{} \geqslant 0[/tex] and the only
possibility for [tex]f[/tex] is [tex]f ( x ) = 0[/tex] but then [tex]f[/tex] is irreducible. Thus for
[tex]x_0 = 1[/tex] we have or [tex]f ( 1 ) \neq 0[/tex] or [tex]f[/tex] irreducible. Contradiction

If [tex]x_0 = 3[/tex] then [tex]f ( 3 ) = \sum_{j = 0} a_j \cdot 3^j > 0[/tex]. Thus [tex]f ( 3 )
\neq 0[/tex]. Contradiction.

P.S: I'm pretty sure I made an error so I'll be happy if you point it out :)

Irreducibility means that you cannot factor polynomial $f$, i.e. there are no $g,h\in\mathbb{Z}[x]$ s.t. $f=gh$. So, it is not necessary $f$ has an integer root.

Are ou sure it works for base 2? As far as I know, this is the most difficult part in establishing the right inequality for the roots.

As I know the problem for $n=2$ is open. Look at [url=http://planetmath.org/encyclopedia/ACohnsIrreducibilityCriterion.html]here[/url].

But it is the problem that I posted, given at oral ENS and the solution presented above is fine to me, I cannot find a mistake. So, I don't understand your assertion.

Also you could see : "http://en.wikipedia.org/wiki/A._Cohn's_irreducibility_criterion". I still think the problem is open. I'll check the solutions posted here.

Did you look on the solution? It seems fine to me, but I may very well be blind.

The reference in PlanetMath about Cohn's irreducibility criterion is outdated, the criterion also holds for base 2. The reference in Wikipedia also was outdated, but I've made an anonymous contribution providing an article in American Mathematical Monthly by Ram Murty with a proof.
I'm trying to edit PlanetMath's article also, but this "object" has an owner and I do not have enough privileges; I've mailed the owner, but no answer until now. Do you have any idea to edit the article?

today I have heard the following result:
if $P \in \mathbb{Z}[X]$ satisfies $\deg(P) \leq 31$, $P$ has nonnegative coefficients and $P(10)$ is a prime number, then $P$ is irreducible ! any proof for that ?


Edit. Really dumb mistake. sorry ignore the below, dumb dumb mistake

I have this webpage, it says that "Find the equation of the line which passes through the point (-2,-6) and is parallel to the line 4x-3y=24. The answer it states is y=4/3x-10/3.
Is this correct? It didn't seem correct to me.
The second to last step it stated was that y+6=4/3(x+2), then it went on to say to simplify to y=4/3x-10/3.

Is this correct? If so, how did y+6=4/3(x+2) turn into y=4/3x-10/3? There must be something I'm missing.

Thank you.

[hide]Well, in an equation of the form $Ax+By=C$, the slope is $\frac{-A}{B}$. So $m=\frac{4}{3}$.

Now to find $b$, the y-intercept, we plug in values for x and y from the point mentioned. Solving we find $b=\frac{-10}{3}$, so now we have the equation to be $y=\frac{4}{3}x-\frac{10}{3}$.[/hide]

just make sure they have the same slope and plug in values


linear algebra
matrix
superior algebra
superior algebra unsolved

Hi how would I prove that two elements of an inverse semigroup, say e and f, are mutually inverse?
I know it is something to do with efe = e and fef = f but I just can't figure it.

Any help would be hugely appreciated.

What's the context? This is too generic to say anything about.

I,m trying to show idempotents commute, I have that if a is the inverse of ef then a is self inverse so i have a = ef. I now need to show a = fa to show that ef and fe commute, but to do this i need to show that ef and fe are mutually inverse....... i think.

Idempotents commute? In what structure?
They don't in a matrix ring: if $ A\equal{}\frac12\begin{bmatrix}1&1\\1&1\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}1&0&\\0&0\end{bmatrix}$, $ A^2\equal{}A,B^2\equal{}B$, and $ AB\equal{}\frac12\begin{bmatrix}1&0\\1&0\end{bmatrix}$, $ BA\equal{}\begin{bmatrix}1&1\\0&0\end{bmatrix}$.

If you have a group, the cancellation laws imply that the identity is the only idempotent.


In a research study conducted by a wildlife reserve agency, 1001 monkeys were randomly captured in a large forest, tagged, then released into the same forest. A week later, the agency went back into the forest, and captured 1001 monkeys at random. 77 had tags. What is the best estimate for the number of monkeys in the forest?

