A group of 10 Caltech students go to Lake Street for lunch. Everyone eats at either Chipotle or Panda Express. How many ways are there for them to divide themselves between the restaurants?

Each student has $ 2$ choices, so our answer is $ 2^{10}\equal{}\boxed{1024}$.

I tried it a different way and got a different answer: I put the problem into casework for (0, 10), (1, 9), ... (10,0). For (0, 10), there is only one possibility. For (1, 9), there are 10 because you choose one person out of ten to go to one of the restaurants. Then, it should be 45 to choose two people out of ten. Then 120 and then 210. When it gets to (6, 4), it should go back to 120, and so on, i.e. 1, 10, 45, 120, 210, 120, 45, 10, and 1. But when I add this up, I only get 562. Can you please tell me what I did wrong?

Thanks!

P.S. I understand why the shown explanation is correct, but I want to know why mine is correct.

[quote="thecmd999"] Can you please tell me what I did wrong?
[/quote]
Hi thecmd999!
You left out the cases $\binom{10}{5}+\binom{10}{6}$
Your casework is a correct way to approach the problem, but you did not get the correct numbers for the cases, which is the one downside of casework.
[hide="Another way to approach this problem"]
For $(10,0)$ we have 10 people, and we have to [b] chose [/b] 0 people to not be in that group of $\binom{10}{0}$
For $(9,1)$ we use the same logic as above and get $\binom{10}{1}$
For $(8,2)$ we have $\binom{10}{2}$
For $(7,3)$ we have $\binom{10}{3}$
For $(6,4)$ we have $\binom{10}{4}$
For $(5,5)$ we have $\binom{10}{5}$
For $(4,6)$ we have the same thing as $(6,4)$ or $\binom{10}{6}=\binom{10}{4}$
For $(3,7)$ we have $\binom{10}{7}$
For $(2,8)$ we have $\binom{10}{8}$
For $(1,9)$ we have $\binom{10}{9}$
For $(0,10)$ we have $\binom{10}{10}$

We have two ways to attack the casework now.  Adding up all our cases, or knowing a formula.
[b]Way 1: Adding the cases[/b]
$\binom{10}{0}=1$
$\binom{10}{1}=10$
$\binom{10}{2}=45$
$\binom{10}{3}=120$
$\binom{10}{4}=210$
$\binom{10}{5}=252$
$\binom{10}{6}=210$
$\binom{10}{7}=120$
$\binom{10}{8}=45$
$\binom{10}{9}=10$
$\binom{10}{10}=1$

$1+10+45+120+210+252+210+120+45+10+1\implies 2+20+90+240+420+252\implies 112+660+252\implies 112+912\implies \boxed{1024}$

[b]Way 2: Knowing a formula[/b]
$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}.......\binom{n}{n-1}+\binom{n}{n}=2^n$

Therefore
$\binom{10}{0}+\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{5}+\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}=2^{10}\implies \boxed{1024}$

Also, note how all the numbers in the first case are the numbers from row 10 of pascal's triangle. [/hide]

There is actually a really simple solution here.
[hide=Sol]10 people. 2 choices. So, it's choices^people. So, 2^10 = 1024.[/hide]

the answer is actually 2, because the questions asks COLLECTIVELY which means the group as a whole has 2 choices to choose from, considering they are eating collectively. :noo:


Hello,

I have noticed that, even when I feel lazy when I wake up early to go to the Gym and do something before going to work, and I think that I shouldn't go (I usually go at night to martial arts anyway), if I go I spend most of the day with an unusual good mood.

I mean, it is not that I am always angry, totally the other way, I think that I generally am in a good mood, but if I workout in the morning, then I feel like too happy all thay long, totally without stress! And I know that I am not the only one. Most of the people that I know who go early to do some excercise (jogging or whatever) spend a better day when they do than when they don't, even when that means that we probably slept ninty minutes less and spend more energy (should be more tired), we feel more energetic all day long. What do you think about it? Do you have a "scientific" explanation to this? I have a theory, but I want to see what people think about it.

Best,

i don't exercise much, especially not in the morning. i don't know if it puts me in a good mood when i do exercise, but i can't recall ever being in a bad mood.

I do track and cross country which are after school so I usually don't workout in the morning.  However, I have gotten up a few times during the offseason to workout early in the morning.  I remember one time when I did this that it was really cold outside but I still went running.  I was also a little sick at the time.  After I finished the run and took a shower I felt so much better and had a great day.

I think that one feels good after a morning workout because it gets them breathing and pumping blood.  Consequently I think this rids their body of some of the wasteful materials inside of it.  There have probably been biological studies undertaken before to study this though.  It would be interesting to hear what the experts say.

i'm in track and tennis. tennis is no camp cupcake. we have to run everywhere and do sprints, exercises, etc. for 3 hrs. track is okay cause all you do is run. running is fun :D exercise doesn't make me feel great, but its good cause you get to meet other people and talk (although not when you're doing sprints) and you get a sense of accomplishment (much like solving a math problem)

I do pushups and sit ups every night.

And it makes me feel good doing them, just exerting physical energy and pushing myself to do as many as I can. I don't have the willpower in the morning, cause I'm usually waking up late anyways, but doing them at night helps me sleep better and I feel better in the mornings.

I agree.  I always feel better after working out in the morning, even if I have to sacrifice some sleep to do so.  (Typically I feel less tired during the day than if I hadn't worked out.)

[quote="filletwho"]Hmm...I go with the sleep and I'm happy too.[/quote]

I agree that sleeping is wonderful, but I would like to be able to sleep a lot less without having bad consecuences. And this make me remember a quote that I read somewhere, but don't remember who said that.

[quote]Never believe all what you hear, eat all what you can or sleep all what you want[/quote]

Sleeping is necesary, but think that every minute you are oversleep is a minute that you are not spending doing things that you like, let's say, some sport, math, dating, talking nonsence on these fora!  :D .

i have a routine of sit-ups and crunches that i do every night b4 i go to bed that work all the parts of my abs. one day i just felt like doing them b4 bed, and now it's become a habit. i dunno, it just makes me feel better or something. but it also helps me sleep better...lol

Haha...I'm not gonna discuss my workout schedule because I really need to renew my membership at the gym...but ANYWAYS...

girls shouldn't try to build muscles. seriously. ripped girls just...just no...no...and no...

[quote="MithsApprentice"]girls shouldn't try to build muscles. seriously. ripped girls just...just no...no...and no...[/quote]

Well MithApprendice, I disagree with you. I mean, I agree that those women with a 15'' or 20'' circumference arms, and little squares in their abs look too masculine for a woman (in my personal opinion), but any way I think that a girl with a nice figure, shouldn't have extra skin in the triceps area, neither in the legs, and some times even skiny girls have, and that is because there is no tention in the muscles, because they are too little and too weak to have nice tension.

Also, women in general shouldn't worry too much about doing too much weights. I have heard too many women saying that they don't do any weights because they don't want big muscles. Well, the problem would be if you did want them. The chemestry in the male and female bodies is very different. Males have tons of testostenone, I believe in average twenty times more than females (it has been eight years since the last time I studied something related), that is why women have almost no hair in the face, arms, chest and legs, in comparation with males, and also, that is why they have a soft voice, nevertheless, it is also why it is a lot harder for women to grow muscle. Therefore, girls, I do recomend you to do sit-ups, some light weights, and why not, some push-ups!

Best,

[quote="cats..."]exercising makes the body release endorphins, which improves your mood.[/quote]

Sleeping makes your body release... en-dolphins, which uh... improves your sleeping...skills.

But it is true with the excersize. Running makes me happier. sorta.

I do Tea Kwon Do three evenings a week and thats pretty cool, I always sleep better those nights...

I play football and basketball so i dont exercise outside much...but i do sometimes and when i do i dont get in a bad mood or anything.....

I dont exercise in the mornings, because then I would have to wake up really early, because school for me already starts too early. I like exercise when I play games, i.e. soccer and frisbee. I always feel great after I have given my all out on a field, because it gives you a feeling of accomplishment, and though you are tired, pushing to the limit makes you stronger and for me, I feel better if I completely tire myself.

I usually play in the afternoons, in fact, I will play ultimate frisbee in about 1/2 half hour for 3 hours. :D

Yet, I do not like straight out running, because (dare I say this!) it is boring, because you are alone and I feel less of an accomplishment as oppossed to a game. Although I do beleive it is necessary for training, I would not just do running.

[quote="djimenez"][quote="MithsApprentice"]girls shouldn't try to build muscles. seriously. ripped girls just...just no...no...and no...[/quote][/quote]

i agree with djimenez on this one. i don't do my routine to get "buff" or anything like that. i do it to slim down a little bit, and to remain in good shape and not have that extra flab that no one likes. now, i do think that some people overdo it a little bit. when the girls go all out and get more muscle than most guys, that's just wrong, but that's not what most girls are aiming for.

i spend about half an hour at night running, along with the usual warm ups and cooldowns...push ups/sit ups. Used to do it in the morning, but then I got too lazy...

i, (apparently like mcalderbank) love ultimate frisbee, play at school a lot...
(me and some friends played three 6 inch deep snow....)

I lift weights pretty much every day. On day 1, I train upper body like chest, biceps and all those stuff. On day 2, lower body, like legs. Then I repeat the pattern of upper body and then lower body to let the muscle to restore.

Sometimes at night I also go lift weights. To get the extra energy off so I can go to sleep...


limit
trigonometry
logarithms
real analysis
real analysis unsolved

Find the limit  :

 $ A\equal{} \lim \sqrt[n] { 1 \plus{} \sin n}$

 as $ n \rightarrow \plus{}\infty$

 Babis

Intuitively it must exist and it is 1.
I mean $ \sin(n)$ is never +1 or -1. Let $ n(\epsilon)$ be the least natural number such that $ |\sin(n) \minus{} 1| < \epsilon$. The idea is to show that $ n(\epsilon)$ grows sufficiently fast when $ \epsilon$ decresases to 0. I think this must be easy enough.

hello, we have $ \lim_{n \to \infty}\sqrt[n]{1\plus{}\sin(n)}\equal{}1$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have $ \lim_{n \to \infty}\sqrt [n]{1 \plus{} \sin(n)} \equal{} 1$
Sonnhard.[/quote]

Yes :) , but how to demonstrate this ?! I think we must find something simple to explain it to high school students.
 This is my main attempt.

 Babis

In other words, you want to show that $ |\pi\minus{}\tfrac mn|\ge c_\delta e^{\minus{}\delta n}$ for all $ \delta>0$. Alas, I do not know any proof of that that would be easy to explain to high school students though Beuker's and Ball's approaches come pretty close to fitting this bill. The standard "school proof" of irrationality gives you something like $ \frac 1{n!}$, which is enough for showing that $ \lim_{n\to \infty}\sqrt[n^2]{1\plus{}\sin n}\equal{}1$, say, but not for resolving your original question.

We say that a real number $ \alpha$ has an  irrational measure $ \mu(\alpha) \geq 2$ provided that,
for any $ \epsilon > 0$, there exist a positive integer $ q_{0}(\epsilon)$ satisfying

$ \left| \alpha \minus{} \frac {p}{q} \right| \geq q^{ \minus{} \mu \minus{} \epsilon}\,.$

for any integer  $ q > q_{0}(\epsilon)$.
\\
note  that :
$ 2^{1/n} > (1 \plus{} \sin n )^{1/n} \geq (1 \minus{} |\sin n |)^{1/n}$

In 1953 K.Mahler was able to show for the first time that $ \mu(\pi)$ is finite (thus, $ \pi$   is not Liouvillian). More precisely, he showed that $ \mu(\pi) < 42$.
Unfortunately, explicit construction for rationals approximating  $ \pi$ is very hard to find. In fact,
it was only in 1993 that Hata Has found
 rational numbers , which yields the best irrationality measure for $ \pi$ today.

in your problem interesting is when $ \sin n \approx 1$ or when $ n \approx (k \minus{} 1/2)\pi$, but since $ \mu (\pi) < \infty$
thus $ |\pi p \minus{} n| \geq C_{\epsilon} \frac {1}{p^{\mu (\pi) \minus{} 1 \plus{} \epsilon}}$
for such $ p$ and $ n$ for which $ n/p$ is close to $ \pi$ we can assume that $ n/p > 1$ thus
$ (1 \minus{} |\sin n|)^{1/n} > \left( \frac {C_{\epsilon}}{n^{\mu (\pi) \minus{} 1 \plus{} \epsilon}} \right)^{1/n}$ for all $ n > n_{0}$ thus
$ \lim \left( \frac {C_{\epsilon}}{n^{\mu (\pi) \minus{} 1 \plus{} \epsilon}} \right)^{1/n} \to 1$
in other words, you can solve this problem if and only if you know that  $ \pi$ is not liouvillian https://www.artofproblemsolving.com/Wiki/index.php/Pi

$ \sqrt[n]{1\plus{}\sin n}\equal{}\exp \left\{ \frac{\ln (1\plus{}\sin n)}{n} \right\}\equal{}\exp \left\{ \frac{\sin n}{n}\cdot \frac{\ln (1\plus{}\sin n)}{\sin n} \right\}\to 1$ as $ n\to\infty$ since $ \frac{\sin n}{n},\,\frac{\ln (1\plus{}\sin n)}{\sin n}\to 0$ as $ n\to\infty.$

[quote="Croata"]$ \frac {\ln (1 \plus{} \sin n)}{\sin n}\to 0$ as $ n\to\infty.$[/quote]
this is not true  :(
try to show that ,set $ \left\{\frac {\ln (1 \plus{} \sin n)}{\sin n} \right\} _{n \equal{} 0}^{\infty}$ is dense on $ [\ln 2 ,\infty)$

I think if we can proof $ \forall c\in\mathbb{R}^{\plus{}},\lim_{n\rightarrow\infty}\sqrt[n]{c}\equal{}1$, then the problem is solved. Isn't it :?:

No, no , no ..  it's not enough.
 $ \lim_{n \to \infty} (c)^{1/n} \equal{} 1$
rewrite it as : $ \lim_{n \to \infty} e^{\ln c /n}\equal{}1$ and   it is same as to prove $ \lim_{n \to \infty}\frac {\ln c}{n} \equal{} 0$ this is obviusly. isn't it ?

what fedja has written, it's more weak  condition on $ \pi$ than what I have written  about "$ \pi$ is not Liouvillian". but if you can to prove fedja's condition, then you   can   prove  simply that limit is equal to $ 1$.

