(paraphrased) There is a polygon with 223 sides having length 1776, 1777, ... , 1998. What is the approximate radius of the smallest circle that can be circumscribed about this polygon?

The answer choices were 25000, 50000, 75000, and 100000.

Anybody got a good way to do this?

[hide]

The best method I've heard so far is to approximate the perimeter of the polygon to the circumference of the circle, but even that leaves alot of room between the answer you get and the answer given.

[/hide]

[hide="What I would Guess"]Well, doing a series calculation gives you 420801. Dividing by pi gives you something around 130000, dividing by 2 gives around 65000? Since the circumference of the circle must be greater than the perimeter of the polygon, I'd choose 75000.[/hide]


what is the formal definition of a limit ?

I don't know the actual textbook definition, but I always viewed it as the value that f(x) approaches as x approaches a specific value.

Let [tex]f[/tex] be a function defined on an open interval containing [tex]c[/tex] except possibly at [tex]c[/tex] and let [tex]L[/tex] be a real number.

[tex]\displaystyle{\lim_{x\rightarrow c}f(x)=L}[/tex]

means that for each [tex]\epsilon > 0[/tex] there exists a [tex]\delta > 0[/tex] such that if

[tex]0 < |x-c| < \delta[/tex], then [tex]|f(x)-L| < \epsilon[/tex]

Basically it means that you can always find a range for x to be in close to y so that f(x) is always less than epsilon away from your limit L.  Note that x can't simply pass through L before reaching y and then go somewhere else because then the values closer to y won't satisfy the definition.

[quote="Spoon"]Later tonight, I don't have time now... and frankly, it doesn't make too much sense, and I consider this definition (in single var calc) moot and somewhat useles. It has some more valuable applications later on in calculus, but for someone just learning, I think it's a bad thing to teach. Just my personal opinion.[/quote]

Yeah, but it's one of the foundational ideas upon which analysis is built.  Yes, it's nice to be able to apply calculus, but you also need to understand it.


It is $a_i = \frac{a_1 b_i}{b_1}$ for all $i$. Set this in to and use distributivity to get a trivial equation (this shows that we don't need to restrict to $3$ variables).

[quote="ashwinrk_jain"]if the problem is easy then please tell the detailed solution[/quote]

Let $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}=k\ (k\neq 0),$ we have $a_i=b_ik\ (k=1,2,3).$

Thus
$p_1a_1^n+p_2a_2^n+p_3a_3^n=p_1(b_1k)^n+p_2(b_2k)^n+p_3(b_3k)^n$
$=(p_1b_1^n+p_2b_2^n+p_3b_3^n)k^n$

$\Longleftrightarrow k^n=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}.\ Q.E.D.$

Here's my approach. I'd normally use equation array, but I just finished typing this on the AoPS Wiki, so copying and pasting was much faster. :D

[hide="Alternate Solution"]Instead of proving the two expressions equal, we prove that their difference equals zero.

Subtracting the LHS from the RHS,
$0= \frac{p_{1}a_{1}^{n}+p_{2}a_{2}^{n}+p_{3}a_{3}^{n}}{p_{1}b_{1}^{n}+p_{2}b_{2}^{n}+p_{3}b_{3}^{n}}-\frac{a_{1}^{n}}{b_{1}^{n}}.$

Finding a common denominator, the numerator becomes
$b_{1}^{n}(p_{1}a_{1}^{n}+p_{2}a_{2}^{n}+p_{3}a_{3}^{n})-a_{1}^{n}(p_{1}b_{1}^{n}+p_{2}b_{2}^{n}+p_{3}b_{3}^{n})$
$=p_{2}(a_{2}^{n}b_{1}^{n}-a_{1}^{n}b_{2}^{n})+p_{3}(a_{3}^{n}b_{1}^{n}-a_{1}^{n}b_{3}^{n})=0.$
(The denominator is irrelevant since it never equals zero)

From $ a_{1}/b_{1}=a_{2}b_{2},$ $ a_{1}^{n}b_{2}^{n}=a_{2}^{n}b_{1}^{n}.$ Similarly, $ a_{1}^{n}b_{3}^{n}=a_{3}^{n}b_{1}^{n}$ from $ a_{1}/b_{1}=a_{3}/b_{3}.$

Hence, $ a_{2}^{n}b_{1}^{n}-a_{1}^{n}b_{2}^{n}=a_{3}^{n}b_{1}^{n}-a_{1}^{n}b_{3}^{n}=0$ and our proof is complete.[/hide]

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1969Problem1

Vo Duc Dien

We have the property that if $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=K$
Then $K=\frac{a+c+e}{b+d+f}$
We can use this and solve this question very easily.

[hide=Solution]Suppose that $$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}=k.$$ Substituting the values of $b_1,b_2,b_3$ into the equation, we find that $$\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}=\frac{k^n(p_1b_1^n+p_2b_2^n+p_3b_3^n)}{p_1b_1^n+p_2b_2^n+p_3b_3^n}=k^n=\left(\frac{a_1}{b_1}\right)^{n},$$ as desired.[/hide]

Let $a_1/b_1=k$. Then we have $k^n=\frac{p_1(b_1k)^n+p_2(b_2k)^n+p_3(b_3k)^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}$, which is clearly true, since the numerator is just the denominator scaled by k^n.


How many ordered integer pairs $ \left(x,y\right)$ are there satisfying following equation:
\[ y \equal{} \sqrt{x\plus{}1998\plus{}\sqrt{x\plus{}1998\plus{}\sqrt{x\plus{}1997\plus{}\sqrt{x\plus{}1997\plus{}\ldots \plus{}\sqrt{x\plus{}1\plus{}\sqrt{x\plus{}1\plus{}\sqrt{x\plus{}\sqrt{x} } } } } } } }.\]

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ 1998  \qquad\textbf{(D)}\ 3996  \qquad\textbf{(E)}\ \text{Infinitely many}$

Since an irrational plus rational is irrational, and the root of an irrational is rational, we must have $ x$ as a perfect square, and $ x \plus{} \sqrt{x}$ as a perfect square also. But the difference between consecutive integer squares is $ (x \plus{} 1)^2 \minus{} x^2 \equal{} 2x \plus{} 1$ which is always larger than $ \sqrt{x}$ so $ x \plus{} \sqrt{x}$ cannot be a perfect square and so there are no solutions


Find the length and direction (when defined) of u × v.

u = 4i + 2j + 8k, v = -i - 2j - 2k

===================

u.v=-24

[quote="curioustoknow"]u × v[/quote]

you should indicate what $ \times$ is.