[hide]This is obviously not going to be exact, but the best way to do this problem is a proportion.  If about $ 77/1001$ of the monkeys are tagged, and there were $ 1001$ monkeys tagged, then we set up a proportion so that $ 77/1001 \equal{} 1001/x$.  When you cross multiply, you get that $ 1001^2 \equal{} 77x$.  So the best estimate is $ 1001*1001/77 \equal{} 13013$.[/hide]

i think that if only 77 monkeys had tags the second time that would mean that 924 mokeys didnt get tagged

soo 924 + 1001 = 1925  :idea:
That my best estimate for the # of monkeys

Ilike monkeys

1925 because you would have 1001 taged monkeys, and 924 untaged.

[hide="Answer"]$ 77/1001$ monkeys were tagged, so we can set up a proportion.
$ 77/1001 \equal{} 1001/x$ so
$ 77x \equal{} 1001 * 1001 \equal{} 1002001$ so
$ 77x \equal{} 1002001$ is $ x \equal{} 13013$[/hide]

I am not sure how you get 1995

The problem is implying a completely random way of choosing monkeys, so a random sampling shows that about $ 77/1001$ of the monkeys were tagged.  So if $ 77/1001$ of the population $ \equal{}1001$, then set up a proportion.

[quote="Funkymunk"]

I am not sure how you get 1995[/quote]

who said 1995?

I think AIMEman and Funkymunk are right. The problem with everyone else's solution is that there could (and most likely) be other untagged monkeys out in the forest other than the 924 randomly picked.

[quote="jackdillon"]after thinking about there solution is better, although either would work because we are just estimating

EDIT: never mind, I just read the question and it asks for the best possible estimate.[/quote]

Please don't look at the following content.

[hide]You looked. you fail.[/hide]

Please don't look at the following content.

[hide]Yeah, it's 13013. You work by proportion. 77/1001 monkeys could be tagged, which means that the proportion gets you to 13013. And by the way, how many times do I have to tell you until you stop looking?[/hide]

PLEASE PLEASE <--the magic word Don't look At the FOllowing Content!

[hide]Do you normally follow your teacher's directions??![/hide]

Make this a lesson to you: you don't look at stuff you're not supposed to, especially if it's stuff that's not done by a weird kid who likes tricking people.


Prove that for all $x \ge 3$, $x^{x}> f(x)$ where $f(x)$ is a power tower with $n$ $2$'s.  Find a function for $n$ that satisfies this inequality.

where is the $n$ in this problem?

if you mean $x \geq 3$ then the identity doesn't hold

$5^{5}= 3125$

$2^{2^{2^{2^{2}}}}= 2^{65536}$

Sorry, I wrote the question wrong.
Edited.

[hide="problem doesnt seem right"]
$n=1$
$f(x) = 2$

$x^{x}\geq 2$ for all $x \geq 3$

I think your question is still worded wrong
[/hide]

I mean a function that describes the highest number of $2$'s that can be stacked to satisfy the inequality.

I think the problem would make more sense if you removed the "Prove that" in the beginning. Then it makes it clear that you're asking for $t(x)=n$, i.e. the function that gives $n$ in terms of $x$.

I bet $t(n)=\left\lfloor \frac{p(x)}{q(x)}\right\rfloor$ where $p,q$ are very messy functions :P


Hi all, when r u writing BITS? Just wanted to know  :D   :maybe:

For me too  :(  Yippee even shridhar is writing on the 16th  :D   :D

Have you registered??? If it goes at this rate, I'll miss all the good dates.

[quote="pardesi"][quote="madness"]im writing on the 16th of may :D . Pardesi, arent u conforming the date today?  :huh:[/quote]
is it necesary to confirm the dates by today :)[/quote]

No you have loads of time.

Rohit, have you done it or not?

so u have to confirm a ...when is the last date for that :?:

March 11th is the last day
And its still not come for me da shreyas  :)

10th of March. Nee Porumaiya english exam ezhudithu varalam.

dei its loading only upto 4 pts da i.e. that green progress thingy at the bottom. What for u da?  :maybe:

amam thanks.....won't they send u any onfo about this...i mean how willl people get to know about this... :?:

and what haven't u got yet madness

the homepage of the sight da  :) . And its clearly there on the bits website da and all might be knowing that's why the server is sooooooooo slow  :mad:

gr8 :D
yes i wud prefer 1 AM to 4 AM :D  :D

[quote="Cyber-Shreyas"][quote="pardesi"]gr8
yes i wud prefer 1 AM to 4 AM   :D[/quote]

Hell Yes!! Anytime after 11 night is okay for me. :D

Owls rule![/quote]

That would be morn 5:30 also da  :P  :D . And i would rather prefer morn 6:00 bcos every night bird would flop and i can manage  :rotfl:

[quote="shruthibaskaran"]I cant study in the morning for nuts :D Prefer staying up late, as very evident by my highly regular sleeping patterns :D

And I managed to get my BITS Date and Time Registration done in a record time of 24 mins :D (Magic fingers rule)[/quote]
yours fingers are really magical. i have been online for 2 hours and havent even got close to getting it regeiistered. probably will try tomorrow, too pissed off now.