[quote="Timestopper_STG"]I think if we can proof $ \forall c\in\mathbb{R}^{ \plus{} },\lim_{n\rightarrow\infty}\sqrt [n]{c} \equal{} 1$, then the problem is solved. Isn't it :?:[/quote]

No, because for any $ c > 0$ we have that $ 1 \plus{} \sin n$ will sometimes fall into the interval $ (0, c)$.

Oops :blush:  Now I understand it. Thx JBL :lol:


Propose a synthesis of 2-methylcyclohex-2-enone from cyclopentanone.

1. Cyano hydrin formation
2. Reduction to amine
3. Tiffeneau-Demjenov rearrangement to cyclohexanone
4. Crossed aldol condensation with formaldehyde giving 2-(hydroxymethyl) cyclohexanone
5. Dehydration using conc. sulphuric acid at 170 C

That looks ok, but isn't there anything shorter?


Does anyone know of any good sources of carpet theorem (if two carpets which cover a room entirely are moved then the area of overlap is the area not covered by either) problems or some other Area/Volume problems?

There's a complete section on carpets theorem in Mathematical Olympiad Treasures by Titu Andreescu and Bogdan Enescu. It has some nice problems as well.  :huh:

Those problems are WAAAAY too easy though.


analytic geometry
calculus
inequalities
calculus computations

Find the dimentions of the rectangular box having largest volume which lies in the first octant with thtre of its sides in the coordinate planes, their common vertex at the origin, and the opposite vertext on the plane with equation x + 3y + z = 3.

I figured that the point on the plane is simply a constraint and so I subbed it into the simple volume formula V = xyz. for x I set y and z to zero, for y I set x and z to zero...etc. From that I found the volume to be 9 units, looking at the plane equations and and z are the same and y is 3 times larger. I devided the volume by 5 units and used the value for x and z dimentions and 3 times the value for y. Im am curious if my method even works as I do not have a solution to refer to. Thanks.[/code]

The maximum volume is $\frac13$, and I didn't use any calculus to get that. Your method is nonsense.

Well If you want a calculus solution, here it is, you want to maximize $xyz$ with a constraint of $x+3y+z=3$
$\nabla f = <yz,xz,xy> = \lambda \nabla G = \lambda<1,3,1>$.
So now we have a systems of equations
$yz=\lambda$
$xz=3\lambda$
$xy=\lambda$

From the first and second equation, we see that $x=z$, from the second, we now have $x^2=3\lambda \Rightarrow \lambda = \frac{x^2}{3}$.  Plugging this into the first equation gives us $yx=\frac{x^2}{3} \Rightarrow y = \frac{x}{3}$.
We now put these values into the constraint
$x+3\left(\frac{x}{3}\right) + x = 3$.
$x=1 \Rightarrow z=1 \Rightarrow y=\frac{1}{3}\Rightarrow xyz = \frac{1}{3}$.

I think what Jmerry did was see that the plane intersects the axes at the points $(3,0,0)$, $(0,1,0)$, and $(0,0,3)$ and took the centroid of the triangle $(1,\frac{1}{3},1)$ and multipled the coordinates because that is where the vertex of the box would touch.

Also, it is well known that inorder to maximize a box like this you set $x=3y=z$.

thanks for that answer.  It really helped.

Somewhat more nifty:

By AM-GM,

$xyz=xy(3-x-3y)=\frac {1}{3}\cdot x\cdot 3y\cdot (3-x-3y)\leq \frac {x+3y+3-x-3y}{9}=\frac {1}{3}$

We have equality iff $x=3y=3-x-3y$, that is, if $x=1,y=1/3,z=1$

could you explain how you set up that inequality, and explain how it really works.  Thanks.

The AM-GM inequality basically states that the arithmatic mean of a set of numbers is always greater than or equal to the geometric mean of the set, and equality holds when every element in the set is the same.


Prove that the fraction
\[\frac{n^{3}+2n}{n^{4}+3n^{2}+1}\]
is always irreducible.

$(n^{4}+3n^{2}+1, n^{3}+2n)=(n^{2}+1,n^{3}+2n)=(n^{2}+1, n)=(1,n)=1$

using the euclidean algorithm....so the numerator and denominator share no common factor, hence that fraction is irreducable


trigonometry
ratio
geometry
perimeter
area of a triangle
Heron\u0027s formula
trig identities

I forgot how to do Law of Sines. Can you teach me?

[quote="alanchou"][url=http://www.artofproblemsolving.com/Wiki/index.php/Law_of_Sines]Law of Sines[/url][/quote]
I still don't get it :cursing:

[quote="zapi2007"]In any triangle,
the ratios of the sin to the opposite side are all equal
$ sinA / a \equal{} sinB / b \equal{} sinC / c$[/quote]

ok.. so their all the same number?

SinA/a= sinB/b= sinC/c

sinA is sin of angle A and a is the side opposite of angle A .. same with b and c

Which are the same number?

A and a

or sinA/a and sinB/b and sinC/c  ?

[quote="airyork6=)"]Huh? I'm confused  :maybe:  :huh:
My computers so messed up I cant download the attachment...[/quote]

looks like someone isnt studying hard at school or doing there homework..... you copied my user....

Oh alright
Well all it showed was that side a is opposite to angle A. Haha
Okay

 $ \dfrac{sinA}{a} \equal{} \dfrac{sinB}{b} \equal{} \dfrac{sinC}{c}$

A, B, and C are the angles.
a, b, and c are the sides.

Side a is opposite from angle A, side b is opposite from angle B, and side c is opposite from angle C.
Now, say you have:

Angle A= 75 degrees
angle B= 26 degrees
and side c= 12

To find angle C: $ 75 \plus{} 26 \plus{} c \equal{} 180$ ..... and $ C \equal{} 79 \circ$
Now, use the formula
$ \dfrac{sinA}{a} \equal{} \dfrac{sinB}{b} \equal{} \dfrac{sinC}{c}$
$ \frac {sinC}{c} \equal{} \frac {sin(79)}{12}$
$ \frac {sinA}{a} \equal{} \frac {sin(75)}{a}$
Now remember that $ \frac {sinC}{c} \equal{} \frac {sinA}{a}$
So...
$ \frac {sin(79)}{12} \equal{} \frac {sin(75)}{a}$
Then solve for a... (to make it a little easier.. take the reciprocal of both so a is on the top)
which should give you $ \approx11.81$
I think... Meh calculator is screwed up.

ok.. thx! =)
I know how to find area..I think..
How do you find x or y then?

No problem =)

Yeah area is just $ \frac{1}{2}bh$

What x or y?

opps never mind...
I was thinking wrong...now..
How do I change a scientific notation number into a regular number?
I'm trying to show my work to how to get x and y..

left,right,down
3 angles=100,42,38

left,right,up
sides=29,x,y

x=3.15x10(to the first)
y=4.64x10(to the first)
area=4.50x10(to the second)

I'm tying to solve for those, even though I know the answer...
I bet its your turn to be confused  :lol:

Haha. Yes, I am confused now. =]
Alright.
Say you have $ 1.2 \times 10^3$
You move the decimal 3 places to the right. Which gives you...: $ 1,200$
$ 1.45 \times 10^5 \equal{} 145,000$
So, whatever power 10 is to... that is how many places you move the decimal.
Oh. $ 3.15 \times 10^1\equal{} 31.5$
$ 4.64 \times 10^1 \equal{} 46.4$
$ 4.50 \times 10^2 \equal{} 450$
Ah.. confused

[quote="Cyrazeno"]Haha. Yes, I am confused now. =]
Alright.
Say you have $ 1.2 \times 10^3$
You move the decimal 3 places to the right. Which gives you...: $ 1,200$
$ 1.45 \times 10^5 \equal{} 145,000$
So, whatever power 10 is to... that is how many places you move the decimal.
Oh. $ 3.15 \times 10^1 \equal{} 31.5$
$ 4.64 \times 10^1 \equal{} 46.4$
$ 4.50 \times 10^2 \equal{} 450$
Ah.. confused[/quote]
ahh ok =)
yes.. I don't know how to make the picture so..
I made you confused   :D

[quote="Cyrazeno"]Haha yeah

Download inkscape, save the picture, add attachment on this, and there it is =P[/quote]

ok  =D

[quote="airyork6=)"]I know how to find area..I think..[/quote]

Directly related to the Law of Sines, you can use area $ \equal{} \frac{1}{2}ab \sin C$

You can also use Heron's Formula ($ \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$, where $ s$ is the semi-perimeter, and $ a$,$ b$, and $ c$ are the sides) to find the area.

Also: The Law of Sines are used with triangles that only give you AAS, or SSA. (I might have missed one).

[quote="azjps"][quote="airyork6=)"]=) I know this one =) theres also SAS =)[/quote]

You use the law of cosines for that (I suppose you could use law of sines to get an expression in one variable, but it would be kind've ugly).[/quote]
hmmm... now can someone teach me law of cosines?  :(  I don't get it

For SAS, you are given two sides and the angle between them. Let the two given sides be $ a$ and $ b$, and the angle be $ \angle C$. Then the law of cosines says that
\[ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2 \cdot a \cdot b \cdot \cos \angle C
\]
After you find $ c$, you can use the law of sines to solve the remaining angles.

Edit: The answer to your question is no, it also works for SSS if you're trying to find an angle.


induction
algebra
polynomial
vector
superior algebra
superior algebra unsolved

Let $K$ be a field, $charK = p\geq 2$ and $m,n$ natural numbers with $(m,p)=1$. Let $A,B\in M_n(K)$ such that $A^{p^{m}}=A$ and $B^p=B+I_n$. Show that it exists $X\in M_n(K)$ such that $AX+XB=I_n$

Here are the ideas: it is classical that the map $f(X)=AX+XB$ is injective (thus bijective) if and only if $A$ and $-B$ have a common eigenvalue. I think you can even find on the forum some proofs of this fact. If not, I will supply one. Anyway, with this observation it follows that it suffices to prove that $A$ and $-B$ have no common eigenvalues. But if $x$ were such an eigenvalue, we would have $x^{p^m}=x$ and $(-x)^p=1-x$. Suppose $p>2$. Then $x^p=x-1$ and since we work in characteristic $p$ we have $x^{p^2}=(x-1)^p=x^p-1=x-2$ and in general (trivial induction), $x^{p^k}=x-k$. If you put $k=m$ you will find that $p$ divides $m$, false. The case $p=2$ is identical.

[quote="harazi"] $f(X)=AX+XB$ is injective (thus bijective) if and only if $A$ and $-B$ have a common eigenvalue. [/quote]
1) If they have or have not a common eigenvalue ?
2) Could you show me the proof please? I can't find it on the forum.
3) Why do you say that if it is injective, it is also surjective? K is not finite field.

Let A,B in M(n,K) the equation AX+XB=I has a unique solution X in M(n,K)
iff (A and -B has non common eigenvalue)

I like your formula. Is the one discovered by the Chudnovski brothers? Anyway, the proof of the fact that the application $f(X)=AX-XB$ is injective if $A,B$ do not have common eigenvalues is really simple. And then, it is not necessary for the field to be finite, all I used is that in finite dimensional vectorial spaces, injectivity and bijectivity of linear operators are the same thing. Here is a sketch of a solution for the injectivity: assume that $X$ is not zero and let $r$ its rank. Then $X$ is equivalent to $J_r$. It is immediate (by writing $X=PJ_rQ$ with $P,Q$ invertible and then considering $P^{-1}AP$ and $Q^{-1}BQ$)  to see that in fact we can assume that $X=J_r$. Now, write each of $A,B$ as block matrices and perform the computations. The problem will solve itself from this point.

First we show that the polynomials $X^{p^m}-X$ and $X^p-X-1$ are coprime.

Suppose they aren't and let $a$ be a root of both of them.
Then $a^p=a+1\Rightarrow a^{p^2}=a^p+1=a+2\Rightarrow a^{p^3}=a^p+2=a+3\Rightarrow\ldots\Rightarrow a^{p^m}=a+m=a$. Contradiction with $(m,p)=1$.

So this polynomials are coprime.

We will prove that the map $f(X)$ is injective. Suppose there is an $X\in M_n(K)$ not null, so that $AX=-XB$.

We get then that $O_n=(A^{p^m}-A)X=-X(B^{p^m}-B)=-XmI_n$. So $X=O_n$.

We have thus that the map $f: V\to V$  is injective ($V=M_n(K)$) and because it is morphism and $dim_KV=n$ finite results it is bijective. So the conclusion follows.

Also what harazi said that if $A,B\in M_n(K)$, $A$ and $B$ have no common eigenvalues then $f: M_n(K)\to M_n(K), f(X)=AX-XB$ is injective(thus bijective).

Now if $P$ and $Q$ are the characteristic polynomials of $A$ respectively $B$ and because $A$ and $B$ have no common eigenvalues it follows that $(P(X),Q(X))=1$, so there exist polynomials $u, v$ so that $uP+vQ=1$.
Now suppose $f$ is not injecitve. There is a$X\neq O_n$ so that $AX=XB$. Then for any polynom  $g$ we have that $g(A)X=Xg(B)$, then $O_n=P(A)X=XP(B)$ but $XQ(B)=O_n\Rightarrow XP(B)=O_n=XQ(B)$ then $XP(B)u(B)=O_n=XQ(B)v(B)\Rightarrow$ $O_n=X(P(B)u(B)+Q(B)v(B))=XI_n=X$ contradiction. So $f$ is injective. and because it is a morphism of finite dimensional vector spaces results it is bijective.