That's the [url=http://en.wikipedia.org/wiki/Cross_product]cross product[/url]. Just calculate it.

Okey "cross product". Moreover,

[quote="curioustoknow"]u.v=-24[/quote]

he/she should write $ u \cdot v$. That's "dot product" http://en.wikipedia.org/wiki/Dot_product

hello.
sorry about the confusion it's a the dot product u.v

But your original post said "find the length and direction." The dot product is a scalar and does not have a direction. The cross product is a vector and does have length and direction. With the equation phrased that way, you did mean the cross product.

that s what it says on the homework quiz

So $ u\times v$=12i - 6k
so $ |u \times v| \equal{} \sqrt {180}$ and direction is just the unit vector of u x v.
so option (c) is correct no. :)

[quote="curioustoknow"]how did you get u.v= 12i-6k  ??

Dot product rule says that : u.v= u1.v1 + u2.v2[/quote]

Cross product, not dot product.

got it
the length is 6 square root 5
direction is 2(square5/5)i-(square root 5/5)k

Thanks for clearinf this up to be cross product and not dot product


Assume that the polynomial \[t^{n}+a_{n-1}t^{n-1}+...+a_{1}t+1\]
has positive coefficients and $n$ real roots.

Show that $f(2) \geq 3^{n}$

Let's hope this is not too easy.
Other than in the topic concerning the system in three variables, I solved this problem myself first so I can assure you that it has a nice solution  :P
(It comes from the same course)

Nice.
Someone with less inspiration than you could also use Hölder to prove that \[\prod_{i=1}^{n}(x+t_{i}) \geq (x+\sqrt[n]{t_{1}t_{2}...t_{n}})^{n}\]
and then fill in $2$.

[hide="other solution"]By Vieta's theorem $a_{i}\geq\binom{n}{i}$, therefore $P(2)\geq (2+1)^{n}=3^{n}$(By binomal theorem).[/hide]


number theory
prime numbers
number theory open

I have a question ( may be i have read somewhere in Mathlinks but i can't remember)
Does there exist infinitely many positive integer k such that :
2^k + 1 is a prime number

well it seems that you are asking whether there are infinitely many primes in the fermat's series 2^2^n   +1,since 2^n +1 can be prime only if n is a power of 2 to obtain that suppose that n=p.2^s,where p is odd then  2^(p.2^s) +1=(2^(2^s) +1)(   1-2^(2^s) +2^(2.(2^s)) -2^(3.(2^s))+......+2^((p-1).2^s)   ) which clearly cant be prime. the question about the fermat series is open and it seems that we know hardly anything about it

The number $2^k+1$ can be prime only if $k=2^l$. It is known that $2^{2^0}+1$, $2^{2^1}+1$, $2^{2^2}+1$, $2^{2^3}+1$ and $2^{2^4}+1$ are prime numbers, but $2^{2^5}+1$ is composite. No other prime numbers of form $2^{2^l}+1$ are known (so called Fermat prime numbers).


AMC
USA(J)MO
USAMO
Olimpiada de matematicas

[color=blue]La secuencia ($a_{n}$) es definida por: $a_{1}= \sqrt{2}$, $a_{2}= 2$
y $a_{n+1}=a_{n}a_{n-1}^{2}$ para $n_{1}\geq 2$.

Probar que para cada $n \geq 1$, tenemos que:

\[(1+a_{1})(1+a_{2}) \cdots (1+a_{n}) < (2+\sqrt{2})a_{1}a_{2}\cdots a_{n}\][/color]

Carlos Bravo
Lima -PERU

Por ahora voy a a hallar explicitamente $a_{n}$, si alguien se anima completa la desigualdad, si no completo la solución después.

Hallando los primeros valores de $a_{n}$ nos damos cuenta que son potencias de 2.
 Sea $a_{n}=2^{f(n)}$, luego $f(1)=\frac{1}{2}$, $f(2)=1$ y $f(n+1)=f(n)+2f(n-1)$.
Podríamos probar defrente que $a_{n}=2^{2^{n-2}}$ (por inducción) dandonos cuenta la forma de $a_{n}$ para valores pequeños, pero vamos explicar de donde sale está fórmula: La recurrencia $f(n+1)=f(n)+2f(n-1)$ tiene ecuación característica $x^{2}=x+2$ que tiene raíces $x_{1}=2$ y $x_{2}=-1$, luego $f(n)=a.2^{n}+b.(-1)^{n}$ , para ciertas constantes $a$ y $b$, para hallar estas constantes usamos los valores iniciales, resolviendo el sistema: $f(1)=2a-b=\frac{1}{2}$, $f(2)=4a-b=1$ obtenemos que $a=\frac{1}{4}$ y $b=0$, luego:
\[f(n)=\frac{1}{4}2^{n}=2^{n-2}\]
$Tipe$


Uhh...You can divide 42 by 7 and get 6, divide that by 3 to get 2, and multiply that by 43 to get 86.. :roll: lol

Probably a computer mistake...

Because computers randomly mess up on one graph out of 6?

probably due to different incraments on the y-axis

[quote="happyme"]Dude who entered the stats was *cough* under the influence *cough*

 :?:

Or you're reading it wrong?[/quote]

I won state without studying, give me some credit.

[quote="zserf"]...  if you add up the amount of people that got each question right in the sprint round it is nearly 500 less then the totals of the individuals in the sprint round when they should be equal.