Shre, you don want afternoon slot coz ull sleep off.. Don sleep in the 430 slot da...
@pardesi: Mr mod, uve gotta delete this post coz its got double meanings if u did understand..

wat double meanings? Where?  :huh:  :huh:

@Varunarasu: Thank you! I rather pride my luck a lot :)

Edit: Babies cant find double meanings even if they want to...

P.S - 333rd post! Yipee!

And see our gr8 forum, 108 posts in a thread started today only  :D  :D   :rotfl:

[quote="Chapplikka"]@pardesi: Wait for shre to look at that post.. then delete it...[/quote]
sure...
rohit endhe janmethela unnale mudiyathu da :P  :rotfl:  :rotfl:

Chappli, semma signature :D

Vazhkai oru vattama da?

Btw chappli isnt that DK's orkut id? Did he get inspired by u?  :maybe:   :D

[quote="shruthibaskaran"]
Edit: Babies cant find double meanings even if they want to...

[/quote]

I swear neither can an adult. (me)

[quote="Cyber-Shreyas"]

I swear neither can an adult. (me)[/quote]
yen da ....thookerdhu ode angilam version ode double meaning :huh:


Let $p(x)=x^{n}+a_{1}x^{n-1}+.....+a_{n-1}x+1$ where $a_{i}$ s are non-ve real numbers. Assume that $p(x)=0$ has $n$ real roots. Show that
$p(2) \geq 3^{n}$.

[quote="RohanSingh"]non-ve real numbers.[/quote]

What does this mean?

[quote="Farenhajt"]I would say it's "non-negative".[/quote]

In that context:
[hide="some progress"]
since all the coefficients are positive, all $n$ roots must be negative by the descarte's rule of signs.

Then, $p(x) = (x+r_{1})(x+r_{2})\dots(x+r_{n})$ with $r_{1}, \dots, r_{n}$ being positive.  Since the last term is $1$, we know that $r_{1}r_{2}\dots r_{n}= 1$.

Therefore, this reduces to the following inequality: if we have $n$ positive reals with a product of 1, prove that:
$(r_{1}+2)(r_{2}+2)\dots(r_{n}+2) \ge 3^{n}$

since I have no idea at all about inequalities, someone can solve this...
[/hide]

[quote="Farenhajt"]Taking on where pkerichang left off:

[hide="Completion of the proof"]By AM-GM we have $r_{i}+2=r_{i}+1+1\geqslant 3\sqrt[3]{r_{i}\cdot 1\cdot 1}=3\sqrt[3]{r_{i}}$, hence $(2+r_{1})(2+r_{2})\dots(2+r_{n})\geqslant 3\sqrt[3]{r_{1}}\cdot 3\sqrt[3]{r_{2}}\dots 3\sqrt[3]{r_{n}}=3^{n}\sqrt[3]{r_{1}r_{2}\dots r_{n}}=3^{n}$, since $r_{1}r_{2}\dots r_{n}=1$. The equality holds only for $r_{1}=r_{2}=\dots=r_{n}=1$.[/hide][/quote]

The equality part:
[hide]
in other words, $p(x) = (x+1)^{n}$.
[/hide]
Just like to keep it in context of the polynomial problem :)


In a shooting match, 12 clay targets are arranged in 3 hanging columns of 2 each and 2 columns of 3 each, with the 2 columns of 3 each on the sides of the shooting panel. A marksman is to break all 12 targets accodring to the following rules:

1. The marksman first chooses a column from which a target is to be broken.
2. The marksman must breakt he lowest remaining unbroken target in the chosen column.
3. The targets on the column on the far right must all be broken before he breaks the targets on the column on the far left.
Given these conditions, in how many ways can he break all 12 targets.

Part 3. is the spin off the original, which came from the 1990 AIME problem 8.

[hide]This can be treated as 4 "columns": one for each of the center columns w/ 2 each, the fourth being [i]both[/i] outside columns w/ 6. Since the targets in each "column" must be shot in a specific order, this is the number of permutations of aabbccdddddd

which is $\frac{12!}{6!(2!)^3}=\boxed{83160}$[/hide]

if you shot down the rightmost column so that a new column is now the rightmost, would you have to shoot all of those before you shoot the leftmost one?

Good question, but I mean if the rightmost is column A, column A must be eliminated before shooting targets on the leftmost, column B, if we label the columns with letters.


geometry
LaTeX
inradius
trapezoid
rectangle
area of a triangle
similar triangles

http://mcis.jsu.edu/mathcontest/geometryans05.pdf

I did some of these, but I didn't have a good solution. As in the shortest, and concise,  and easy way.