I approached this as a distance rate problem. First call the amount typed by both of them D and you get Daniel types D=30t and Paul types D=70t. The time that Daniel typed is D/30=t and the time for Paul is D/70=t. Then add these together to get 10D/210=2t, which simplifies to D/21=2t. Then multiply both sides by 21 to get [hide="answer"]D=42t, where 42 is the rate[/hide]

Sorry, I don't have latex.

You dont need latex to use it online.
Just use dollar signs.
You only need LaTeX downloaded to your computer if you're trying to write an article or something.

Well, I think it is too complicated. As long as there is nothing too complicated, I can express it with regular keyboard keys.

You can also just say that the article is a convenient number of words. Let's say it's 420 words, so they each type 210 words. It takes daniel 7 minutes and paul 3 minutes. So the total is 10 minutes for 420 words or [b]42 wpm[/b]

[hide="<20 second problem"]um it is simply the harmonic mean between 70 and 30 so it is $ \frac{2}{\frac{1}{30} \plus{} \frac{1}{70}}$ = $ \frac{2}{\frac{1}{21}}$ = $ \fbox{42}$[/hide]

[quote="cognos599"][hide="<20 second problem"]um it is simply the harmonic mean between 70 and 30 so it is $ \frac {2}{\frac {1}{30} \plus{} \frac {1}{70}}$ = $ \frac {2}{\frac {1}{21}}$ = $ \fbox{42}$[/hide][/quote]

k, thank u guys,

[quote="cognos599"][hide="<20 second problem"]um it is simply the harmonic mean between 70 and 30 so it is $ \frac {2}{\frac {1}{30} \plus{} \frac {1}{70}}$ = $ \frac {2}{\frac {1}{21}}$ = $ \fbox{42}$[/hide][/quote]

this fact comes from what shentang did, but it's pretty helpful to know

I saw that problem where you used harmonic mean to find the rate where that guy typed 30 words per min and that other guy typed 70 words per min.  I never knew what the harmonic mean was useful for until now.  I'm going to look online for uses of it.  You have inspired me, ant!
Time to go look for some knowledge...

[quote="verizonwires"]Daniel types 30 words per minutes and Paul types 70 words per minutes. They each type one half of an article. In words per minute, what is the rate at which the artcile was typed?

I think this question is leaving out sum key parts[/quote]

There is not enough information ethier :(


algebra
polynomial
superior algebra
superior algebra unsolved

Hi, I'm having trouble to understand the piece of explaination from book [i]Topics In Algebra[/i] By Herstein. page 157.
[quote="Herstein in Topics In Algebra"]
   Let $F$ be the field of rational numbers and conseider the polynomial $p(x)=x^3-2$ in $F[x]$. As easily verified, it is irreducible over $F$, whence $F[x]/(x^3-2)$ is a field. What do its element look like? Let $A=(x^3-2)$, the ideal in $F[x]$ generated by $x^3-2$.

    Any element in $F[x]/(x^3-2)$ is a coset of the form $f(x)+A$ o f the ideal $A$ with $f(x)$ in $F[x]$. Now, given any polynomial $f(x) \in F[x]$, by the division algorithm, $f(x)=t(x)(x^3-2)+r(x)$, where $r(x)=0$ or $\text{deg } r(x) < \text{deg } (x^3-2) =3$. Thus $r(x)=a_0+a_1x+a_2x^2$  where $a_0$, $a_1$, $a_2$ are in $F$; consequently $f(x)+A=a_0+a_1x+a_2x^2+t(x)(x^3-2)+A=a_0+a_1x+a_2x^2+A$ since $t(x)(x^3-2)$ is in $A$, hence  by the addition and multiplication in $F[x]/(x^3-2)$, $f(x)+A=(a_0+A)+a_1(x+A)+a_2(x+A)^2$. If we put $t=x+A$, then every element in $F[x]/(x^3-2)$ is of the form $a_0+a_1t+a_2t^2$ with $a_0$, $a_1$ , $a_2$ in $F$. What about $t$? Since $t^3-2=(x+A)^3-2=x^3-2+A=A=0$ (since $A$ is the zero element   of $F[x]/(x^3-2)$ we see that $t^3=2$.
[/quote]
I don't understand why $t(x)(x^3-2)$ is in $A$. what are elements of $A$ looks like?
isn't it integer multiple of $x^3-2$ (i.e $2(x^3-2)$)?

And how to get $(a_0+A)+a_1(x+A)+a_2(x+A)^2$ (please explain it in details)?

and about $a_0+a_1t+a_2t^2$
i'm doing the subtitution $t=x+A$ but i get as usual $(a_0+A)+a_1t+a_2t^2$ what next?

Thanks in Advance  :oops:  :)

[quote="Beginner"]Hi, I'm having trouble to understand the piece of explaination from book [i]Topics In Algebra[/i] By Herstein. page 157.
[quote="Herstein in Topics In Algebra"]
   Let $F$ be the field of rational numbers and conseider the polynomial $p(x)=x^3-2$ in $F[x]$. As easily verified, it is irreducible over $F$, whence $F[x]/(x^3-2)$ is a field. What do its element look like? Let $A=(x^3-2)$, the ideal in $F[x]$ generated by $x^3-2$.

    Any element in $F[x]/(x^3-2)$ is a coset of the form $f(x)+A$ o f the ideal $A$ with $f(x)$ in $F[x]$. Now, given any polynomial $f(x) \in F[x]$, by the division algorithm, $f(x)=t(x)(x^3-2)+r(x)$, where $r(x)=0$ or $\text{deg } r(x) < \text{deg } (x^3-2) =3$. Thus $r(x)=a_0+a_1x+a_2x^2$  where $a_0$, $a_1$, $a_2$ are in $F$; consequently $f(x)+A=a_0+a_1x+a_2x^2+t(x)(x^3-2)+A=a_0+a_1x+a_2x^2+A$ since $t(x)(x^3-2)$ is in $A$, hence  by the addition and multiplication in $F[x]/(x^3-2)$, $f(x)+A=(a_0+A)+a_1(x+A)+a_2(x+A)^2$. If we put $t=x+A$, then every element in $F[x]/(x^3-2)$ is of the form $a_0+a_1t+a_2t^2$ with $a_0$, $a_1$ , $a_2$ in $F$. What about $t$? Since $t^3-2=(x+A)^3-2=x^3-2+A=A=0$ (since $A$ is the zero element   of $F[x]/(x^3-2)$ we see that $t^3=2$.
[/quote]
I don't understand why $t(x)(x^3-2)$ is in $A$. what are elements of $A$ looks like?
isn't it integer multiple of $x^3-2$ (i.e $2(x^3-2)$)?

No, it is any multiple of $x^3-2$ (any elements in $F[x]$)

And how to get $(a_0+A)+a_1(x+A)+a_2(x+A)^2$ (please explain it in details)?

Since it's polynomial multiple it should be clear.

and about $a_0+a_1t+a_2t^2$
i'm doing the subtitution $t=x+A$ but i get as usual $(a_0+A)+a_1t+a_2t^2$ what next?

basically A is the additive identity in the quotient ring, so..
Thanks in Advance  :oops:  :)[/quote]

Thanks Soarer.

I should realize that when i typed $A$ an Ideal  :oops: . Now the other two questions are also answered.

i need futher explaination (sorry if I distrub anybody, but i don't know where to ask about this material since it is not my course material)

If $A$ is a multiple of polynomial then i can understand why $f(x)+A=a_2x^2+a_1x+a_0+A=a_2(x+A)^2+a_1(x+A)+a_0+A$.

How to get that using addition and multiplication on $F[x]/(x^3-2)$?  I have verified that by expanding that expresion (and then we can work backward since $a_0,a_1,a_2$ is in a Field).
It seem that it's not a usual way to obtain that (the book state that directly as if it is really trivial). can we proceed like this

$a_2x^2+a_1x+a_0+A=(a_0+A)+(a_1x+A)+(a_2x^2+A)$ (by addition of coset in $F/(x^3-2)$)
$(a_0+A)+(a_1x+A)+(a_2x^2+A) = (a_0+A)+(a_1+A)(x+A)+(a_2+A)(x^2+A)$ (by multiplication in $F/(x^3-2)$)
$(a_0+A)+(a_1+A)(x+A)+(a_2+A)(x+A)^2$ also by multiplication.  How to obtain $a_2(x+A)^2+a_1(x+A)+a_0+A$ ?

About $a_2t^2+a_1t+a_0+A$ since $A$ is additive identity in $F/(x^3-A)$ and $a_2t^2+a_1t+a_0$ in $F/(x^3-A)$ we have $a_2t^2+a_1t+a_0+A=a_2t^2+a_1t+a_0$, right?

But it is using $a_2t^2+a_1t+a_0 \in F/(x^3-A)$,  that's why we make the subtitution for, to show that every element in  $F/(x^3-A)$ is of the form  $a_2t^2+a_1t+a_0$. so any other explaination?

thanks again.

1. It's not surprising that you don't get to the formular, because the following definition is ... was ;) unknwown to you:

$a (p + A) : = ap + A$

You're invited to check, that this is well-defined :-).

2. Of course, $a_2 t^2 + a_1 t + a_0 + A = a_2 t^2 + a_1 t + a_0$ is wrong. It's right, that $A$ is additive identity in $F[x]/(x^3-2)$ (by the way, you have often changed the notation of this field ;)), but this just shows us $(p+A)+A = p+A$ for all $p \in F[x]$.

However, it is justified to [i]identify[/i] $a_2 t^2 + a_1 t + a_0 + A$ with $a_2 t^2 + a_1 t + a_0$.

By the way, the whole thing can be immediately generalized: If $F$ is a field and $f \in F[x]$, then $F[x]/(f)$ is represented by the polynomials of $F[x]$ of lower degree than $f$. The factor ring $F[x]/(f)$ is also a $F$-algebra and is a field if and only if $f$ is irreducible. This is the formal way of adjoining a root of a polynomial to a field - the first step in field theory.


modular arithmetic
pigeonhole principle
number theory
prime factorization
Ring Theory
number theory unsolved

can any one give a proof of the well known lemma- " if a|bc and g.c.d(a,b)=1 then a|c" without using the theorem " if g.c.d(a,b)=1,then there exist integers x,y such that ax+by=1" ?  If not, then i've found a mistake in the book 'problem solving strategies' of 'arthur engel'. i give the sketch of the proof. consider the integers a,2a,3a,...,(b-1)a. the book claims that no number between them is divisible by b (why?). if no number between them leaves the remainder 1 then by php there exist two of them which leaves the same remainder mod b. then b divides the difference, contradiction (why?). so atleast one of them leaves remainder 1, and there exist integers x,y such that ax+by=1.

                                   i've given two 'why' above. because both of them is used the lemma. but how can we use them? if anyone can give the proof of the lemma which is independent to the theorem, then it's all right. otherwise this is a foolish mistake, which should not be done in any serious mathematics book.

Why?  Do you think the argument is circular?  I'm afraid I don't see what you mean by "mistake."  (I also don't know what you mean by "php.")

Probably Pigeonhole.

I don't see the mistake. If I'm reading this correctly, he first assumes that none of the $ b\minus{}1$ numbers $ a, 2a, ... (b\minus{}1)a$ are congruent to $ 1\pmod{b}$. Then, as $ (a, b)\equal{}1$, none of those $ b\minus{}1$ numbers are divisible by $ b$ and thus all must be congruent to some number from $ 2$ to $ b\minus{}1$, inclusive, in $ \pmod{b}$. There are $ b\minus{}2$ residues possible but $ b\minus{}1$ numbers, so by pigeonhole, there are at least two numbers that are congruent $ \pmod{b}$.

When subtracting these two numbers, say $ na$ and $ ma$, you find $ b|a(m\minus{}n)$, so $ b$ divides $ m\minus{}n$. Thus $ m\equiv{n}\pmod{b}$, clearly a contradiction. Thus, some number is congruent to $ 1\pmod{b}$, so there exists integers $ x, y$ so that $ ax\plus{}by\equal{}1$.

[quote="udyam.bnghs"]you all didn't understand my point. you said "as (a, b)=1, none of those b-1 numbers are divisible by b" and " b|a(m-n), so b divides m-n". why???
in both cases you have used the fact that if a|bc and g.c.d(a,b)=1 then a|c. but how can you use them? because this is a immediate consequence of the theorem which you have proved and the book of arthur engel have done. i hope now you have understood my point. so can you give a proof of the lemma "if a|bc and g.c.d(a,b)=1 then a|c" which is independent of the theorem? i don't know such a proof.[/quote]
It is a bit confusing. The lemma $ a \mid bc, \gcd(a,b) \equal{} 1 \Rightarrow a \mid c$ follows immediately from writing $ a,b,c$ primefactors.

absolutely not. after factorising you have to assume the fact that if p is a prime and p|ab then p divides at least one of them. this is a result of our lemma. and ultimately you have to use the fact that (a,b)=1 implies that there exist x,y such that ax+by=1. and multiplying both sides by c you'll get a divides left side so a|c. this is the proof. but this uses the theorem what the book has proved by using the lemma. which is a stupid mistake if anyone will not be able to give the proof of the lemma which is independent to the theorem.

Of course it follows from writing $ a$, $ b$, $ c$'s primefactors.

PLEASE THINK MORE. ULTIMATELY YOU HAVE TO USE THE FACT THAT IF  P IS A PRIME AND P|ab THEN P DIVIDES AT LEAST ONE OF THEM. ISN'T IT? THE LEMMA IS SO OBVIOUS. PERHAPS THAT IS THE CAUSE OF OUR PROBLEM. IF YOU ARE STILL THINKING THAT THE LEMMA CAN BE PROVED USING ONLY PRIME FACTORISATION WHICH IS COMPLETELY INDEPENDENT FROM THE THEOREM, THEN PLEASE GIVE ME THE PROOF, (IN DETAIL,OF COURSE). I'M SO STUPID THAT I COULD NOT UNDERSTAND!!!

But that last fact you gave follows from writing out $ a,b$ in terms of it's prime factors.

$ ab\equal{}kp$, but $ ab$ has no prime factor of $ p$ in it's prime factorization since neither $ a$ nor $ b$ do, contradiction!