Can any one think of an explanation?[/quote]
Yes.  I know a few reasons this could happen.  Let's not speculate.  If you are concerned, point out the inconsistency to your coach.  If your coach doesn't already know the reason, ask them to ask your State Coordinator.

[quote="jli"]how do u study for mathcounts?[/quote]

I rarely study math in general, normally I just do my homework.

by the way, the word is spelt y-o-u

Computers aren't always reliable. My state made a mistake with the result sheet too. They listed wrong names for the top 20 individuals. Luckily they have a correct list for the countdown round (or did they :roll: ... just kidding).

When only one graph is screwed up it makes you wonder.

With the whole studying thing I might be like that chess player who said he never studied but flunked out of college because he played chess too much.

[quote="kstan013"]from wi, there was a kid named christian, and they listed him as christina :rotfl:  :rotfl:

exaggeration scenario: what if everyone aced the sprint and got 0 on the target? They wouldn't have to be equal...[/quote]

LOL.  He was from my school, and he wasn't too happy about that...

[quote="kstan013"]exaggeration scenario: what if everyone aced the sprint and got 0 on the target? They wouldn't have to be equal...[/quote]

There are 6 graphs total listed as (X,Y) one graph is (Individual sprint score, number of students) and a second one is (sprint round question, number correct)

If you multiply the number of students that got a specific score by that score for all the scores (If X students got a score of Y and A students got a score of B then XY+AB) it should equal the sum of the correct questions (I.E. if XY+AB=20 then there should be 20 correct answers to questions)

If you do the same thing for the target round it equals out.

[quote="zserf"]before mods start yelling at me this discusses the number of students, not any scores

according to the nice graphs they gave out after state there were 42 individuals, 42 people took the target round, but 86 people took the sprint round. Also if you add up the amount of people that got each question right in the sprint round it is nearly 500 less then the totals of the individuals in the sprint round when they should be equal.

Can any one think of an explanation?[/quote]

don't worry

on the thing at chapters

it said i got a 16 on sprint

and it goes like 110100101010010101001 (not actual, just showing)

so there were 25 1's, but it said 16 next to it :rotfl:

and i still get 6th written

and with target they had me as a 5, but 6 1's

jorian

[quote="kstan013"]...and why isn't my scenario possible?

when you say "number" they got right do you mean problem number or quantity number[/quote]

if on the sprint round you take the quantity if students who got each question right it should equal the combined total of the sprint round scores. and it doesn't

for example say 4 people are taking a 3 question test

1 person gets 1 point=(1*1)=1
2 people get 2 points=(2*2)=4
1 person gets 3 points=(3*1)=3
1+4+3=8 therefore there were 8 correct answers.

one possibility is that
3 people got question one
3 people got question two
2 people got question three

3+3+2=8

seriously. it's not like fixing the graph wouldv'e made any difference at states...or would it :huh: . my state doesn't make graphs (idk we're probably too underfunded...or not)

[quote="kstan013"]WHY DOES IT MATTER???[/quote]
Besides the whole "integrity of the results" thing?
I'd like to know how it happened because I am a naturally curious person.


OK I kept getting pm's from this one user whose pms I am not interested in reading. Would it be possible to set up a filter for just me?

EDIT: [b]never mind about this. i got it settled out with the person[/b]

able to filter PMS? quick, tell me how to do it too!

OK So you can't filter PMs. I just sent him a reply saying never to PM me again and he stopped.


advanced fields
advanced fields unsolved

Supppose that $G$ is a topological group which is First countable. Prove that if $G$ is seperable or if it's Lindelöf, then $G$ has countable basis.

I suppose that First countable means it has the first axiom of countability...
If $G$ is separable, let $A$ be a subset of $G$, countable, so that $A^_=G$. For each $x\in A$ we take the neighbourhoods of $x$ wich are countable because it has the first axiom of countability. The reunion of all these subsets when $x\in A$ is a basis for $G$.

If $G$ is Lindeloff it's not necessary separable.
Example:
$\mathbb R_l$: the open sets are $[a,b)$.


What is the only month that can ever begin with the same day of the week as the month before it?

We notice that February can have $ 28$ days, the only multiple of $ 7$, so every non-leap year, the first day of February is the same as the first day of $ \boxed{\text{March}}$.

When there is a problem asking for a "unique" month asking for stuff like this, remember that the only month with unique days is February. So, remember, if the problem has something to do with the end of the month, remember to put March, and if it asks for like the beginning, put February. (In this case, the answer is March because it had something to do with the end)


Hi all,

I am just doing some review for my analysis exam tomorrow, I have a question on Fourier coefficients.

I know that a necessary condition for f to be continuous on S^1 is that f have square summable Fourier coefficients, and a sufficient condition is that they are absolutely summable. However, I came across the following question in a previous year's exam that relates to this but I can't solve the second part:

a) Prove the map T: little el-one(Z) to C(S^1) defined as T({a_n}) = sum over n of (a_n*exp(2*pi*i*n*t)
is a bounded linear operator (wrt. uniform norm on C(S^1)).

No problems with part (a), then part b) says: Prove there exists a function f in C(S^1) st. the sequence of its Fourier coefficients is not in little el-one(Z).

Now I know such a function must exists b/c in class it was mentioned about that there were some continuous functions that only had square summable coefficients, but we never proved it and I am not sure how to use part (a) here to prove existence, I thought perhaps assuming no such f existed and then using open mapping theorem to show the two spaces are isomorphic but then I got stuck.

Anyways I would really appreciate any insight, as I have an exam tomorrow and want to be as prepared as possible. Thanks.

Did you go over an example of a continuous function whose Fourier series doesn't converge at a point? I'm asking because most courses include it and that function cannot have an absolutely convergent Fourier series for sure.

No we did not, is there any easy example of one that you know?


Please take the time to make your posts make sense.

eren_nike copied my post from another topic asking for explaining I suppose

Oh, I know that.  There was nothing wrong with your post.  But his post was a) a direct copy of your post into a new thread, which wasn't really warranted, and b) not clear at all.

I think that is a good proof. If i will find i send.