49
48
44
32

20

18 and 23 and 36 down, 5 more elegant solutions to go  :lol:
I can't believe I forgot how to do 18. I got it right on vestavia, a similar problem.  :rotfl:

I think I got half of these right, and I got a total of 45/50 because it is an easy test that almost anyone can do.
I did some really ugly stuff, and guessed on a few.

[hide="18"]ok so draw CB. this is a sector of a circle with radius 4. so CB=AB=4. now draw AC. likwise, this is 4, so that means ABC is equilateral and CBA=60 and CAB=60. Now we see it's just the area of an equilateral triangle with side length 4 plus two circular segments. Now to find the area of the circular segments, conider sector ACB with arc AC. this is a 60 degree sector with radius 4. so [circular segment]=[sector ACB]-[triangle ACB]. so find the area of the sector and subtract the equilateral triangle to get the circular segment, then multiply by 2 to get the total area.[/hide]

[hide="20"]woah I'm actually gonna use latex for this

ok so let the leg length be $x$, thus the hypotenuse is $x\sqrt{2}$. so the perimete is $x+x+x\sqrt{2}=2x+x\sqrt{2}=k$ solving for x yields $x(1+\sqrt{2})=k\rightarrow x=\frac{k}{1+\sqrt{2}}$ rationalizing the denominator results in $x=\frac{k(2-2\sqrt{2}}$
the area is $\frac{x^{2}}{2}$ so this is $(k(2-\sqrt{2}))^{2}\times \frac{1}{2}=k^{2}(3-2\sqrt{2})$  [/hide]

[hide="23"]recall the area of a triangle is $rs$ where r is the inradius and s is the semiperimeter. writing in terms of T, we have $\frac{T}{2}\times s=T\rightarrow s=2$[/hide]

32- I have a solution, but it is very messy and ugly

[hide="49"]
let $s = \text{side length of square}$
since $AF \parallel BC$ and $m\angle AGF = m\angle CGB$, $\triangle AGF \sim \triangle CGB$.
which means $\frac{AF}{FG}=\frac{BC}{BG}=\frac{1}{3}$
$AB \parallel DE$ and $m\angle AFB = m\angle DFE \Longrightarrow \triangle AFB \sim \triangle DFE$
since $AF=\frac{s}{3}$ and $AD = AF+FD = s$, $FD = \frac{2s}{3}$
$\triangle AFB \sim \triangle DFE$ implies $\frac{AF}{FD}= \frac{BF}{FE}= \frac{1}{2}$.
so $FE = 2\cdot BF = 2(3+9)=\boxed{24}$
[/hide]

[hide="Hint for 44"] Similar triangles to find radius of $O_{2}$. [/hide]
[hide="44 Solution"] Draw vertical lines through $O_{2}$ and $O_{3}$, and a horizontal line through $O_{1}$. We have 2 similar right triangles. $r$ is the radius of $O_{2}$. $\frac{2r+8}{2}=\frac{3+r}{r-3}\iff r^{2}-15=0\Longrightarrow r=\sqrt{15}$.
Result follows. [/hide]

[hide="Hint for 32"]We can partition the trapezoid into a rectangle and two triangles (you should know how to do this by now). Let the height be $x$. Then the area of the trapezoid is $52x+\frac{13}{2}x = \frac{117}{2}x$. All you need to do is find $x$, the height. From the Pythagorean theorem, the height satisfies the equation: $\sqrt{144-x^{2}}+\sqrt{25-x^{2}}= 13$. Solve![/hide]

32

[hide]Drop an altitude from B to AD at E and another from C to AD at F.

Let the altitudes be $h$.  We know that $AE+FD=13$.

$AE^{2}+h^{2}=144$
$FD^{2}+h^{2}=25$
------------------
$AE^{2}-FD^{2}=119$
$AE^{2}-(13-AE)^{2}=119$
$26AE=288$
$AE=\frac{144}{13}$

Plugging this back in, $h=\frac{60}{13}$.

[BCFE]=$\frac{60}{13}\times 52=240$.

Scoot triangle CDF so that BE is on CF.  We get a 5-12-13 triangle.  The area is $\frac{1}{2}13\times \frac{60}{13}=30$

240+30=$\boxed{270}$[/hide]

[quote="davidlizeng"]the radii of the circles in #44 form a geometric
progression. would this be true for all problems
like these??
or is it just a coincidence?

if it is true, then these type of problems would be super easy[/quote]

im pretty sure they do

[quote="davidlizeng"]the radii of the circles in #44 form a geometric
progression. would this be true for all problems
like these??
or is it just a coincidence?

if it is true, then these type of problems would be super easy[/quote]

They should by similar triangles I believe.

But with 2 triangles you still have to show the work.

3 triangles or more you have to find the commmon ration.