Similarly for the general one, $ ab\equal{}kc$, take a prime dividing $ c$, it must divide $ a$. Divide through by the prime and repeat.

udyam.bnghs, please don't write in CAPS. Your points are correct, but where does Engel even prove that stuff¿ I just took a short look and he only seems to mention those facts without proof at the beginning of chapter 6.

[quote="udyam.bnghs"]but the unique factorization theorem(UFT) comes from the last two facts, isn't it? you can't prove the uniqueness of UFT without using them. though here , i think you don't have to assume the uniqueness. i just want to say that many standard number theory books first give the proof of the "ax+by=1" theorem, then they prove a|bc and (a,b)=1 implies that a|c, then if p|ab implies p(prime) divides at least one of them and at last they prove UFT. this is the standard arrange of this system. in fact any standard book of  modern number theory, which requires some basic concepts of abstract algebra agree with this system. this is natural because they want to define g.c.d. using ideals and they carry on all the theorems to the principal ideal domain(PID). so if someone want to violet the system and want to prove a theorem by using the next one(in the standard system) he must have to arrange the system first, properly, then he can do. otherwise if he just prove a theorem by using the next one, that is bogus(doesn't matter how beautiful the proof is). specially this should not be done by any well known mathematics book. if they do so, they must have to mention all of the points and about the arranging of the system.     isn't it?[/quote]
Assume that $ m \in \mathbb{N}$ doesn't have unique factorisationn. So for some primes $ p_1,p_2,...,p_r,q_1,q_2,...,q_s$ where $ p_i \neq q_j, p_i,q_j > 0$ we have:
$ m \equal{} p_1p_2...p_r \equal{} q_1q_2...q_s$. Further more assume that $ m$ is the smallest such number.
wlog $ p_1 > q_1$
$ m' \equal{} (p_1 \minus{} q_1)p_2p_3...p_r \equal{} q_1(q_2...q_r \minus{} p_2...p_r)$, and $ 0 < m' < m$. Then $ m'$ doesn't have unique factorisation either and we obtain a contradiction since $ m$ was assumed to be the smallest such number.

the proof was in chapter 4, (E5). He proved the theorem in that way. i'm not writing this without seeing it properly. :roll:         please check it out.

 the proof what Mathias_DK has given is correct, i think.  may be the theorem can be proved in many ways, but one should not violet the system to give a proof of a theorem without restate the chain what he follows. he must have to clear the chain of the system. then he can give any beautiful proof. that was my point.

[quote="udyam.bnghs"]Also there is a drawback of Mathias_DK's proof. this proof is only valid for the natural numbers, not for the ring of integers. because one cannot assume m to be the smallest number with this property when we work in $ \mathbb{Z}$ .[/quote]
Where is the challenge in extending it from $ \mathbb{N}$ to $ \mathbb{Z}$? If $ m<0$ then $ m \equal{} \minus{}p_1p_2...p_r \equal{} \minus{}q_1q_2...q_s$, and we can use the same argument again. Or: we could just look at the smallest $ |m|$ instead.

yeah, you can do. now the chain is complete, i think. we have done another chain of the system. but arthur engel didn't mention this! :!:  :o


function
real analysis
real analysis unsolved

A diffeomorphism is bijective, differentiable, and its inverse is differentiable. The second and third are strictly local properties- see the Inverse Function Theorem. The first isn't- we can easily miss points in the codomain, or cover points more than once. The example is periodic; $ f(x\plus{}2\pi)\equal{}f(x)$, and $ f$ is not injective.

[quote="math beginner"]In my textbook, it said: A typical example of a local diffeomorphism that is not a global diffeomorphism is the map f: R-S1, defined by f(t) = (cost, sint).

Can someone tell me what's a global diffeomorphism?? and Why the example shows that it's not?[/quote]

what text book are you using? what is the name of the course you are taking?

It seems interesting to me, especially that embedding problem about a circle in $ R^4$.


A certain state issues license plates consisting of six digits (from 0 to 9). The state requires that any two license plates differ in at least two places. (For instance, the numbers 027592 and 020592 cannot both be used.) Determine, with proof, the maximum number of distinct license plates that the state can use.

Just like the other problem with subsets, we should try to do this with induction.

Testing out some small cases, when n=2 digits, the maximum number of distinct license plates is 10=10^1, and it can be achieved with the set {(01), (12), (23),...(90)}.

For the n=3 digits case, we notice that we can keep the last two digits the same as from the set in the n=2 case.  Then we can add 0s as the hundreds digit to each case, 1s as the hundreds digit to each case..., 9s as the hundreds digit to each case.

Intuitively, we notice that the maximum number of license plates in the n digit case is $ 10^{n \minus{} 1}$.  However, we need to prove that (1) this is indeed the maximum, (2) this is achievable.

The maximum is kind of hard and I think there may be something wrong with my logic, but Ill give it by best shot.
Basically we look at some license plate with $ N \equal{} x_1x_2x_3...,x_n$ where each $ x_i$ is a digit.  We need to take out all of the license plates that differ in exactly 1 place.  So for a given N, we look at its first digit.  If the first digit is different, there can be 9 different other possibilities for the first digit.  Then for the rest, it has to be the same, so there are $ 10^{n \minus{} 1}$ remaining possibilities for $ x_2, x_3,...,x_n$.  The problem with this is that I think its true, but I dont know how to say that we only need to look at the first digit (we dont have to consider all the $ x_i$).  Once we have removed the other plates based on the first digit, we do not need to consider the other digits.

For example: say we have three plates- (123456), (223456), (233456).  We remove (223456) because it differs in only 1 place.  After we remove (223456), we do not need to remove (233456) because the (223456) is already removed.  And this is circular, so I think it should turn out nicely.  Can someone prove this elegantly?

Anyways, that means that we remove $ 9*10^{n \minus{} 1}$ license plates, leaving us with $ 10^{n} \minus{} 9*10^{n \minus{} 1} \equal{} 10^{n \minus{} 1}$ license plates as the maximum.

Dang...I thought I had a set for (2), but it seems like it wouldnt work.  Anybody else have thoughts on how to do this?  Most of my proofs so far have been partial (maybe good enough to earn 1/7 or 2/7, perhaps 6/7 if I get lucky), but I know there are other Indiana people better than me and can solve complete USAMO problems.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Describe all groups of order $p{\cdot}q$,where $p$ and $q$ are primes.

[hide="the famous answer"]Direct or (when $p | (q-1)$ or vice versa) semi-direct product of cyclic groups of order $p$ and $q$. [/hide]

forgive me my ignorance

i am really unfamiliar with semidirect products

In our course we saw, that if N is normal subgroup of G, and HN=G and H and N intersect trivially,
G is semi direct product by definition of H and N , and H is isomorphic with G/N

i also know that if q is the biggest prime, by sylow, if p does not divide q-1, the group is cyclic

how does this all fit in, and is your answer easily proven?

sorry for asking me so much i would definitely be happy already with a link or reference to some theorem so i can do further inquiries on my own :ninja:

All right- how to construct groups with semidirct products.
To construct the semidirect product of $H$ and $N$, we need a homomorphism $\phi: H\mapsto\mathrm{Aut }N$. If this homomorphism is trivial, the product is direct. Multiplication on pairs in $N\times H$ is defined by $(a,b)\times(c,d)=(a\cdot\phi(b)(c),bd)$.

In the case where $H=\mathbb{Z}/p\mathbb{Z}$ and $N=\mathbb{Z}/q\mathbb{Z}$, we have $\mathrm{Aut }N=\mathbb{Z}/q\mathbb{Z}^*$, which is cyclic of order $q-1$. If $p\not|q-1$, the only homomorphism is trivial. If $p|q-1$, there are two kinds of homomorphisms- the trivial one and the injective homomorphisms. The injective homomorphism can easily be seen to lead to isomorphic groups, as they are equivalent under automorphisms of $H$.
The direct product case is also the cyclic group here.

Actually, there's one more detail: show using Sylow theory that the subgroup of order $q$ is normal, assuming $q>p$.


algebra
polynomial
symmetry
complex analysis
calculus
calculus computations

x_0=0
x_1=1
x_2=2
P(x_0)=2
P(x_1)=-1
P(x_2)=4

This is data given for a polynomial P(x) of unknown degree.

What is the coeffecient of x^2 in P(x) if all third order differences are one.

the answer is 3.5

Im sure its quite an easy problem, this isnt really complex analysis, but i didn't know where else to put this.

Can nayone give me nay hints on how that answer is derived.

Thanks for any help,

-sean

If all third order differences are 1, the polynomial is $P(x)=\frac{x^3}6+Q(x)$, where $Q$ has degree at most two. We could set up a system and solve for the coefficients of $Q$.

Alternately, we use symmetry to make things easier. The second difference centered at zero depends only on the even degree terms with exponent at least 2, so it is twice the coefficient of $x^2$ in this case. A calculation:
$\begin{array}{cccc}-1&0&1&2\\ ?&2&-1&4\\ ?&-3&5&\,\\ ?&8&\,&\,\\1&\,&\,&\,\end{array}$
Filling in the blanks, we have
$\begin{array}{cccc}-1&0&1&2\\ 12&2&-1&4\\ -10&-3&5&\,\\ 7&8&\,&\,\\1&\,&\,&\,\end{array}$

The entry we care about is the 7, so the coefficient of $x^2$ is $\frac72$. The full polynomial is $\frac16x^3+\frac72x^2-\frac{20}3x+2$.

thanks a lot, this was extremely helpful.


geometry
trigonometry
calculus
trig identities
Law of Sines

Is given a triangle $ ABC$ such that $ \angle B\equal{}7x$ and $ \angle C\equal{}2x$, if $ D$ is the midpoint of $ BC$. Ascertain the maximum
value of $ \angle CAD$

[hide]The answer is very close to $ 20$ degrees[/hide]

Using $ \text{Area}\equal{}\frac{1}{2}ab\sin{C}$, we get $ b\sin{CAD}\equal{}c\sin{BAD}$.
Then by Law of Sines $ \frac{\sin{B}}{\sin{C}}\equal{}\frac{\sin{BAD}}{\sin{CAD}}$.
from here I'm more or less clueless.

[hide="Algebra bashes"]Applying Laws of Cosine into triangle $ CAD$,we get:
$ cos(CAD) \equal{} (AC^2 \plus{} AD^2 \minus{} DC^2)/2AD.DC$ (1)
Since AD is the median, $ AD \equal{} \sqrt {2(AC^2 \plus{} AB^2) \minus{} BC^2}/2$.

Plugging in this formula into (1), we have:
$ cos(CAD) \equal{} (AC^2 \plus{} [2(AC^2 \plus{} AB^2) \minus{} BC^2]/4 \minus{} CD^2)/(AC\sqrt {2(AC^2 \plus{} AB^2) \minus{} BC^2}) \equal{} 6AC^2 \plus{} 2AB^2 \minus{} 2BC^2/AC\sqrt {2(AB^2 \plus{} AC^2) \minus{} BC^2}$
Dividing both denominator and numerator by $ AC^2$ and using the property:
$ AB/AC \equal{} sin(7x)/sin(2x)$ and $ BC^2 \equal{} AC^2 \plus{} AB^2 \plus{} 2AC.AB.cos(9x)$, we get the following equation:
$ cos(CAD) \equal{} 4 \minus{} 4(sin2x/sin7x).cos9x/\sqrt {1 \plus{} sin^2(2x)/sin^2(7x) \plus{} 2sin(2x)/sin(7x).cos9x}$
Now I don't see any better ways to deal with this expression than using Calculus with notice that $ 0 < x < 20$[/hide]


superior algebra
superior algebra solved

$R$ is a ring with $1$. $F$ is a subfield of $R$. $1\in F$. Show that any non-zero homomorphic image of $R$ contains a subfield which is isomorphic to $F$.

Let $f:R\to R'$ be an epimorphism. Then $R'$ is isomorphic to $R/I$ with $I=\ker f$. Clearly, $F\cap I=\emptyset$, so the restriction of $f$ to $F$ has a field isomorphic to $F$ as its image.


A purse contains a collection of quarters, dimes, nickels, and pennies. The average value of the coins in the purse is 17 cents. If a penny is removed from the purse, the average value of the coins becomes 18 cents. How many nickels are in purse?

a) 2   b) 5   c) 0   d) 1    e) 8

Please help me to solve this question.

Let $x$ be the value of all $y$ coins.

$\frac{x}{y}=17$

$\frac{x-1}{y-1}=18$

By substitution:

$\frac{17y-1}{y-1}=18$
$17y-1=18y-18$
$y=17$
$x=289$

So 17 coins with total value of 289 cents must be found... 11 quarters, $\boxed{\text{2 nickles}}$, 4 pennies is the only solution i think.

[hide]$\frac{m}{c}=17\implies m=17c$
$\frac{m-1}{c-1}=18\implies m=18c-17$.
Number of coins=c=17.
Dollar Value of coins=m=289.

Then we find the coins are: 11q, 2n, 4p, so 2 NICKELS.[/hide]


Got a question that I'm not sure how to begin.
Im supposed to evaluate the following matrix where a,b,c are the sides of triangle ABC.
|  a     b     c   |
|sinA sinB sinC|
| 24    16   5   |

I tried doing something with the formula to calcuate the determinant, but I doubt that it was actually useful.
Any ideas greatly appreciated.

Thanks

[hide]Not sure about this, but, as $a = 2R\sin\alpha$, we get that
\[\left|\begin{array}{ccc}a & b & c \\ \sin\alpha & \sin\beta & \sin\gamma \\ 24 & 16 & 5\end{array}\right| = 2R\left|\begin{array}{ccc}\sin\alpha & \sin\beta & \sin\gamma \\ \sin\alpha & \sin\beta & \sin\gamma \\ 24 & 16 & 5\end{array}\right|\]and because row 1 = row 2, the determinant equals 0.[/hide]


[quote="clinkmath"]Find all rational solutions to the system: xx+x=yy and xx+1=zz[/quote]

Do you mean $x^2+x=y^2$ and $x^2+1=z^2$ or ?

conceive

Yes x^2+x=y^2 and x^2+1=z^2. Both requirements need to be met.