Let $ABC$ is a right triangle, $\angle C=90^\circ$, $CL$ is a bisector of $C$. Then $S(L)=\frac{2ab}{a+b}=HM$.

Well, let $G$ be such point on $AB$ that $AG:GB=\sqrt{b}:\sqrt{a}$, then $S(G)=\sqrt{ab}$.


For how many real values of x is $\sqrt{120-\sqrt{x}}$  an integer?

[quote="Jhonny Ruiz"]$x$ could equal
$20^2, 39^2, 56^2, 71^2, 84^2,95^2,104^2, 111^2, 116^2, 119^2,$ $or$ $120^2$

That's 11 possible answers[/quote]

Got proof that those are the only ones?????

[hide]
$x$ is a integer only when $120-\sqrt{x}$ is a perfect square. There are 11 perfect squares where $0 \leq p \leq 120$. Therefore, there are $11$ real integer values of $x$[/hide]

Yup, that is exactly my thinking but still Jose... do you have any other solutions to the problem above.....

Nope, that's all.  We know that $0\le x\le 120^2$.  Therefore let $x^2=y$.  Then since y is an integer, $0\le y \le 120$  Then since there are exactly 11 perfect squares less then 120, 11 is the answer.

[quote="Hamster1800"][quote="Ignite168"]Well what about 1? -1 is a square root of 1, 120 - - 1 = 121 which is a perfect square. Since the problem doesn't specify positive square roots, it has infinite solutions.[/quote]

$\sqrt{x}$ is the [i]principle[/i] square root of $x$. The principle root of a number is the most like the original:

If you have the equation $x^2=4$, then $x=\pm2$. HOWEVER, if you have $\sqrt{4}$, you only get $2$, not $\pm2$. If you wanted to get $-2$, you'd have to specify the negative so it would be $-\sqrt{4}$[/quote]
I stand corrected, never learned that. :huh:

The resulting integer can be anywhere from $0$ to $10$.
Thus,
$\boxed{11}$

[quote="Ignite168"]Well what about 1? -1 is a square root of 1, 120 - - 1 = 121 which is a perfect square. Since the problem doesn't specify positive square roots, it has infinite solutions.[/quote]

That was my example, I think that if the problem does not specify, you must consider the negative roots too

That's exactly what I used to think several years ago. However, it is well known that a square root equation only has ONE answer, namely the positive root, whereas a quadratic equation has two. Think of it this way. If the opposite were true, then there would be many contradictions. For example $1=-1$ could easily be proved:
$-1=-1$
Square both sides to get:
$1=1$
Take the square root of both sides:
$\sqrt{1}=\sqrt{1}$
Since $\sqrt{1}$ is, by the original supposition (yours) equal to $1$ as well as $-1$, then we can take one of the roots for each side of the equation and get:
$1=-1$
Which we all know is not true!
  :)

Yes, that proof is incorrect (we all know that)
We know that $\sqrt{a}$ as two solutions, named $\pm b$. However, $+b \ne -b$.
Those proofs seemed correct, but we all know that's untrue.

[quote="kevyy"]:P  too many[/quote]

??????????????

well...i got 11 just because the integers have to be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. (0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100)
and we can subtitute any value of x to manipulate it to make a $120 - n^2$equal a square.

..before i was confused bcuz i thought there were an infinite number of solutions given that since we can take something like $\sqrt{n^2}$ and get $\pm{n}$ but i had no clue about the "principle square root"  :o


AMC
USA(J)MO
USAMO
induction
geometry
3D geometry
pyramid

To what extent do you think that someone had to write a proof for any of the questions to get partial credit, for example number five; I think that proving the cases for all rational r_i was trivial.  How many points would that get?  None?

[quote="Sean Gu"][quote="worthawholebean"][quote="Sean Gu"]I don't think for proving rational u get any points.....or else everybody would have gotten at least 1 point... :D[/quote]Not really. Some people happened to spend 4:20 on their solution for #4.[/quote]
what about for #6 I first considered the guys with no friends..... does that get 1 point ...hopefully  :D[/quote]

No, not non-trivial.

[quote="Sean Gu"][quote="worthawholebean"][quote="Sean Gu"]I don't think for proving rational u get any points.....or else everybody would have gotten at least 1 point... :D[/quote]Not really. Some people happened to spend 4:20 on their solution for #4.[/quote]
what about for #6 I first considered the guys with no friends..... does that get 1 point ...hopefully  :D[/quote]

All mathematicians have no friends, so I'd expect seven points if you did that.

:)

[quote="JAM"][quote="Sean Gu"][quote="moogra"]If you want to do it that method, you'll have to do all cases. I tried that and found that my 4 pages of work was crap and I started over.[/quote]
no, I mean only considering the guys that have no friends...(like the first step in the official solution)... :D[/quote]
I think that's probably too trivial to be worth anything.[/quote]
That's what I thought too,,,, :lol:

[color=green]Please stop the pyramid quoting. -mod[/color]
[color=blue]Aww...pyramid quoting is fun. Well, I'll just put the line that I really need to quote since it isn't the post above me[/color]

[quote="Lazarus"]
All mathematicians have no friends, so I'd expect seven points if you did that.

:)
[/quote]

Hmm...I've heard various accounts of this solution. Some are
"All mathematicians are friends."
"A mathematician usually does not have more than two friends".

Which one is true? :)

In problem 3, I reverse the x and y axis(x become vertical and y becomes horizontal) but a lot of times when I define a point, I forget and use the coordinate system which is not reversed (for example, when defining odd frame and even frame as in official solution). But I draw the figure so I think grader know what my actually thought is. Also, at last, I say that every path originally will have at most 1  point difference in both region (odd and even), add the n special place where path can go through two points in the same region, so the points in two region will be differ at most by 2n-1, contradict to the fact that they should differ by 2n. Unfortunately, because of extreme time limit, I didn't even quite specific where the n special places are. So though I had the correct idea, the solution is a messy. How much points do you think I can get (1? 2?..)

[quote="matt276eagles"]You'll probably get a point for that, it's not so trivial that every single person taking the USAMO would have figured it out.  Besides, it's progress towards a solution.