[quote="davidlizeng"]the radii of the circles in #44 form a geometric
progression. would this be true for all problems
like these??
or is it just a coincidence?

if it is true, then these type of problems would be super easy[/quote]
[hide="Proof"] $r_{1}$ is the radius of circle $O_{1}$, etc. $O_{1}O_{2}=r_{1}+r_{2}$, $O_{2}O_{3}=r_{2}+r_{3}$. Let the feet of the perpendiculars from $O_{2}$ and $O_{3}$ to the horizontal line through $O_{1}$ be $A_{2}$ and $A_{3}$ respectively. $O_{2}A_{2}=r_{2}-r_{1}$, $O_{3}A_{3}=r_{3}-r_{1}$.
By similar triangles,

\begin{eqnarray*}\frac{O_{1}O_{2}}{O_{2}A_{2}}&=&\frac{O_{1}O_{3}}{O_{3}A_{3}}\\ \frac{r_{1}+r_{2}}{r_{2}-r_{1}}&=&\frac{r_{1}+2r_{2}+r_{3}}{r_{3}-r_{1}}\\ r_{1}r_{3}-r_{1}^{2}+r_{3}r_{2}-r_{1}r_{2}&=&r_{1}r_{2}+2r_{2}^{2}+r_{3}r_{2}-r_{1}^{2}-2r_{1}r_{2}-r_{1}r_{3}\\ 2r_{2}^{2}-2r_{1}r_{3}&=&0\\ r_{2}&=&\boxed{\sqrt{r_{1}r_{3}}}\end{eqnarray*} [/hide]


[hide]You jog at 6.25 mph.

$t=\frac{d}{j+5}$
$t=\frac{\frac{1}{2}d}{j}+\frac{\frac{1}{2}d}{j+50}$
$\frac{\frac{1}{2}d}{j}+\frac{\frac{1}{2}d}{j+50}=\frac{d}{j+5}$
simplify that and you eventually get $j=6.25$[/hide]

[quote="mtms5467"]I bike 5 miles per hour faster than I jog. I go by car 50 miles per hour faster than I jog. I bike to school. Then I jog halfway back and drive halfway back. It takes the same time both ways. How fast do I jog?[/quote]

[hide]$\frac{25}{4}$ mph. Since distance and time are both the same, the average speed back is the same as the speed going. So x= jogging speed and $x+5=\frac{1}{\frac{\frac{1}{x}+\frac{1}{x+50}}{2}}$ and you solve to get $\frac{25}{4}$.[/hide]


Solve for n:

The square root of [1+the square root of (2+the square root of n)]=2

Too familiar, if I may say so.

it helps to read the problem out loud and try to interpret it on paper

[quote="LynnelleYe"]If anyone wants to see it in LaTeX:
Problem--solve for n in equation
$\sqrt{1+\sqrt{2+\sqrt{n}}}=2$
Solution--
$1+\sqrt{2+\sqrt{n}}=4$
$\sqrt{2+\sqrt{n}}=3$
$2+\sqrt{n}=9$
$\sqrt{n}=7$
$n=49$[/quote]

great! i need to learn more about this. i tried, but it didn't work

[quote="marucho8d7"]

great! i need to learn more about this. i tried, but it didn't work[/quote]

You can always go to the $\text{\LaTeX}$ section of this site.

In the meantime here's how you would $\text{\TeX}$ this example.  First of all you need to put dollar signs around what you are $\text{\TeX}$ing.  On this board it doesn't matter if you use just one on each side or two on each side...it would matter if you were doing something yourself.  Next, you put in between the dollar signs what you want to be shown.  There are commands for certain things that you can find on the $\text{\LaTeX}$ page.  For yours you only need square roots.  So you would type  b/w the dollars signs:  \sqrt{ your expression here }  .  and that s it.  Additionally if you wanted to have say cube root you would do:  \sqrt[3]{ your expression here }.

That should be enough to get you started.  Also, if you place your pointer over someone else's $\text{\TeX}$ed stuff you should see the code for what they typed...that is a great way to learn new things and such.


abstract algebra
search
function
superior algebra
superior algebra unsolved

suppose the R is an abelian ring and with1.and f(x)= $ %Error. "display" is a bad command.
\sum_{i = 0}^n\ a_i x^i$ is the member of the following ring: R[x]={$ %Error. "display" is a bad command.
\sum_{i = 0}^n\ b_i x^i$ $ %Error. "display" is a bad command.
\mid \ b_i \in R$ }.it is clear that R[x] is ring.if f(x) has inverse in R[x],show $ %Error. "display" is a bad command.
\ a_1\ a_2\ ...\ a_n$ are nilpotents?

use the search function, or do it by yourself, hint: x nilpotent <=> x lies in every prime ideal


Let $ x,y,z$ be the positive reals satisfying $ xyz\minus{}x\minus{}y\minus{}z\equal{}2$.Find the minimum value of $ [x]\plus{}[y]\plus{}[z]$

[quote="SUPERMAN2"]Let $ x,y,z$ be the positive reals satisfying $ xyz \minus{} x \minus{} y \minus{} z \equal{} 2$.Find the minimum value of $ [x] \plus{} [y] \plus{} [z]$[/quote]

I guess $ [x]$ should mean floor function.