You can easily show that denominators of $x, y, z$ are the same, so problem can be
reduced to this one: Find integers $a, b$ s.t. $(a, b)=1$ and numbers $a^2+b^2$ and $a^2+ab$ are both perfect squares, but this is as far as I went :(

From $x^2+1=z^2$ we have $x=2k$ for some $k$. Now $y^2=4k^2+2k,$ thus $\nu_2(y^2)=2$ so the equation doesn't have a solution.

Didn't you ask yourself, why is that so easy... that would be correct of course if we are looking for integer solutions  ;)

Read the text again!!

P.S. From $x^2+1=z^2$ you get $1=(z-x)(x+z)$ which is posible only for $z+x=z-x=\pm 1$ in other words $z=\pm 1$ and $x=0$  :lol:  .


$f: \Re\rightarrow\Re$ such that$|f(x)-f(y)|\leq \sqrt{|x-y|}$.
Prove that:
1) $\exists a>0$ such that $|x|<a \Rightarrow |f(x)|< a$
2) $\exists u \in \Re$ such that $f(u)=u$.

1) $|f(x)|\le |f(x)-f(0)|+|f(0)|\le |\sqrt{x}|+|f(0)|$
taking $a$ sufficiently large we will have that $|f(x)|\le |\sqrt{x}|+|f(0)|\le |\sqrt{a}|+|f(0)|\le a,\forall |x|\le a$.
2) Assume that $f(x)\ge x,\forall x\in \mathbb{R}$, then $|\sqrt{x}|\ge |f(x)-f(0)|=f(x)-f(0)\ge x-f(0)$ for all sufficiently large $x$. Which is false. Similarly for $f(x)<x$.

The only case remaining is $\exists u, f(u)=u$.

[quote="alekk"]draw a picture and it is straightforward.[/quote]
could you explain please?

[quote="zhaobin"][quote="xirti"]
2) Assume that $f(x)\ge x,\forall x\in \mathbb{R}$, then $|\sqrt{x}|\ge |f(x)-f(0)|=f(x)-f(0)\ge x-f(0)$ for all sufficiently large $x$. Which is false. Similarly for $f(x)<x$.

The only case remaining is $\exists u, f(u)=u$.[/quote]
all these I think you should prove that $f$ is continuous first,though it is easy ;)[/quote]
I omited that thing. Since $f$ is a contraction it is continuous ;)


I would like to have those ifs being horizontally at the same level
[code]\tilde{\sigma}x:=\begin{cases}\varphi\left (\sigma\right )x\quad\text{if}\quad x\in E_s\\
\sigma x\quad\quad\quad \text{if}\quad x\in F_s.
\end{cases}
[/code]
\[ \tilde{\sigma}x: =\begin{cases}\varphi\left (\sigma\right )x\quad\text{if}\quad x\in E_{s}\\ \sigma x\quad\quad\quad \text{if}\quad x\in F_{s}. \end{cases}\]

& vertically aligns the text as you want it. Also if you put a space inside \text so you have \text{if } then there's no need for any \quad (unless you prefer them). [code]\tilde{\sigma}x:=
\begin{cases}
 \varphi\left(\sigma\right )x & \text{if } x\in E_s\\
 \sigma x & \text{if } x\in F_s.
\end{cases}[/code] gives
$ \tilde{\sigma}x: = \begin{cases}\varphi\left(\sigma\right )x & \text{if }x\in E_{s}\\ \sigma x & \text{if }x\in F_{s}. \end{cases}$

I'm having a slight problem with cases as well.

[code]
\begin{equation*}
x^{n} =
\begin{cases}
1 & \text{if } n = 0 \\
\underbrace{x \times x \times x \times x \cdots }_{x \; \textrm{number of terms}} & \text{if } n > 0\\[20pt]
\underbrace{\frac{1}{x \times x \times x \times x \cdots}}_{x \; \textrm{number of terms}} & \text{if } n < 0
\end{cases}
\end{equation*}
[/code]

Produces this:

$ x^{n}= \begin{cases}1 & \text{if }n = 0 \\ \underbrace{x \times x \times x \times x \cdots }_{x \; \textrm{number of terms}}& \text{if }n > 0 \\[20pt] \underbrace{\frac{1}{x \times x \times x \times x \cdots}}_{x \; \textrm{number of terms}}& \text{if }n < 0 \end{cases}$

Any suggestions on perhaps improving the aesthetics of the above? Also, how would I give the entire case a single equation number (preferably one that works in conjunction with your aboxed command you created here : [url]http://www.mathlinks.ro/viewtopic.php?t=157721[/url]).

Do you mean something like this: [code]\begin{equation}
\boxed{x^{n} =
\begin{cases}
1 & \text{if } n = 0 \\
\underset{n\text{ times}}{x \times x \times \dots \times x}  & \text{if } n > 0 \\[20pt]
\underset{n\text{ times}}{\dfrac{1}{x \times x \times \dots \times x}}  & \text{if } n < 0
\end{cases}}
\end{equation}[/code]
\[ \begin{equation}\boxed{x^{n}= \begin{cases}1 & \text{if }n = 0 \\ \underset{n\text{ times}}{x \times x \times \dots \times x}& \text{if }n > 0 \\*[20pt] \underset{n\text{ times}}{\frac{1}{x \times x \times \dots \times x}}& \text{if }n < 0 \end{cases}}\end{equation}\]
It looks better than this in a document. You only need aboxed in the align environment.

Counter isn't working properly. See attachment.

That is odd. Somehow the equation counter must have been redefined. Can you post the tex document?

By the way, equation 14 isn't true. You cannot divide by 0 under any circumstances whatsoever. Furthermore, infinity is not a number so cannot equal anything let alone something that doesn't exist. See http://www.mathpath.org/concepts/division.by.zero.htm

Fine, fine I'll mark 14 as undefined.

Anyway, here's the tex document.

Ah yes, we were using a different command \eqnumber to number the equations. I can't remember why :) but you can use the same command for equation 15 as well:[code]\begin{equation*}
\boxed{x^{n} =
\begin{cases}
1 & \text{if } n = 0 \\
\underset{n\text{ times}}{x \times x \times \dots \times x}  & \text{if } n > 0 \\[10pt] \eqnnumber \label{algebradef15}
\underset{n\text{ times}}{\dfrac{1}{x \times x \times \dots \times x}}  & \text{if } n < 0
\end{cases}}
\end{equation*}[/code]

Do you think it's possible to get the eqnnumber command to use the regular numbering with the 1.1.1 style, or maybe something customizable? I'm not too good with programming latex commands  :oops:

Easy to do both. Look for the line where \eqnnumber is defined [code]\newcommand{\eqnnumber}{\stepcounter{equation}\tag{\arabic{equation}}}[/code]The counter you are actually seeing is \arabic{equation}. You can put the Chapter, section and/or subsection before it using for example \thesection.\arabic{equation} so you get [code]\newcommand{\eqnnumber}{\stepcounter{equation}\tag{\thesubsection.\arabic{equation}}}[/code]
You can even go to extremes (but don't it looks dreadful!) [code]\newcommand{\eqnnumber}{\stepcounter{equation}\tag{\thechapter/\thesection/\thesubsection.\arabic{equation}}}[/code]but I give this just to show what is possible.

As a general rule try to limit the depth of the numbering to a.b.c as a.b.c.d gets difficult to follow.

Also, you can reset the equation number for each chapter or section or subsection. So suppose you want to reset it every subsection then use the line in the preamble [code]\numberwithin{equation}{subsection}[/code]If you use for example [code]\numberwithin{equation}{subsection}
\newcommand{\eqnnumber}{\stepcounter{equation}\tag{\thesubsection.\arabic{equation}}}[/code]then note how the numbering starts again in 1.1.2 Multiplication and Division.


calculus
integration
induction
combinatorics solved
combinatorics

hey ppl
here's a problem
there are three sticks each of integral length \geq n such the the sum of their lengths is n(n+1)/2.prove that the sticks can be divided into n pieces of length 1,2,3,.....n .
help me with it
best regards
ps:i'm sure this is combinatorics :) .so i guess this'll be moved to its rightful place

The ideea is induction.
For n+1 \geq 11.
Det l1,l2,l3 be the lengths of the 3 sticks.
We have l1+l2+l3=(n+1)(n+2)/2, so there is one of them, WLOG l1 \geq (n+1)(n+2)/6 \geq 2n+1.
So if we remove a n+1 length piece from l1, we get that l1'=l1-n-1, l2'=l2, l3'=l3 are all  \geq n and add up to n(n+1)/2 by our presumptive induction hypothesis, we finish the problem, if only we prove it for n \leq 9.

I think the real problem is proving it for n<=9  ... how do you do that ?

This is a specific case of a problem known as the "cutting sticks" problem: [url]http://home.wxs.nl/~faase009/cutsticks.html[/url]


Let $\mathbb{G}$ be a large set such that $\mathbb{G} \subset \{0, 1\}^*$
I want to find a one-way function $f : \{0, 1\}^* \times \{0, 1\}^* \mapsto \mathbb{G}$ such that

1) $f(a, f(b, c)) = f(f(a, b), c) \forall a, b, c \in \{0, 1\}^*$ (i.e. $f$ is associative)
2) $f$ is easily computable (i.e. $f$ is computable in P-time.)
3) $f$ is non-invertible with respect to one argument. That is for any randomly selected $a,b \in \{0, 1\}^*$, given $c=f(a, b)$ and $a$ (that is $f(a, b)$ is computed but the value of $b$ is kept secret or destroyed), it is hard to find an equivalent $\hat{b}$ such that $c = f(a, \hat{b})$ (however it may be entirely possible to find two random values $\hat{a}, \hat{b}$ such that $c = f(\hat{a}, \hat{b})$ which is not a problem. What I need is a function that is associative and non-invertible with respect to [i]one[/i] (left or right or both) arguement.)

The smart people here can possibly come up with a working model of such a one-way function  :?  :roll:

[EDITED: I made a typo in the original version]

It sounds interesting but I don't understand about half of the problem statement. (More exactly, what is $\{0,1\}^*$ and the non-invertable part.) Could you explain a little bit more?

{0, 1}* is the set of all strings composed of 0s and/or 1s, which we can interpret as the set of all binary number representations. By non-invertible with respect to one argument, i think there's a typo and he means that if we have:
. both a value a0, and a value c0 where c0 = f(a0, b0)
then it's not trivial to find either b0, or another bi such that c = f(a0, bi).

My question is, by polynomial time, do you mean with respect to the number of bits in max(a, b)? Also, are you looking for a mathematical expression or is an algorithm alright?

Anything will do... such functions are known as "strong Associative One-Way Functions (AOWFs)"

By "non-invertible" I mean that it is "computationally" infeasible. That is, even if an inverse exists, computing this inverse is not a P-time problem.

Yes I mean polynomial w.r.t the number of bits. A mathematical expression is ok as long as it can be converted to an algorithm...

you are right, I did make a type (stupid me.. I just noticed it)

It should have been $c = f(a, b)$ instead of $a=f(a, b)$. I have corrected it in the original post.

[quote="yongqli"]For a hard-to-invert hash function, any crytographycally secure hash function should do (SHA-256, etc). If you need the associative property, I think one-way-accumulators will do the job.[/quote]

Can you show how to construct an associative one-way function.. becasuse this has been a long-standing open problem I think..

All properties are trivial, the  associativity is the hardest. if we use some sort of muliplicative function that is associative then it is not "strongly" non-invertible.. given one argument it is easy to find the other.


HCSSiM
trigonometry
geometry
geometric transformation
rotation
perpendicular bisector
trig identities

Dude you gotta be cooler like David and play more piano instead of guitar...

[quote="archimedes1"]In $ \triangle ABC$, $ \angle B \equal{} 2 \angle C$.  Line $ l$ is the perpendicular bisector of $ BC$.
$ P,Q \in l$ such that $ AP, AQ$ trisect $ \angle BAC$.  Prove that
"$ \triangle ABC$ is obtuse <=> $ PQ < AB$."[/quote]
argh stop going off topic

Taylor: haha dude keep the peace bro

WAHOOOO DAERMON POSTED YAY
DUDE DAERMON YOU ARE PRETTY AWESOME

dude Matt: have you met a dude called Mohit at New Hampshire during the summer?
he went to Hampshire too and he is one of my friends from NY

hmm... arch: doing TSTs?  2001 number 5 to be particular...
I kinda am not a fan of my solution as it is essentially a trig bash but here we go:
[hide="Solution"]
Let $ l$ meet $ AC$ at $ R$ and let $ \angle C \equal{} \gamma$.  Then, $ \angle ABC \equal{} 2\gamma$, so $ \angle BAC \equal{} 180 \minus{} 3\gamma$, so $ \angle PAB \equal{} \angle QAP \equal{} \angle QAC \equal{} 60 \minus{} \gamma$.  Furthermore, as $ BR \equal{} RC$, we have that $ \angle RBC \equal{} \gamma$ and thus $ \angle ABR \equal{} \gamma$.  This means that $ \angle PRC \equal{} 90 \minus{} \gamma$, so $ \angle AQR \equal{} \angle QRC \minus{} \angle QAR \equal{} 30$.  Thus, by the Law of Sines on $ \triangle APQ$, we have that $ \frac {PQ}{AP} \equal{} \frac {\sin (60 \minus{} \gamma)}{\sin 30}$.  Law of Sines on $ \triangle APR$ gives that $ \frac {AP}{AR} \equal{} \frac {\sin (90 \minus{} \gamma)}{\sin (\gamma \minus{} 30)}$, so $ PQ \equal{} \frac {\sin (60 \minus{} \gamma)AR\cos \gamma}{\sin 30\sin (\gamma \minus{} 30)}$.  Now, Law of Sines on $ \triangle ABR$ gives that $ \frac {AB}{AR} \equal{} \frac {\sin 2\gamma}{\sin \gamma} \equal{} 2\cos \gamma$.  This means that $ AR \equal{} \frac {AB}{2\cos \gamma}$, so\[ PQ \equal{} \frac {\sin (60 \minus{} \gamma)\cdot \frac {AB}{2\cos \gamma}\cdot \cos \gamma}{\sin 30\sin (\gamma \minus{} 30)} \equal{} \frac {AB\sin (60 \minus{} \gamma)}{\sin (\gamma \minus{} 30)}\implies \frac {PQ}{AB} \equal{} \frac {\sin (60 \minus{} \gamma)}{\sin (\gamma \minus{} 30)}\]Now, $ PQ < AB\iff \sin (60 \minus{} \gamma) < \sin (\gamma \minus{} 30)$.  Clearly, $ \gamma\ge 0$, so $ \gamma \minus{} 30 > 60 \minus{} \gamma$, so $ \gamma > 45$, so $ \angle ABC > 90$, so $ \triangle ABC$ is obtuse, as desired.
[/hide]
yo arch: what's your solution?