Take a look at last year's #5 - although it's fairly obvious to try induction, the graders gave at least one point (some people even thought two points after seeing their total scores) for just setting up the induction.[/quote]
I heard from zabel that it was worth 2 points

Wow that was worth two points?  Hehe...that's actually pretty depressing for me since it shows how little I got for the rest of the test.

For last year's #4, didn't they give out a point for giving the numerical answer and setting up induction? (so, for this year's #3, do you think it will be similar if one does some easy work and sets up an induction?)

[quote="matt276eagles"]Sure, but that work has to lead toward a solution.  If you prove some completely irrelevant stuff you're probably not gonna get much credit, but if you make the first step or two towards a legitimate solution you can easily earn a point.[/quote]
I think mine was actually a step toward a legitimate solution~~ :D  :blush:

thats pretty ambigious...i can say drawing a good picture is a legitimate step towards a solution but they will never reward any points for that  (even if you look at the USAMO solutions they always have a really good picture)


limit
real analysis
real analysis unsolved

Find a series $ \sum_{n = 1}^{\infty}a_{n}$ satisfy:

$ 1.$ $ a_{i} > 0$, for $ \forall$ i.

$ 2.$ $ \overline{\lim_{n \rightarrow \infty}}na_{n} = 1$

$ 3.$ $ \sum_{n = 1}^{\infty}a_{n}$ converges    [color=red](Edit, I have forgot to add this condition)[/color] :D

[color=red]I think this series:[/color]

$ a_{n} = \{\begin{array}{cc}\frac {1}{n^2}, & \mbox{ if }n \not = 2^m \\
\frac {1}{n}, & \mbox{ if }n = 2^m\end{array}$

where $ m \in \mathbb{N^{ + }}$

will be ok.

but i can't give a prove to it  :maybe:

I think the harmonic series, $ \sum_{n\equal{}1}^{\plus{} \infty} \frac1n$ also qualifies.

[quote="Carcul"]I think the harmonic series, $ \sum_{n \equal{} 1}^{ \plus{} \infty} \frac1n$ also qualifies.[/quote]

sorry ,I forget to add the condition: $ \sum_{n\equal{}1}^{\infty}a_{n}$ must be converges :D

The sequence defined by
$ a_{n}=\{\begin{array}{cc}\frac{1}{n^{2}}, &\mbox{ if }n\not = 2^{m}\\ \frac{1}{n}, &\mbox{ if }n = 2^{m}\end{array}$
satisfies, for infinitely many $ n$
$ na_n = 1$
and, for all $ n\in\mathbb{N}$ we have $ na_n \leqslant 1$. So, $ \limsup na_n = 1$

We have
$ \sum\limits_{n} a_n < \sum\limits_{n} \frac{1}{n^2} + \sum\limits_{m} \frac{1}{2^{m}} < +\infty$

since, our series converges.


Let $a,b,c$ be real numbers such that  $\frac{c}{a}+\frac{a}{b}+\frac{b}{c}=-\frac{3}{2}$.

Prove that  $\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}\geqq 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)$

Let s = $\sum_{cyc} b/a$

$s^2 = LHS + \sum_{cyc} c/a = LHS - 3/2$

so:

$LHS \geq 2s$
$\Leftrightarrow s^2 -2s + 3/2 \geq 0$

But the discriminant is -2.

Yes, just similar solution to the previous post

Squaring both sides of $\frac{c}{a}+\frac{a}{b}+\frac{b}{c}=-\frac{3}{2}$
we get $\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{b^2}{c^2}+2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)=\frac{9}{4}$

So we are to prove

$\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}\geqq \frac{9}{4}-(\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{b^2}{c^2})$

This is equivalent to
$\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{b^2}{c^2}\geqq \frac{9}{4}$, which is obvious because by AM-GM

$\frac{b^2}{a^2}+\frac{a^2}{b^2} \geq 2$, $\frac{c^2}{b^2}+\frac{b^2}{c^2} \geq 2$ and $\frac{a^2}{c^2}+\frac{c^2}{a^2} \geq 2$ and hence $\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{b^2}{c^2}\geqq 6> \frac{9}{4}$

good to mention that having the condition be = -1 is stronger


Write down the negations of the following assertions (where m, n, a, b are natural numbers):

a) if Coke is not worse than Pepsi then nothing Mandelson says can be trusted.
b) $ \forall m \exists n\forall a\forall b (n >\equal{} m)$ /\ $ [(a\equal{}1)$ \/ $ (b\equal{}1)$ \/ $ (ab \ne n)]$

[quote="Pythagorean12"]Write down the negations of the following assertions (where m, n, a, b are natural numbers):

a) if Coke is not worse than Pepsi then nothing Mandelson says can be trusted.
b) $ \forall m \exists n\forall a\forall b (n > \equal{} m)$ /\ $ [(a \equal{} 1)$ \/ $ (b \equal{} 1)$ \/ $ (ab \ne n)]$[/quote]

a) Coke is not worse than Pepsi and everything Mandelson says can be trusted.
b) $ \exists m \forall n \exists a \exists b (n < m)$ ∨ $ [(a \neq 1)$ ∧ $ (b \neq 1)$ ∧ $ (ab \equal{} n)]$

In general, the negation of a conditional is iff. the former statement is true but the latter statement is false.

For the second one, switch $ \forall$ and $ \exists$ and negate the statement.
In this case, the statement is a combination of things. you have to negate each term, keep parentheses, and switch "and" and "or."

The negation of "nothing" is certainly not "everything"!

Useful LaTeX: \vee, \wedge, \geq, \leq.


Since maybach revived the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293298]1 vs. 100[/url] thread, I thought maybe I could make another one on a similar vein.

Once again, the questions will be about math. However, some the questions might be more about mathematical trivia, instead of the math itself. There will be one contestant at a time, and that contestant will have any amount of time to answer a series of 15 multiple-choice questions with four answers, advancing him or her to the ultimate goal of 1 million dollars.