Then $ xyz\equal{}2\plus{}x\plus{}y\plus{}z \ge 2\plus{}3 \sqrt[3]{xyz} \Rightarrow (\sqrt[3]{xyz}\plus{}1)^2(\sqrt[3]{xyz}\minus{}2) \ge 0$

So $ xyz \ge 2^3\equal{}8$, $ [x] \plus{} [y] \plus{} [z]\plus{}3>x\plus{}y\plus{}z\equal{}xyz\minus{}2 \ge 6$, so $ [x] \plus{} [y] \plus{} [z] \ge 4$

Now it's easy to find an example, e.g. $ (x,y,z)\equal{}(\frac{5}{3}, \frac{9}{5},\frac{41}{15})$ such that $ [x] \plus{} [y] \plus{} [z]\equal{}4$

So the min. value is $ 4$


There is a list of seven numbers. The average of the first four is 5, and the average of the last four is 8. If the average of all seven numbers is $6$ $4/7$, then the number common to both sets of numbers is

$A.$ $5$ $3/7$
$B.$ $6$
$C.$ $6$ $4/7$
$D.$ $7$
$E.$ $7$ $3/7$

[hide]As the average of the first four numbers is 5,  their sum is 20. The average of the last four numbers is 8, so their sum is 32. Thus, altogether the sum of the 7 numbers plus the middle number is 52. The average of all 7 numbers is $6 \frac{4}{7}$, so the sum is 46. 52-46=6, answer B.[/hide]

[hide]
The first 4 numbers add up to 20.
The last 4 numbers add up to 32.
The sum of all of the 7 numbers is 46.
52-46=6, B[/hide]

[quote="pascal_1623"][hide]
The first 4 numbers add up to 20.
The last 4 numbers add up to 32.
The sum of all of the 7 numbers is 46.
52-466=6, B[/hide][/quote]

You mean 52-46=6 right?


[quote="ifai"]有一句話：「共患難易，共享福難。」如果你是一個公司的大老闆，這句話適合不過了。全體員工為公司賺了錢，怎麼把盈利分配給各位員工才算公平呢？
　　有甲、乙、丙、丁四個人，甲是一間工廠的老闆，乙、丙、丁是他的僱員。甲不能獨自經營這工廠，於是就聘請乙當經理，他們為工廠賺了100萬元。如果甲聘請丙或丁做工人，工廠就可賺500萬元。如果丙和丁都被聘請做工人，工廠就可賺2000萬元。但是如果沒有乙當經理，或者甲不再做老闆，工廠就不能運作。請問：甲、乙、丙、丁怎樣分配這2000萬元盈利才算公平呢？[/quote]

考虑到甲和乙同等重要，丙和丁同等重要，
当只有甲和乙时，每人分50万；
甲，乙和丙 或者 甲乙和丁时多挣400万，由于没考虑是丙还是丁被聘请，所以四个人分这400万，甲400/3,乙400/3，丙和丁各200/3；
当四个人时，又多挣1500万，每人分1500/4=375万，
所以，最后甲乙丙丁分别分到：1675/3万，1675/3万，1325/3万，1325/3万

[quote="Dreamer"][quote="ifai"]
甲，乙和丙 或者 甲乙和丁时多挣400万，由于没考虑是丙还是丁被聘请，所以四个人分这400万，甲400/3,乙400/3，丙和丁各200/3；[/quote][/quote]
不明白. :?:

[quote="ifai"][quote="Dreamer"][quote="ifai"]
甲，乙和丙 或者 甲乙和丁时多挣400万，由于没考虑是丙还是丁被聘请，所以四个人分这400万，甲400/3,乙400/3，丙和丁各200/3；[/quote]
不明白. :?:[/quote][/quote]

我想是这样的，甲乙两个人时挣100万。甲，乙和丙 或者 甲乙和丁时挣500万，即多挣400万，但不知道是丙还是丁被聘用，所以把这400万先分三分，甲乙先各得三分之一为400/3，剩下400/3让丙丁平分。

不知还有没有更好的办法。


Let be given a sequence of [b]positive real [/b]numbers $a_0,a_1,a_2,...,a_n$ satisfy : $a_{i-1}+a_{i+1}\le 2a_{i} \forall i=1,2,3,..$ $,...,n-1$.Prover that:
$(\sum_{i=0}^na_i)^2\ge\frac{n-1}{4}\sum_{i=0}^na_i^2$


So, I created this problem a long time ago and I was thinking that I have posted it here. However, after looking for it here for an hour I gave up that idea :huh: . Anyway, I decided to post it now, and I am sorry if this is a repost.