why you goin off topic quatto? that aint good forum etiquette

[quote="The QuattoMaster 6000"]Taylor: haha dude keep the peace bro

WAHOOOO DAERMON POSTED YAY
DUDE DAERMON YOU ARE PRETTY AWESOME

dude Matt: have you met a dude called Mohit at New Hampshire during the summer?
he went to Hampshire too and he is one of my friends from NY

hmm... arch: doing TSTs?  2001 number 5 to be particular...
I kinda am not a fan of my solution as it is essentially a trig bash but here we go:
[hide="Solution"]
Let $ l$ meet $ AC$ at $ R$ and let $ \angle C \equal{} \gamma$.  Then, $ \angle ABC \equal{} 2\gamma$, so $ \angle BAC \equal{} 180 \minus{} 3\gamma$, so $ \angle PAB \equal{} \angle QAP \equal{} \angle QAC \equal{} 60 \minus{} \gamma$.  Furthermore, as $ BR \equal{} RC$, we have that $ \angle RBC \equal{} \gamma$ and thus $ \angle ABR \equal{} \gamma$.  This means that $ \angle PRC \equal{} 90 \minus{} \gamma$, so $ \angle AQR \equal{} \angle QRC \minus{} \angle QAR \equal{} 30$.  Thus, by the Law of Sines on $ \triangle APQ$, we have that $ \frac {PQ}{AP} \equal{} \frac {\sin (60 \minus{} \gamma)}{\sin 30}$.  Law of Sines on $ \triangle APR$ gives that $ \frac {AP}{AR} \equal{} \frac {\sin (90 \minus{} \gamma)}{\sin (\gamma \minus{} 30)}$, so $ PQ \equal{} \frac {\sin (60 \minus{} \gamma)AR\cos \gamma}{\sin 30\sin (\gamma \minus{} 30)}$.  Now, Law of Sines on $ \triangle ABR$ gives that $ \frac {AB}{AR} \equal{} \frac {\sin 2\gamma}{\sin \gamma} \equal{} 2\cos \gamma$.  This means that $ AR \equal{} \frac {AB}{2\cos \gamma}$, so
\[ PQ \equal{} \frac {\sin (60 \minus{} \gamma)\cdot \frac {AB}{2\cos \gamma}\cdot \cos \gamma}{\sin 30\sin (\gamma \minus{} 30)} \equal{} \frac {AB\sin (60 \minus{} \gamma)}{\sin (\gamma \minus{} 30)}\implies \frac {PQ}{AB} \equal{} \frac {\sin (60 \minus{} \gamma)}{\sin (\gamma \minus{} 30)}
\]
Now, $ PQ < AB\iff \sin (60 \minus{} \gamma) < \sin (\gamma \minus{} 30)$.  Clearly, $ \gamma\ge 0$, so $ \gamma \minus{} 30 > 60 \minus{} \gamma$, so $ \gamma > 45$, so $ \angle ABC > 90$, so $ \triangle ABC$ is obtuse, as desired.
[/hide]
yo arch: what's your solution?[/quote]Exactly like yours  (even used the same triangles and the same labeling of $ R$. ):( I was hoping you would have some beast solution with rotations or something.  But I remember hearing one where you project $ A$ onto $ l$, call the point $ D$.  Then $ 2AD \equal{} AB$ or something and it makes for no trig...

@Quatto:
OMG YES!
We called him Big Bad Moe, and he dominated at Ultimate. His arm-span is something like 6'6"... I'm pretty tall and that's STILL hard to throw over.

Yeah, Mohit Tiwari, right?
He's really cool.

Matt: yeah Mohit is truly awesome
I knew him since I was like born... well more like since I was like 9 or something but whatever

arch: yeah I don't really know why they put that on the TST
I looked at the problem and got that idea in like less than a minute and the computations should take another 3 (maybe 5 max)...

it was the first thing that I could come up with, and it was kinda late, so I didn't really look at many of the synthetic aspects of the problem, but there is probably a nice solution given all the nice 60 degree angles

there is indeed a nice solution here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=371492#371492

I think you guys should make a weight loss plan for me.

How about...dream about Meggy Meg in place of 15th dinner.


quadratics
algebra
number theory proposed
number theory

Maybe you first precise in what set are you looking for solutions. Anyway substitute: $a=x+y$, $xy=b$ and you get quadratic equation in $a$

of course x and y are complete, like all diophantic equations :D
but what's next?

$(a-1+\sqrt{1+3b})(a-1-\sqrt{1+3b})=0$

hallo
i have a very ungy solution  :blush:

[quote="molina"][quote="mapedd"]could anybody help me with solve of this problem?

$2(x+y)+xy=x^2+y^2$

i'll be glad if somebody do anything[/quote]

$\large{(x+y)^2=3xy+2(x+y)\,\,\Longrightarrow{(x+y)^2-2(x+y)-3xy=0\,\,\,\frac{\Delta}{4}\,=\,\sqrt{(x+y)^2-3xy}=\sqrt{(x-y)^2-xy}\,\in{\,\mathbb c{N}\,\,}}}$

$\Longleftrightarrow{(x-y)^2-xy=k^2\,|\,k\in\,\mathbb c{Z}\,\,\Longrightarrow{xy\equiv{-1,0\,\,mod\,\,4\,\,\Longrightarrow{x=-1,y=1\,\,\textrm{or}\,\,x=y=0\,\textrm{or}\,x=0,,y\not=}\,\,\textrm{or}\,\,y=0,\,\,,x\not=0}}}$

PS.: See that $\large{(x-y)^2+xy}$ is homogenuos.

The solutions : $\large{\,(\,0\,2\,)\,,\,(\,2\,,\,0\,)\,(\,0\,,\,2\,)}$[/quote]

i don't understrand well your solutoin, can explain again easily?

[hide="Hints"]Write equation as $x^2-(y+2)x+(y^2-2y)=0$. Discriminant must satisfy $\Delta_x^2(y)\ge0$, so find range of integers of $y$ satisfying this discriminant condition. After you find the range of values for $y$, which ones give $\Delta_x^2(y)$ a perfect square? After you find those $y$ values, find $x$ by using quadratic formula. Final answer should give you 6 ordered pairs.[/hide]

maybe help you
x=2a, y=2b

and now solution

a+b+ab=a2+b2

and then foun de solution

x=4 y=4

first
x,y=0

if xor y is0 then the other is 2

now, if x,y>0 then i can substitute
x=2a      y=2b
before

a+b+ab=a*a+b*b

1=(a-1)/b + (b-1)/a

if
(a-1)/b=0 then     a=1   b=2         (2,4)

the other case is the same thing     (4,2)

if
(a-1)/b,(b-1)/a>0

a,b>1

and a=b

a=2a-2

then               a=2 b=2                 (4,4)

if a=b+r   r>0

1=(b+r-1)/b + (b-1)/(b+r)

because
(b-1)/(b+r)>0

and
(b+r-1)/b>1
inthis case we have no solution

then the solutions are

(0,0)
(0,2)
(2,0)
(4,2)
(2,4)
(4,4)


Let S_k(n) denote the sum of digits of natural number n in base k (example S_2(3) = 2, because 3 = (11)_2).
Determine all natural numbers m such that for each natural n that is divisible by m, also m divides S_2003(n).

I have only a particular result:
Let's imagine that we have some suitable n, so for 0 <= a_i < 2003 (for i < j) and 0 < a_j < 2003 we have that:
Let n = (a_j.2003j + a_(j-1).2003j-1 + ... + a_1.2003 + a_0),
then m|(a_j.2003j + a_(j-1).2003j-1 + ... + a_1.2003 + a_0) and also
m|S_2003(n) - obviously m|(a_j + a_(j-1) + ... + a_1 + a_0). By this
m|(n-S_2003(n)) that is the same as
m|(a_j.(2003j- 1) + a_(j-1).(2003j-1- 1) + ... + a_1.2002)
m|2002.(a_j.(2003j-1 + (a_j + a_(j-1)).2003j-2 + ... + (a_j + a_(j-1) + ... + a_2).2003 + (a_j + a_(j-1) + ... + a_1))
Now it's clear that m can be any divisor of 2002, hence m = 1;2;1001 and 2002. But I can't solve whether m can have another value.
Are there another m? And if there aren't - how to prove it?

Thanks.

Lets look at m, not at n for now.
If m has the property in the problem text, then since m|m, we get m|s(m), so m \leq s(m), so m is a digit. So m \leq 2002.

Let n=1a be the smallest 2 digit multiple of m in base 2003. This number obviously exists because m<2003. We have a<m, or else m|n-m=1(a-m), which is still a 2 digit multiple of m.
m|n => m|s(n)=1+a. So m|a+1, so m \geq a+1. But a<m, so a+1 \leq m. So m=a+1, so a=m-1, so m|1(m-1), so m|2003+(m-1), so m|2002.

This proof can be generalized from 2003 to any natural.


Two prime numbers have a least common multiple of 35. What is the sum of the two numbers?

$ 35\equal{}5*7$
5 and 7 are the only prime factors, so you do 5+7
$ 5\plus{}7\equal{}12$


inequalities
logarithms
inequalities unsolved

Give a,b,c >0.
a)Prove that:
 [tex]  \sum \frac {a^k}{(b+c)^k} \geq 3(\frac12)^k [/tex] where [tex] k \geq \frac23 [/tex]
b)Find the least value of k satisfying
 [tex]  \sum \frac {a^k}{(b+c)^k} \geq 3(\frac12)^k [/tex]

We note $ M(a,b,c;k)={(\frac{a^k+b^k+c^k}{3})}^{\frac{1}{k}}. $ It is known that k<p implies M(a,b,c;k)<M(a,b,c;p). So for $ k\geq\frac{2}{3} $ we have that $ M(a,b,c;k)\geq M(a,b,c;\frac{2}{3}). $ So it is suficient to prove the inequality for $ k=\frac{2}{3}. $
We have:
$ {(\frac{a}{b+c})}^{\frac{2}{3}}=\frac{\sqrt[3]{2}a}{\sqrt[3]{2a(b+c^2)}}\geq \frac{3\sqrt[3]{2}a}{2(a+b+c)}. $ We apply the mean inequality. So:
$ {(\frac{a}{b+c})}^{\frac{2}{3}}+{(\frac{b}{a+c})}^{\frac{2}{3}}+{(\frac{c}{b+a})}^{\frac{2}{3}}\geq \frac{3\sqrt[3]{2}(a+b+c)}{2(a+b+c)}. $ :D

[quote="Thanhliem"]I don't sure in England,they call him Nesbit,but in VietNam they call him so.
This is his inequality:
[tex] \sum \frac {a}{b+c} \geq \frac 32\ [/tex] :)[/quote]

Hi all,

Somewhere this inequality (and its generalization for n variables) is called Schapiro's inequality.

Namdung

[quote="Namdung"]
Somewhere this inequality (and its generalization for n variables) is called Schapiro's inequality.[/quote]

Could you state this inequality plz?

a1/(a2+a3) + a2/(a3+a4) + ...+ an/(a1+a2) >= n/2.

Not that this inequality is wrong generally. There are many strong results concern to this ineq. But elementarily, we can show this ineq for n<=6.

Namdung

No one have any ideas about ii) :(
I guess [tex] k= \log_{\frac12}^{\frac23} [/tex] :maybe: but I haven't prove it yet. :(


prove that in the dehydraytion of $1-Butanol$ there is a possibilty of obtaining $2-Butene$ without rearrangement or even formation of $n-Butyl Cations$

Step 1: 1-butanol is protonated at the hydroxilic oxygen.

Step 2: a molecule of water (or other base present) participate in a beta-elimination of the protonated butanol to yield 1-butene.

Step 3: 1-butene is protonated at the double bond, yielding sec-butyl cation.

Step 4: that carbocation loses a proton (with the help of another molecule of water or other base) to yield the more stable 2-butene.


I dunno if this is possible, but on another board I've been to, the Fun and Games equivelent doesn't count for post count, and I was wondering if that would work here.  So for example, if person A has 100 posts right now, and 20 posts in Fun and Games, the person' s post count would say 80, since the rest of his posts were is Fun and Games.

I don't think that matters. It's better to leave the spammers in the F&G where they get pride in spamming while raising their post count without bothering the math forums. If we close down F&G for tallied post counts, they may turn to other places (the few who are desperate to raise their post count). Maybe that thought was a little naive - hey but it's possible!

Yes, people intent on raising their post count will do it, even if you close down F&Gs impact on the post count.

But on a similar note, is there a possibility of opening a sub-forum in F&G that does not count towards the post count? We could toss things like Mafia games and "Never-ending story" in there where people won't be offended by 2-word posts.

Perhaps it's just part of maturity that people will eventually quit caring about post counts.  I don't think it's worth thinking too hard about.  After all, the people who are not competitive about post counts...don't care about the totals.

At the very least, the people who do care about their post counts here...are spending time on an educational forum.  It could be much worse.

[quote="MithsApprentice"] But on a similar note, is there a possibility of opening a sub-forum in F&G that does not count towards the post count? We could toss things like the "Never-ending story" in there where people won't be offended by 2-word posts.[/quote]

good idea. Have you read the entire story?  :roll:
There must be a way to make the story a little better otherwise (more word requirements?), but you're the moderator, you have the power on the forum.