This contestant will have three lifelines along the way:
1. 50:50. This lifeline gets rid of two wrong answers, leaving one wrong answer and the correct answer.
2. Ask the Audience. This lifeline gives one day for people to send me PM's about what they think the answer is. No reasoning please, just the answers themselves.
3. Ask an Expert. This lifelines gives another AoPS user that isn't a contestant a chance to post his or her views about the question on this thread. The post may not be longer than 1,000 characters, and comments do count as part of the 1,000.

And an extra 2 along the way:
4. Double Dip. This lifeline allows you to answer a question twice. However, once you use this lifeline, you can't walk away, so be careful. Coupled with the 50:50, this lifeline allows you one free question.
5. Consult Wikipedia. This lifeline allows you to look at one Wikipedia page of the said topic, and use it in your response. (Please don't do so anyway... It just ruins the fun.)

Now, we'll start with a fastest-finger question, but we need 10 contestants first.

The fastest-finger question requires contestants to put a certain group of 4 items in some given order. For example:

[i]Question:[/i] Put these mathematicians in order of their birthdates:
A) René Descartes
B) Pierre de Fermat
C) Sir Isaac Newton
D) Leonardo Da Vinci

The correct answer is DBAC.

[quote="maybach"]I'm joining.  And there's no negative connotation when you said "revival", right?[/quote]

Nope, none at all. *awesome face*

[quote="maybach"]Do we have to PM you what we think is the result to the fastest finger question or do we post it?[/quote]

PM. But we don't have 10 players yet.

I suppose I will join, but this will hopefully be over by the start of school.

Only 1 more person left.

Will multiple games be played or just one?

Multiple, obviously... We'll keep playing until all 10 players have run out.

BTW, in the fastest finger question, the [i]first[/i] person to come up with the correct response is the person who will get to be the contestant.

If nobody registers by 9:00 PM tomorrow, we'll start the game with only 9 players.

Those questions are pretty hard for hundreds of dollars.  That last one should've been the $16,000 question at least.

[quote="maybach"]Those questions are pretty hard for hundreds of dollars.  That last one should've been the $ \$16,000$ question at least.[/quote]

Ah. I'm sorry, I guess my questions are just hard at times...shall I try again?

For $ \$500$:

[i]The Moon takes 29.53 days to orbit the Earth, so there are about 354 days in 12 months. Assuming that a year on Earth lasts exactly 365 days, how many new moons are there in six years, if a new moon occurred on December 31 of the year before the first year?[/i]

A) 36
B) 37
C) 38
D) 39

[quote="maybach"]I'm not going to get out on a question before I at least get a little help.

50/50 please.[/quote]

Alright, the answers to be removed, courtesy of Random.org:

A) 36
D) 39

Leaving two answers: B) 37, and C) 38.

[quote="maybach"]I'm gonna press my luck! [b] D[/b][/quote]

Um, D was one of the eliminated answers...

The answers left are:

B) 37
C) 38

[quote="maybach"]Misread sorry............[b]C[/b][/quote]

Pfft... sorry, it was B.
Was this question too hard again...? I honestly need feedback on the difficulty of my questions.

No, that question was fine.  I realized I made a really careless mistake................

[quote="sdkudrgn88"]I'm sorry about missing the deadline, but I'm going to be re-opening signups until we have 10 players again.[/quote]

I'll join.

[quote="polkadotman1"]ok im joining.

for $ \$$1000:

A)2 trillion

B)2.3124 trillion

C)1.2411 trillion

D)3 trillion

pliz tell me if this is too hard.[/quote]

Oh, um, first of all, I'm the one who makes questions, you're supposed to be a contestant who answers them.

Second of all: I choose B.

Oh, sorry, I didn't know.

Sorry....

And it's C.

But that doesn't count.

[quote="polkadotman1"]And it's C.

But that doesn't count.[/quote]

You know, there wasn't even a question. Anyway, I might not be able to continue the game later on, considering that school starts in about... a week or so.


Does Anyone have any Chapter and State Sprint rounds before 2004? I don't have much to offer though.

I have 2000 state and chapter and 2003 chapter.

Have a look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189]here[/url]. :wink:

Do you have anything to trade? I need MC tests kinda badly.

Well what do you want? You probably have everything that I have though. So ask


geometry
3D geometry
prism
parallelogram
geometry unsolved

delegat wrote:

Interesting

--------------------------------------------------------------------------------

Show how to divide rectangular prizm ino three equal volumes such that vertices of smaller pieces are vertices of prizm.

answer
Let the prism be ABCDA'B'C'D', with ABCD parallelogram and AA', etc edges.
ABCDB', ADD'A'B' and CDD'C'B' is one division (each different face and the vertex opposite to the common one).

Best regards,
sunken rock

A blind man sees better the details


A gentleman whose age is between thirty and forty   years  is found murdered on his bed.
The murder weapon is a pair of scissors but  no marks are visible in his body.
 How did he die?

Is this all the inmformation provided and by teh way what has the agew got to do with this

[hide]It was a waterbed, someone cut it open and he drowned[/hide]


Perform substitution $ x=2/a, y=2/b, z=2/c$ and you'l reduce it to if $ABC=1$ then $\frac{1}{2+A}+...\leq 1$ which has been discussed at least 7 times.

Myth, If I will ever write a book about inequalities, I will ask your permission to use this great method. Hope you will agree.  ;)  :)

Could you explain to me a little bit more about the substitution...

What should I explain? You introduce in the relation and it's done.

[quote="Myth"]IPT again...
And I remember the general inequality :)[/quote]
What does IPT stand for?

For the original one you can just open the brackets to get: $x^{2}+y^{2}+z^{2}\ge{12}$. :D
Can you post the generalization and indeed, what is IPT?

The general inequality must be the one discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=19230 . As for the IPT (= inflection point theorem), I remember it has been posted here, but I can't find it anymore. Maybe it was lost during the hacker attack on ML. Anyway, if I am not mistaken, the IPT is a generalization of the Jensen inequality, and the core of the IPT is that if a function is convex on some interval and concave on another, then, in order to minimize the sum of the values of this function over some numbers, one should move together the numbers in the interval where the function is convex and move asunder the numbers in the interval where it is concave. Still it would be very nice if Vasc or somebody else would post the IPT in its exact formulation.