[color=darkred]Problem: [/color][color=blue]The circle $\omega$ touches the circle $\Omega$  internaly at the point $X$.  $Y$ is an arbitrary point on $\Omega$, and the line $YZ$ touches the circle $\omega$($Z\in{\omega}$). Prove that the value of the fraction $\frac{YX}{YZ}$ does not depand on the choice of $Y$ .[/color]

$YX \cap \omega = \{ K \}\\ YZ^{2}= pow_{\omega}Y = \bar{YX}.\bar{YK}\\ \Rightarrow \frac{YX^{2}}{YZ^{2}}= \frac{YX}{YK}= \frac{R_{\Omega}}{R_{\Omega}-R_{\omega}}=const$
[hide="Image"][img]http://i108.photobucket.com/albums/n9/PDatK40SP/464534534.jpg[/img][/hide]

[quote="PDatK40SP"]$YX \cap \omega = \{ K \}\\ YZ^{2}= pow_{\omega}Y = \bar{YX}.\bar{YK}\\ \Rightarrow \frac{YX^{2}}{YZ^{2}}= \frac{YX}{YK}= \frac{R_{\Omega}}{R_{\Omega}+R_{\omega}}=const$
[hide="Image"][img]http://i108.photobucket.com/albums/n9/PDatK40SP/543643645.jpg[/img][/hide][/quote]

I said that the circles touch each other [b]internaly[/b] and not externaly! :wink:

Sorry, sorry, I have edited my post  :)
( And I will delete this post to :D )


1. $ \ce{CH_{3}COONHCOC_{6}H_{5}} + \ce{OH^{ - }} + \Delta \to ?$

2. $ \text{1 - phenyl 2 - propynol + } \ce{H^{ + }} \to ?$

3. $ \text{N - Nitroso aniline + } \ce{HCl} \to ?$

4. $ \ce{C_{6}H_{5} - N = N - NH - C_{6}H_{4} - CH_{3}} + \ce{HCl} \to ?$
(Methyl is para)

5.
 $ \text{2 - (2' - amino phenoxy) 1,5 - dinitro benzene + n - BuLi (followed by acidification)} \to ?$

[color=blue]Question 1 has been edited[/color]

Are you sure the first one is properly written?

I have purposely written in that order and it's correct.

[hide="Number 3"]N-nitrosoaniline is simply an intermediate species in the diazotization of aniline. The final product is phenyl diazonium chloride.[/hide]

[hide="Number 4"]The final product is 4-methyl-2-phenylazoaniline.[/hide]

In number 2 you are referring to 1-phenylprop-2-yn-1-ol, right?

yes sir. And the first one was wrongly written :blush:  :oops: . I am sorry about the blunder. I have edited it.

That's what I thought.

[hide="Number 1"]This is a Lossen rearrangement. The product is aniline, after hydrolysis of the isocyanate.[/hide]

[hide="Number 2"]This seems to be a Meyer-Schuster rearrangement.[/hide]

Brilliant sir!!!! :10:  :clap:    :coolspeak:


linear algebra
matrix
superior algebra
superior algebra solved

Let A be a matrix with n lines and m columns, n>m, and let A' be its transpose. Prove taht det (AA')=0.

A is matrix with n lines and m columns, A' its transpose

rank(A.A') \leq Inf (rankA;rankA')

but rankA \leq min(n,m)=m, and rankA'  \leq min(n,m)=m

=>

rank(A.A')  \leq m < n
A.A' is a square matrix with n lines and n columns

A.A' is not inversible det(A.A') = 0


prove that $5^{2n+2}-24n-25$ is divisible by 576

[hide]The expression becomes $25*25^n-24n-25$. It is easy to see that this expression is divisible by 24.

Since $576=24^2$, divide by 24 to get $25(1+25+25^2+\cdots+25^{n-1})-n$. Modulo 24 this is $\underbrace{1+1+\cdots+1}_n-n\equiv 0$.

So $\frac{25*25^n-24n-25}{24}$ is divisible by 24 $\Rightarrow 25*25^n-24n-25$ is divisible by 576.[/hide]


linear algebra
matrix
vector
linear algebra unsolved

Hi, am new to this forum, so please forgive any transgressions here. I have this simple & basic problem (actually taken out of finance book I'm reading) whose answer eludes my mind now.