A telephone number has the form $ ABC \minus{} DEF \minus{} GHIJ$, where each letter represents a different digit. The digits in each part of the numbers are in decreasing order; that is, $ A > B > C$, $ D > E > F$, and $ G > H > I > J$. Furthermore, $ D$, $ E$, and $ F$ are consecutive even digits; $ G$, $ H$, $ I$, and $ J$ are consecutive odd digits; and $ A \plus{} B \plus{} C \equal{} 9$. Find $ A$.

$ \textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8$

[hide="solution"]
Note that DEF = 420, 642, or 864, and GHIJ = 7531 or 9753.  Obviously ABC cannot contain 9, so we must have GHIJ = 9753.  This means that ABC must contain 1.  The only way now to make the three digits sum to 9 with the remaining numbers being even is if the other two digits are 8 and 0, or 6 and 2.  Because no selection of DEF will allow ABC to have 6 and 2, we must have ABC = 810, meaning A = 8.
[/hide]

[hide]The telephone number is $810-642-9753$. $\boxed{E}$[/hide]

[hide="Answer"]$G,H,I,J$ are either $1,3,5,7$ or $3,5,7,9$. If the former, then one of $A,B,C$ is $9$, meaning that both of the others are $0$ so that $A+B+C=9$, which is impossible. Therefore, we must have the latter, and one of $A,B,C$ is $1$. $D,E,F$ are either $0,2,4$, $2,4,6$, or $4,6,8$. The two digits not used here must add to $8$, so our only possibility is $8$ and $0$. Therefore, $A=8\Rightarrow \boxed{E}$.[/hide]


For positive integers $ m$ and $ n$, how many integers are greater than the
product $ mn$ and less than the product $ (m \plus{} 1)(n \plus{} 1)$? Express your
answer in terms of $ m$ and $ n$.

The number of integers between integers x and y is abs(x-y) - 1.
Since we know (m+1)(n+1) > mn, we proceed-

$ (m \plus{} 1)(n \plus{} 1) \minus{} mn \minus{} 1$

$ mn \plus{} m \plus{} n \plus{} 1 \minus{} mn \minus{} 1$

$ m \plus{} n$


When the number $n$ is written in base $b$ its representation is the two-digit number $AB$ where $A=b-2$ and $B=2$. What is the representation of $n$ in base $b-1$?

[hide="answer"]In terms of b, n=b^2-2b+2. (b-1)^2=b^2-2b+1, so the first digit is 1. The remainder upon dividing the original term be (b-1)^2 is 1, so the second digit is 0, the third 1. The remainder is now 0, so the # is 101.[/hide]

[quote="perfect628"][hideanswer]In terms of b, n=b^2-2b+2. (b-1)^2=b^2-2b+1, so the first digit is 1. The remainder upon dividing the original term be (b-1)^2 is 1, so the second digit is 1. The remainder is now 0, so the # is 110.[/hide][/quote]
Almost. Pay attention to your place values.


How many numbers are in the list $ 3\frac23,\, 4\frac13,\, 5,\, 5\frac23,\, \ldots,\, 26\frac13,\, 27 ?$

[quote="AIME15"]Multiply all the terms in the sequence by $ 3$ to get $ 11, 13, 15, 17, \dots, 81$.  Add one and divide by $ 2$ to get $ 6, 7, ...,41$.  Subtract 5 to get 1,2,3,4,...,40.  Thus there are 40 numbers in the sequence.[/quote]

Subtracting 5 would get you 1,2,3...36, not 40. So there's 36.
But good job finding a method!

[hide]Multiplying each number by 3, we get $11,13,15,\ldots,79,81$. Then we can subtract 1 and divide by 2 to get $5,6,7,\ldots,39,40$. So there are $40 - 5 + 1 = \boxed{36}$ numbers.[/hide]

[quote=asdfdad]why not just take the difference between the highest and lowest number and divide by the interval, then add 1?
27 - 3 2/3 = 23 1/3 = 70/3
70/3 divided by 2/3 = 35
Add 1 and you have 36 numbers.[/quote]

Nobody said you couldn’t do it. It’s just that nobody thought of it. :thumbup: Good job finding it though!

[quote=Arthurwei][quote=AIME15]Multiply all the terms in the sequence by $ 3$ to get $ 11, 13, 15, 17, \dots, 81$.  Add one and divide by $ 2$ to get $ 6, 7, ...,41$.  Subtract 5 to get 1,2,3,4,...,40.  Thus there are 40 numbers in the sequence.[/quote]

[quote=pi.3.14]Multiplying each number by 3, we get $11,13,15,\ldots,79,81$. Then we can subtract 1 and divide by 2 to get $5,6,7,\ldots,39,40$. So there are $40 - 5 + 1 = \boxed{36}$ numbers.[/quote]

[quote=asdfdad]why not just take the difference between the highest and lowest number and divide by the interval, then add 1?
27 - 3 2/3 = 23 1/3 = 70/3
70/3 divided by 2/3 = 35
Add 1 and you have 36 numbers.[/quote]

Also, [size=200][color=#ff0]PLEASE HIDE ANSWERS!!![/color][/size][/quote]

I do not think they will see your post, since they are pretty old posts.

They’re not that old but yeah I don’t think they will see it. At least you tried though...


vector
function
Functional Analysis
advanced fields
advanced fields unsolved

If $ S$ is a closed vector subspace of Banach space $ L^1(\mu),$ where $ \mu$ is $ \sigma$-finite measurement on a space $ X$. Let that $ f\in S$ implicates $ f\in L^p(\mu)$ for some $ p>1$. Prove that exists $ q>1$ so that $ S\subset L^q(\mu).$

Thanks.

I have a hint to use a Baire Category Theorem, but still don't know how.

Upsss... $ \mu$ is not a $ \sigma$-finite measurement on space $ X$. It is finite measurement on space $ X$. Sorry... :(

Another thing that could be related. We know that $ L^p(\mu)\subset L^q(\mu)$ for $ p>q\geqslant 1$. Under what conditions do these two spaces contain the same functions?

Isn't the condition you are given that $ S \subset L^p$?

The last question I meant in general case, not directly connected to the problem (but indirectly, just to find good direction of solving it). I found out that one condition is that - $ \mu$ should be like that it has finite number of atoms. But, as far as I see it doesn't help me to solve the problem.

[quote="Kalle"]Isn't the condition you are given that $ S \subset L^p$[/quote]
No, it is "for every $ f\in S$ there exists $ p\,{\color{red}\equal{}p(f)}>1$..."

Hi again. Here is what I have done.

Because $ S\equal{}\bar{S}$ than $ S$ is complete subspace of $ L^1(\mu)$.

Let $ S^p : \equal{} \{f\in S : f\in L^p(\mu)\}$ for $ p>1.$
From $ \mu (X)<\infty$ we have that - $ L^q(\mu)\subset L^p(\mu)$ for some $ q, p$ set to $ q>p$ and $ q,p>1$. That implicates $ S^q\subset S^p$ for $ q>p$.
Now we can make sequence $ \{S^p\}$ just to use that indexes $ p$ which are in $ \{2,3,4,\dots\}$ and in $ \{1\frac{1}{2}, 1\frac{1}{3}, 1\frac{1}{4}, \dots\}$ (we don't need all $ (1,\infty]$).
So let that be index set $ I.$ $ |I|\equal{}|\mathbb{N}|$.
We now have that $ S\equal{}\cup_{p\in I} S^p$. From this we can even form a disjoint family of subsets $ D^p$so that $ S\equal{}\cup_{p\in I} D^p$.

By the one version of Baire Category Theorem, there exists $ q\in I$ so that closure $ \bar{S^q}$ has an interior point (we can say that for $ D^q$ respectively). Let that interior point (function) be $ g\in S^q$. That means that it has a ball around itself for some $ \epsilon >0$ - $ B(g,\epsilon)\subset \bar{S^q}$ and of course $ B(g,\epsilon)\subset S^q$ too.

Now all elements in $ S^q$ $ (D^q$ if we are looking on it simultaneously)  can be written like linear combination of some element from $ B(g,\epsilon)$.

But now what?

We can assume opposite that exists $ f\in S^p$ where $ 1<p<q$ and $ f\notin S^q$ (same thing can be said for $ D^p$ and $ D^q$). But again, I don't see anything useful with I can do something... So, please help.

Well, you have almost got the solution. The trick is to consider instead of intersections of S with $ L_p$ the intersections with balls of $ L_p$. The advantage is that any closed ball of $ L_p$ is closed as a subset of $ L_1$ as well.

More details: consider  $ p_n \equal{} 1 \plus{} \frac{1}{n}$, and denote $ B_n$ the closed unit ball of $ L_{p_n}$. Then $ L_{p_n}$ equals $ \bigcup_{m \in \mathbb N} m B_n$. Consequently
\[ S \equal{} \bigcup_{n \in \mathbb N} \bigcup_{m \in \mathbb N} (m B_n \cap S).\]
Since this double union is still countable and the sets  $ m B_n \cap S$ are closed, the Baire theorem sais that for some $ n, m$ the interrior in $ S$ of $ m B_n \cap S$ is not empty. But then the linear span of $ m B_n$ contains $ S$ and the job is done.


Let $ \left\{x_n\right\}$, with $ n \equal{} 1, 2, 3, \ldots$, be a sequence defined by $ x_1 \equal{} 603$, $ x_2 \equal{} 102$ and $ x_{n \plus{} 2} \equal{} x_{n \plus{} 1} \plus{} x_n \plus{} 2\sqrt {x_{n \plus{} 1} \cdot x_n \minus{} 2}$ $ \forall n \geq 1$. Show that:

[b](1)[/b] The number $ x_n$ is a positive integer for every $ n \geq 1$.

[b](2)[/b] There are infinitely many positive integers $ n$ for which the decimal representation of $ x_n$ ends with 2003.

[b](3)[/b] There exists no positive integer $ n$ for which the decimal representation of $ x_n$ ends with 2004.

[Moderator edit: This post referred to an older, incorrect version of the problem, where the recurrence relation was given as $ x_{n\plus{}2} \equal{} x_{n\plus{}1} \plus{} x_n \plus{} \sqrt{x_{n\plus{}1} \cdot x_n \minus{} 2}$ instead of the right $ x_{n\plus{}2} \equal{} x_{n\plus{}1} \plus{} x_n \plus{} 2\sqrt{x_{n\plus{}1} \cdot x_n \minus{} 2}$.]

If everything in the statement of the problem is correct:

$ x_3 \equal{} x_2 \plus{} x_1 \plus{} \sqrt {x_2 x_1 \minus{} 2} \equal{} 603 \plus{} 102 \plus{} \sqrt {61504} \equal{} 953$.  But then $ x_4 \equal{} x_3 \plus{} x_2 \plus{} \sqrt {x_3 x_2 \minus{} 2} \equal{} 953 \plus{} 102 \plus{} \sqrt {97204}$ which is not an integer (if Maple is correct).

a) It is easy to show xn+3=2xn+2+2xn+1-xn for all n>=1.
b) Since x4=2003 and xn is invertible (because xn=2xn+2+2xn+1-xn+3) we conclude there are infinitle many n such that xn ends with 2003.
c) xn mod 16 = 11,6,1,3,2,9,3,6,9,11,2,1, and again 11,6,1,...

I don't understand how do ''invertible '' :?:

Esier soloution for part c:
$x_1=3 (mod 4)$
$x_2=2(mod 4)$
$x_3=1(mod 4)$
And it continiues in simillar way
So number 4 doesn't divide  non of these $x_i$'s .

Does anyone have complete soloution for part a?


induction
inequalities unsolved
inequalities

[hide="Long solution"]$ a^{2}_{p}\ge\sum_{i \equal{} p}^{n}a^{2}_{p}\sqrt {p}(\frac {1}{\sqrt {i}} \minus{} \frac {1}{\sqrt {i \plus{} 1}})$

So :

$ 4\sum_{i \equal{} 1}^{n}a^{2}_{i}\ge\sum_{i \equal{} 1}^{n}(\sum_{j \equal{} 1}^{i}a^{2}_{j}\sqrt {j})(\frac {1}{\sqrt {i}} \minus{} \frac {1}{\sqrt {i \plus{} 1}})$

Let's prove that:

$ (\sum_{i \equal{} 1}^{k}a^{2}_{i}\sqrt {i}) (\frac {1}{\sqrt {k}} \minus{} \frac {1}{\sqrt {k \plus{} 1}})\ge \frac {1}{4}(\frac {\sum_{i \equal{} 1}^{k}a_{i}}{k}$

Now use Cauchy-Schwartz :

$ (\sum_{i \equal{} 1}^{k}a^{2}_{i}\sqrt {i})(\sum_{i \equal{} 1}^{k}\frac {1}{\sqrt {i}})\ge(\sum_{i \equal{} 1}^{k}a_{i})^{2}$

 So it'll be enough to prove that :

$ 4k^{2}(\frac {1}{\sqrt {k}} \minus{} \frac {1}{\sqrt {k \plus{} 1}})\ge \sum_{i \equal{} 1}^{k}\frac {1}{\sqrt {i}}$

We have:

$ \sum_{i \plus{} 1}^{k}\frac {1}{\sqrt {i}}\le 1 \plus{} \sum_{i \equal{} 1}^{k \minus{} 1}\frac {2}{\sqrt {i} \plus{} \sqrt {i \plus{} 1}} \equal{} 1 \plus{} \sum_{i \equal{} 1}^{k \minus{} 1}2( \sqrt {i \plus{} 1} \minus{} \sqrt {i}) \equal{} 2\sqrt {k} \minus{} 1$

Now let's prove that $ 2\sqrt {k} \minus{} 1 < 4k^{2}(\frac {1}{\sqrt {k}} \minus{} \frac {1}{\sqrt {k \plus{} 1}}) \equal{} \frac {4k\sqrt {k}}{k \plus{} 1 \plus{} \sqrt {k^{2} \plus{} k}}$.