  Darij

Yes could someone please do that? Concavity/Convexity arguments are always great.

[quote="darij grinberg"]...Still it would be very nice if Vasc or somebody else would post the IPT in its exact formulation.[/quote]
"Somebody else" who invented IPT doesn't want to do it...

[quote="Vasc"]The IPT is Mith’s creation. My creations are RCF-Theorem and LCF-Theorem.

[b]Right-Convex Function Theorem [/b](RCF-Theorem).
Let $f$ be a function on an interval $I \subset R$ and convex for $x\ge s$, $s \in I$. If
$f(x_1)+f(x_2)+...+f(x_n)  \ge nf(\frac{x_1+x_2+...+x_n}{n})$   $(1)$
for all $x_1,x_2,...,x_n \in I$ such that $x_1+x_2+...+x_n=ns$ and $x_1=x_2=...=x_{n-1} \ge s$,
then (1) is true for all $x_1,x_2,...,x_n \in I$ such that $x_1+x_2+...+x_n \ge ns$.[/quote]

Vasc, can you show us one exmaple that may use RCF or LCF?

i really don't know what to say...
totally incredible generalization!!

I revive this topic to ask for the proof of RCF. [b]Vasc[/b] can you provide it or maybe your article is available somewhere on the Web?

Also I'd like to see complete formulation of [b]Myth's[/b] IPT (have you already written an article about it?)but I suppose the author won't share with us his beatiful theorem  ;) (I'm trying to play on an ambition  :P)

I wrote a proof for RCF last autumn for one my good friend (as I remember it was a Karamata's application). Note that I also wrote one proof on the forum, but it uses IPT.
I could share those proof on the forum too, but I am affraid Vasc would killed me.

So I hope Vasc will allow to post proofs of RCF

And what about mystical IPT ? I'd [and surely not only me] really like to see it in full. Myth, did you write or do you plan to write any article/book about it?

The simpliest form of IPT (which I invented long time ago) looks like RCF, but I required that the correspondinf function has an unique inflection point. Vasc replaced it with more weak condition, but... One thing, which makes using of RCF sometimes quite difficult, is that we need to prove one-variable ineq. So I added to IPT some sufficient conditions guaranteing that ineq is true. Unfortunately, it involes a bit calculus (remember, (n-1)EV principle also involves calculus).

[quote="Myth"] Unfortunately, it involes a bit calculus (remember, (n-1)EV principle also involves calculus).[/quote]
That's not a problem. BTW, (n-1)EV can, probably, be proved without calculus (perhaps, Vasc even knows such a proof, it is just we who couldn't find it  :P  :blush: ). But, indeed, Myth, why not to post the full formulation of the IPT? Many people including me would be grateful :). Of course, keeping it a forum legend is a lot of fun but it has been this way for sooo looong... :( Also, recall your own reaction when Vasc didn't immediately disclose his (n-1)EV :P.  As to the proof, you may not tell it if you don't want. It will be even more interesting to try to find it by ourselves.

I. To prove the (n-1)EV principle, I didn’t use Lagrange multipliers' method.
II. I hope an article with RCF and LCF theorems will appear soon in Crux Math, in the issue for Klamkin commemoration. If not, I will post a proof in this forum. Of course, if somebody will post a proof in the forum (for either RCF theorem, or (n-1)EV principle) - no problem.

[b]Myth[/b]: please,  you can post now proof of RCF  :)

My proof is far from optimality (despite of its triviality it takes two pages!), since Vasc said that it will appear in Crux, so I guess his approach is much shorter and nice.

who can tell me what is the caramate's inequality in Myth's solution? :)

well,I think I have a nice solution.(maybe similar with VASC's),and can I post it?

[quote="Vasc"] Of course, if somebody will post a proof in the forum (for either RCF theorem, or (n-1)EV principle) - no problem.[/quote]

thank you. :)
now here is my proof as following:
the main idea:induction.
$lemma:(n-2)f(x)+f(y) \geq (n-1)f(s),if ,(n-2)x+y=(n-1)s,x \geq s$
$proof:let,x'=\frac {(n-2)x+s}{n-1}$,we have $(n-1)x'+y=ns$
by the condition we have $(n-1)f(x')+f(y) \geq nf(s)$
so we should only prove $(n-2)f(x)+f(s) \geq (n-1)f(x')$,It can be proved by largrange theorem easily.the $lemma$ is proved.
return the pro,assume $x_1 \leq  x_2 \leq \cdots \leq x_n$,if $x_1 \geq s$,easy.now assume $x_1<s$
we have $f(x_2)+f(x_3)+\cdots +f(x_n) \geq (n-1)f(\frac{x_2+x_3+\cdots+x_n}{n-1})$
so we should only to prove: $f(x_1)+( n-1)f(\frac{x_2+x_3+\cdots+x_n}{n-1})\geq nf(\frac{x_1+x_2+...+x_n}{n}) $
we have:$f(x_1)+(n-1)f(\frac{ns-x_1}{n-1}) \geq nf(s)$(because $x_1<s$)
so we remain to prove $(n-1)f(\frac{x_2+x_3+\cdots+x_n}{n-1})+nf(s) \geq (n-1)f(\frac{ns-x_1}{n-1})+ nf(\frac{x_1+x_2+...+x_n}{n}) $
and it is easy apply to largrange theorem.
we are done.hope you like it :D  :D

EDIT:because f'(x) is not must be exist.so I can't use the largrange theorem to prove some inequ,but I think they can also be proved by the slope. :)  :)

My proof is similar to that one of Myth, but my solution end is a little simpler.

Indeed Zhaobin, the derivative f'(x) is a problem in your solution. ;)

[quote="zhaobin"]so we should only prove $(n-2)f(x)+f(s) \geq (n-1)f(x')$,It can be proved by largrange theorem easily.[/quote]
Lagrange theorem is Jensen theorem, isn't it ? So we're done.