Let $ T$ be a [i]nxn[/i] real, positive semi-definite symmetric matrix
Let $ T \equal{} \sigma \sigma^T$ where $ \sigma$ is a [i]nxm[/i] matrix. Let us denote the element in the $ i^{th}$ row and $ j^{th}$ column in matrix $ \sigma$ by $ \sigma_{ij}$
Let $ \lambda_i$ and $ a_i , 1 \leq i \leq m$ be the eigenvalues and respective orthonormal eigenvectors of matrix $ T$ with $ a_i \equal{} \{a_{ki}, 1 \leq k \leq n\}$
Let $ A$ be matrix comprised of those orthonormal eigenvectors $ a_i$ as columns

Then, orthogonally diagonalize matrix $ T$:
$ T\equal{}ADA^T$ where $ D$ is a [i]nxn[/i] diagonal matrix with $ \lambda_i$ being the non-zero entries in the diagonal corresponding to the columns of orthogonal matrix $ A$ (that comprise of the eigenvectors of matrix $ T$)
Let [b][i]dz[/i][/b] be a [i]mx1[/i] orthogonal vector

Then, prove that:
$ \sum_{k\equal{}1}^m \sigma_{ik} dz_k \equal{} \sum_{k\equal{}1}^m \sqrt{\lambda_k} a_{ik} dz_k$

Thanks in advance

"a $ m\times 1$ orthogonal vector"? That "orthogonal" makes no sense.

Also, $ A$ needs $ n$ columns to make sense. Let's add in the other $ n\minus{}m$ eigenvalues and eigenvectors at the end.

OK, what could it be, if "orthogonal vector" has no context to make sense in? Let's look at what happens with an arbitrary vector. The left side is simply the $ i$th entry of $ \sigma\cdot dz$. The right side is trickier; the $ a_{ik}$ indicate that we're multiplying by $ A$, but that's the wrong size to multiply with the $ m\times 1$ vector $ dz$. $ A\cdot\sqrt{D}$ has $ ik$ entry $ a_{ik}\cdot\sqrt{\lambda_k}$, since only the $ kk$ entries of $ \sqrt{D}$ are nonzero. Cut off the right $ n\minus{}m$ columns; if $ C$ is $ \sqrt{D}$ with its last $ n\minus{}m$ columns removed, that right side is the $ i$th entry of $ A\cdot C\cdot dz$.

Could $ \sigma$ equal $ AC$, making this true for all $ m\times 1$ vectors? Well, $ ACC^TA^T\equal{}T$ if $ CC^T\equal{}D$. The $ ij$ entry of $ CC^T$ is $ \lambda_i$ if $ i\equal{}j\le m$, and zero otherwise, so that equality is true if the columns we cut off were all zero. Assume that's the case.
Now, let $ S\equal{}A^T\sigma$. We have $ S^TS\equal{}A^T\sigma\sigma^TA\equal{}D\equal{}CC^T$. That's nice, but $ S$ still has lots of freedom; if $ D$ were the identity, any orthogonal matrix would do.

All right, I'm stumped. I have no idea what "orthogonal vector" could mean in order for this to be true.

Thanks jmerry for taking the time out to try solve this even though you had so many doubts on whether the question was correctly stated in the first place.

My fault, apologies: I was trying to extact a linear algebra question from a finance paper. I'll try to remedy this here.

[b][i]dz[/i][/b] is a [i]mx1[/i] vector of orthogonal processes rather than an [i]mx1[/i] orthogonal vector. In the paper, they are actually independent weiner processes (each being $ dz_i(t)$ where the t (dropped by me in my initial question) denotes time, but you don't need to know properties of weiner processes for this. Just that $ dz_i * dz_k \equal{} 0$ [i]if i <> k[/i]

On the symmetric matrix requiring n real eigenvalues and n eigenvectors, you have a good point I must have overlooked. Ok, in my question, assume [i]m = n[/i]. Reason why I say this is that before the relevant equations in that finance paper, this assumption is talked about but it isn't anywhere obvious that the equations start from the assumption. So, let's assume [i]m=n[/i]

Thanks again

Edit: it would be nice if $ MM^T \equal{} NN^T$ implied $ M\equal{}N$ as I could then apply that for $ M \equal{} A\sqrt{D}$ and $ N \equal{} \sigma$ but that isn't always necessarily true (though I haven't seen that explicitly stated in the only undergraduate linear algebra text I have).

[quote="morhpeus19"]Edit: it would be nice if $ MM^T \equal{} NN^T$ implied $ M \equal{} N$[/quote]
Why would that have any chance at all of being true?

For instance, suppose $ N\equal{}MQ$ where $ Q$ is an arbitrary orthogonal matrix, $ QQ^T\equal{}I.$ (And there are a lot of orthogonal matrices.) Then $ NN^T\equal{}MQQ^TM^T\equal{}MM^T.$