It gives:

$ \frac {2k}{\sqrt {k \plus{} 1} \plus{} \sqrt {k}} < k\le{k \plus{} (\sqrt {k} \minus{} 1)^{2}} < k \plus{} \sqrt {k^{2} \plus{} k} \minus{} 2\sqrt {k} \plus{} 1$

$ \Leftrightarrow\frac {2k}{\sqrt {k \plus{} 1} \plus{} \sqrt {k}} < k \plus{} \sqrt {k^{2} \plus{} k} \minus{} 2\sqrt {k} \plus{} 1$

$ \Leftrightarrow{2k}(\sqrt {k \plus{} 1} \minus{} \sqrt {k}) < k \plus{} \sqrt {k^{2} \plus{} k} \minus{} 2\sqrt {k} \plus{} 1$

$ \Leftrightarrow{2\sqrt {k}}(k \plus{} 1 \plus{} \sqrt {k^{2} \plus{} k}) \minus{} (k \plus{} 1 \plus{} \sqrt {k^{2} \plus{} k}) < 4k\sqrt {k}$

$ \Leftrightarrow{2\sqrt {k} \minus{} 1} < \frac {4k\sqrt {k}}{k \plus{} 1 \plus{} \sqrt {k^{2} \plus{} k}}$.[/hide]


geometry rectangle rotation algebra polynomial MATHCOUNTS

can you prove my solutions, or come up with your own???

Q1: How many rectangles are there in an n*n grid with sides parallel to the sides of the grid sides?

[hide="my soluiton"]There are $n^{2}$ 1*1's, there are 2*n*(n-1) 1*2 squares

why: There are n-1 1*2 squares in a row, and we multiply by n for the n rows, and we multiply by 2 for accounting for the vertical and horizontal ones.

There are 2n(n-2) 1*3's, so on to get

$n^{2}+2n(n-1)+2n(n-2)+2n(n-3)+\cdots+2n=n^{2}+2n(\frac{(n-1)n}{2}=n^{2}(n-1)+n^{2}=n^{3}$

We do it for 2*2's and 2*3 and so on to 2*n's to get $(n-1)^{3}$, so I just continued the pattern to get

$1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+\cdots+n^{3}=\frac{n^{2}(n+1)^{2}}{4}$[/hide]

Q2: n people sit in a circle. In how many ways can they do this?

[hide="S2"]There are n! arrangements in the line, but we overcounted, so we divide by n since there are n ways to rotate the circle, so there are (n-1)! ways to do it.[/hide]

Q3: Find a systematic way to find the formula for $1^{x}+2^{x}+3^{x}+\cdots+n^{x}$

{no solution from me}

http://mathworld.wolfram.com/HarmonicNumber.html

the last series does not have a explicit formulae

Here's a general [hide="method"]I don't know the proof of this, but it can always be expressed as a polynomial of degree $x+1$. Set $S_{x}(n)$ equal to ${a}_{x+1}{n}^{x+1}+{a}_{x}{n}^{x}...+{a}_{1}{x}+{a}_{0}$. Now, clearly, $S_{x}(n+1)=S_{x}(n)+{(n+1)}^{x}$. Also, $a_{0}=0$, as $S_{x}(0)=0$. Now, Set these two polynomials equal, set their individual terms equal, and solve the resulting system for the coefficients.[/hide]

Can we not discuss the third question, as it is a bit advanced for MATHCOUNTS level.  One of them employs the hockeystick identity, and the other uses summation-bashing.  AoPS book 2 should have more info.  Thanks, but questions 1 and 2 are ok!

question 3 was on a National Masters round, in case you didn't know.  :wink:

I tried the Bernoulli sum for squares, but I couldn't "find" the formula, but I knew it.

This Bernoulli sum is for nth powers, but that should explain x=2.

Maybe I should post this question in iIntermediate?

[quote="perfect628"]that masters was where i got that method...[/quote]
Really, which masters?
I came up with the exact same method after noticing that $\sum^{k}_{n=1}n^{x}$ was of a polynomial of $x+1$th degree.

well, i mean i did it and derived that method. ironically, it did turn out to be a similar one given in the solution. this was a Q on one i can't remember-i have it, but there's no year labeled.

Dude #1 is so much easier than you're making it

[hide]
Just pick which two of the $n+1$ vertical edges which two of the $n+1$ horizontal edges you're using and be like

BAM

${n+1\choose2}{n+1\choose2}$
$=\frac{n^{2}(n+1)^{2}}{4}$.
[/hide]


Given $ a, b, c > 0$ and $ ab\plus{}bc\plus{}ca\equal{}3$. Prove that: $ \frac{a}{\sqrt[4]{a^2\plus{}b^2\plus{}14}}\plus{}\frac{b}{\sqrt[4]{b^2\plus{}c^2\plus{}14}}\plus{}\frac{c}{\sqrt[4]{c^2\plus{}a^2\plus{}14}} \ge \frac{3}{2}$  :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that: $ \frac {a}{\sqrt [4]{a^2 \plus{} b^2 \plus{} 14}} \plus{} \frac {b}{\sqrt [4]{b^2 \plus{} c^2 \plus{} 14}} \plus{} \frac {c}{\sqrt [4]{c^2 \plus{} a^2 \plus{} 14}} \ge \frac {3}{2}$  :)[/quote]
 $ \left(\sum_{cyc}\frac {a}{\sqrt [4]{a^2 \plus{} b^2 \plus{} 14}}\right)^4\sum_{cyc}a(a^2\plus{}b^2\plus{}14)\geq(a\plus{}b\plus{}c)^5$ helps.


If $ \log(xy^3)\equal{}1$ and $ \log(x^2y)\equal{}1$, what is $ \log(xy)$?

$ \textbf{(A)}\ \minus{}\!\frac{1}{2} \qquad
\textbf{(B)}\ 0 \qquad
\textbf{(C)}\ \frac{1}{2} \qquad
\textbf{(D)}\ \frac{3}{5} \qquad
\textbf{(E)}\ 1$

[quote="worthawholebean"]If $ \log(xy^3) \equal{} 1$ and $ \log(x^2y) \equal{} 1$, what is $ \log(xy)$?

$ \textbf{(A)}\ \minus{} \!\frac {1}{2} \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ \frac {1}{2} \qquad \textbf{(D)}\ \frac {3}{5} \qquad \textbf{(E)}\ 1$[/quote]

Since $ log(xy^3)$ = $ log(x^2y)$ + $ log(y^2/x)$

We know that $ log(y^2/x)$ = 0

Which implies $ x \equal{} y^2$

And then substitution gives $ y \equal{} 10^{1/5}$ and x as $ 10^{2/5}$

So $ log xy$ is $ 3/5$

which is $ D$

@Arrange your tan : Thanks, Edited

[quote="befuddlers"][quote="worthawholebean"]If $ \log(xy^3) \equal{} 1$ and $ \log(x^2y) \equal{} 1$, what is $ \log(xy)$?

$ \textbf{(A)}\ \minus{} \!\frac {1}{2} \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ \frac {1}{2} \qquad \textbf{(D)}\ \frac {3}{5} \qquad \textbf{(E)}\ 1$[/quote]

Since $ log(xy^3)$ = $ log(x^2y)$ + $ log(y^2/x$)

We know that $ log(y^2/x$) = 0

Which implies $ x \equal{} y^2$

And then substitution gives $ y \equal{} 10^{1/5}$ and x as $ 10^{2/5}$

So $ log (xy$) is $ 3/5$

which is $ D$[/quote]

bufuddlers,

you have to include the denominator x INSIDE the parentheses
as I have shown above.  Otherwise what you have shown is
taking the logarithm of an expression, and then dividing it by
the variable x, in two different places.

We can make things [i]look[/i] more systematic by rewriting it as follows:

$ \begin{array}{c} \log x + 3 \log y = 1 & 2 \log x + \log y = 1 \end{array}$
Find $ \log x + \log y$

We can then rewrite it without logs as follows:

$ \begin{array}{c} a + 3b = 1 & 2a + b = 1 \end{array}$
Find $ a + b$

Subtract the second equation from the first to obtain $ a = 2b$. We obtain $ a = \frac {2}{5}$ and $ b = \frac {1}{5}$.

$ \therefore a + b = \frac {3}{5}$

[quote="worthawholebean"]If $ \log(xy^3) = 1$ and $ \log(x^2y) = 1$, what is $ \log(xy)$?

$ \textbf{(A)}\ - \!\frac {1}{2} \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ \frac {1}{2} \qquad \textbf{(D)}\ \frac {3}{5} \qquad \textbf{(E)}\ 1$[/quote]

[hide="... and the answer is..."]

This problem is evidently meant to test your knowledge of the properties of logarithmic functions; namely, you should recognize that:

\begin{eqnarray*} \log(xy^3) & = & 1 \\ \log(x^2y) & = & 1 \end{eqnarray*} \Rightarrow \begin{eqnarray*} \log x + 3\log y & = & 1 \\ 2\log x + \log y & = & 1 \end{eqnarray*}

Solving the above simultaneously we get:

$ \begin{array}{l} \log y  =  \frac{1}{5} \\ \\ \log{x}  = \frac{2}{5}\end{array}$

$ \therefore \log x + \log y = \log(xy) = \frac{3}{5}$[/hide]


Whats up fellow geniuses... lets see if you can figure the pattern out here.

B  W  9  H  L  X  A  Q  6  P

66 87 57 72 76 88 65 81 54 80

B  W  W  O  T  X  E  9  I  K
66 87 87 79 84 88 69 57 73 75

Each number corelates to the letter above it, for referance. See if you can find the pattern here, feel free to ask questions!

73 84-72-73-78-75 73 71-79-84 73-84.

(Dashes indicate that the letters are in the same word.)

the numbers correspond to the posisition in the alphabet + 64, and the numbers is just 48 + the number

It's just ASCII code corresponding to the character.

Yes it is ASCII, what I am trying to see if you guys can figure out the root algorithm ... As well, each is its own combo. Its not something you use together as posted above.

B = W 9 H L X A Q 6 P

66 = 87 57 72 76 88 65 81 54 80

B = W W O T X E 9 I K
66 = 87 87 79 84 88 69 57 73 75

That may help ;)


vector
real analysis
real analysis unsolved

If the sequence $(a_{n})$ does not converge to $0$, then $A$ is not compact: just consider its action on the subspace spanned by basis vectors $e_{n}$ with $|a_{n}|\ge\epsilon$.

Conversely, if $a_{n}\to 0$, then $A$ is a uniform limit of finite rank operators, and is therefore compact.


Our math chat wasn't able to get the answer on Kalva.

3.  A object moves in a sequence of unit steps. Each step is N, S, E or W with equal probability. It starts at the origin. Find the probability that it reaches (2, 2) in less than 7 steps.

Note: the problem has been reworded; the integers between 000 and 999 inclusive restriction doesn't apply.

You...weren't?  Well, Kalva says the answer is [hide]3/64[/hide].  I'll post how I did it in a second.

just make sure you dont overcount and then its easy mac

My vote - count the number of ways you can do it in 4, followed by the number of ways you can do it in 6, and eliminate the overlap.  It can be done without too much brutality.


Yersteday at the central exam for greek Universities there was given the problem(a partial question) :

  [color=blue][b]Five batteries of type A have a lifetime : 20,26,24,22,18 thousand hours.[/b][/color]
[color=blue][b] Five batteries of type B have a lifetime : 26,32,19,20,23 thousand hours .
 If a battery of time A costs 38 euros and a battery of type B costs  40 euros , which type of battery would you buy and why ?[/b][/color]

  As statitistics in Greece makes its first steps in our schools , I would like to have an answer with some  theoritical elements.What would we do , if the two samples had the same mean ?

 Babis

Well you could always bring Economics into it and start talking about marginal utility and stuff.

Heck, even considering only statistics and probability, the problem is still a bit vague. For example, would you want to maximize average battery life years per euro or minimize average euros expended per year? Depending on the distribution of battery lives, these aren't necessarily the same.

I suppose we're supposed to assume they are normally distributed though


limit trigonometry algebra polynomial logarithms integration calculus

[hide="Answer"]$\prod_{k=1}^{n}\sin\left(\frac{k\pi}{2n}\right) = \frac{\sqrt{n}}{2^{n-1}}$.

So, let $L$ be your limit. Then:

$L = \lim_{n\to\infty}\left(\frac{\sqrt{n}}{2^{n-1}}\right)^{1/n}= \lim_{n\to\infty}\frac{n^{1/2n}}{2^{1-1/n}}= \frac{1}{2}$
[/hide]

Carcul : How do you justify the first equality? Please answer.

To: color

Hint: Consider the cyclotonomic polynomial $\Phi_{n}(z) = 1+z+z^{2}+...+z^{n}$

This polynomial can be factorized as $f(z)=\prod_{k=1}^{n}(z-e^{2\pi ki/n})$ in the complex plane. So $n+1=\Phi_{n}(1)=f(1)$

With a little trig work you can show that identity by using the above fact that these factors multiply to $n+1$, or something like that.

We can also consider the equation

$z^{2n}-1 = 0$

with n natural. What are the roots of this equation? Then factorize $z^{2n}-1$.

You're wrong. Each term is equal to $\infty$, since we have that $\tan\frac{\pi}{2}$ in there. If we left off the last term in the product, each term would be exactly 1.

One approach:

Let $x_{n}=\left(\prod_{k=1}^{n-1}\tan \frac{k\pi}{2n}\right)^{1/n}$.

Then

$\ln x_{n}=\frac{1}{n}\sum_{k=1}^{n-1}\ln \tan \frac{k\pi}{2n}\rightarrow \int_{0}^{1}\ln \tan \frac{\pi x}{2}dx=0$, and hence $x_{n}\rightarrow 1$.

Actually, you have to be careful with that- this is an improper integral, after all. Riemann sums don't always converge to an improper integral, since you can always choose a point that's too close to the singularity. This happens in the original version of the sum, and we get $\infty$ in that case.

If you do look at it carefully, you notice that no limit is needed, since the product is always exactly 1 by symmetry.