[quote="zhaobin"]so we should only to prove: $f(x_1)+( n-1)f(\frac{x_2+x_3+\cdots+x_n}{n-1})\geq nf(\frac{x_1+x_2+...+x_n}{n}) $[/quote]
This ineq. is just the hypothesis for $P(n)$, isn' t it ? Then, it is true.

[quote="Vasc"]Relating my first note, I am right.
Relating my second note, I am wrong,and you are right.  ;)
The last ugly ineq

$\displaystyle (n-1)f(\frac{x_2+x_3+\cdots+x_n}{n-1})+nf(s) \geq (n-1)f(\frac{ns-x_1}{n-1})+ nf(\frac{x_1+x_2+...+x_n}{n}) $

follows by the Karamata Inequality.[/quote]
 :) ,but I don't like to use karamate inequality(because I can't prove this inequality now) ;) ,
I think this can be proved directly. :)

[quote="zhaobin"]I think this can be proved directly. :)[/quote]
I think the same as. ;)

[quote="warut_suk"]If $ xyz \equal{} 8$, then $ \frac {2}{2 \plus{} x^2} \plus{} \frac {2}{2 \plus{} y^2} \plus{} \frac {2}{2 \plus{} z^2}\geq1$.[/quote]Perform substitution $ x^2\equal{}u^3,y^2\equal{}v^3,z^2\equal{}w^3$, and you'l reduce it to

$ \frac {vw}{vw \plus{} 2u^2} \plus{} \frac {wu}{wu \plus{} 2v^2} \plus{} \frac {uv}{uv \plus{} 2w^2}\geq1$

for all positive numbers $ u,v,w$.

[b]Proof[/b] $ \frac {vw}{vw \plus{} 2u^2} \plus{} \frac {wu}{wu \plus{} 2v^2} \plus{} \frac {uv}{uv \plus{} 2w^2}\minus{}1$

$ \equal{}\frac{uvw(u\plus{}v\plus{}w)\left[(v\minus{}w)^2\plus{}(w\minus{}u)^2\plus{}(u\minus{}v)^2\right]}{(2u^2\plus{}vw)(2v^2\plus{}wu)(2w^2\plus{}uv)}\geq0$.

[quote="warut_suk"]If $ xyz \equal{} 8$ then prove that $ \frac {2}{2 \plus{} x^2} \plus{} \frac {2}{2 \plus{} y^2} \plus{} \frac {2}{2 \plus{} z^2}\geq1$.......(1)[/quote]
[b]Proof [/b]  Put  $ x \equal{} 2\sqrt {\frac{{abc}}{{a^3 }}} ,y \equal{} 2\sqrt {\frac{{abc}}{{b^3 }}} ,z \equal{} 2\sqrt {\frac{{abc}}{{c^3 }}} (a,b,c > 0)$,

$ (1) \Leftrightarrow \frac{{a^3 }}{{a^3  \plus{} 2abc}} \plus{} \frac{{b^3 }}{{b^3  \plus{} 2abc}} \plus{} \frac{{c^3 }}{{c^3  \plus{} 2abc}} \ge 1$.........(2)

By Cauchy inequality

$ \frac{{a^3 }}{{a^3  \plus{} 2abc}} \plus{} \frac{{b^3 }}{{b^3  \plus{} 2abc}} \plus{} \frac{{c^3 }}{{c^3  \plus{} 2abc}} \ge \frac{{(\sqrt {a^3 }  \plus{} \sqrt {b^3 }  \plus{} \sqrt {c^3 } )^2 }}{{a^3  \plus{} b^3  \plus{} c^3  \plus{} 6abc}}$
and
\[ \begin{array}{l}
 (\sqrt {a^3 }  \plus{} \sqrt {b^3 }  \plus{} \sqrt {c^3 } )^2  \\
  \equal{} a^3  \plus{} b^3  \plus{} c^3  \plus{} 2\left( {\sqrt {(ab)^3 }  \plus{} \sqrt {(bc)^3 }  \plus{} \sqrt {(cb)^3 } } \right) \\
  \ge a^3  \plus{} b^3  \plus{} c^3  \plus{} 2 \times 3\sqrt[3]{{\sqrt {(ab)^3 }  \cdot \sqrt {(bc)^3 }  \cdot \sqrt {(cb)^3 } }} \\
  \equal{} a^3  \plus{} b^3  \plus{} c^3  \plus{} 6abc \\
 \end{array}
\]

Well, to prove the original problem, just need to notice that if we take $ x\equal{} 2\frac{\sqrt{bc}}{a}, y\equal{}2\frac{\sqrt{ca}}{b}, z\equal{}2\frac{\sqrt{ab}}{c}$, we have

$ \frac{2}{2\plus{}x^2} \equal{} \frac{a^2}{a^2\plus{}2bc} \ge \frac{a^2}{a^2\plus{}b^2\plus{}c^2}$

And similarly get another two inequalities, then sum up.


[quote="Skyward_Sea"]a,b,c are positive reals, prove that
$\Sigma{a \sqrt{a^{2}+2bc}}\geq \Sigma{ab}$[/quote]
$\sum_{cyc}a\sqrt{a^{2}+2bc}\geq\sum_{cyc}a^{2}\geq\sum_{cyc}ab.$

[quote="Skyward_Sea"]a,b,c are positive reals, prove that
$\Sigma{a \sqrt{a^{2}+2bc}}\geq \Sigma{ab}$[/quote]
Is this true: $\Sigma{a \sqrt{a^{2}+2bc}}\geq \sqrt{3}\Sigma{ab}$

http://www.mathlinks.ro/Forum/viewtopic.php?t=34108

  darij

[quote="Skyward_Sea"]a,b,c are positive reals, prove that
$\Sigma{a \sqrt{a^{2}+2bc}}\geq \sqrt{3}\Sigma{ab}$[/quote]
I have an ugly solution  :blush: :
$\sum \sqrt{a^{4}+2a^{2}bc}\ge\sqrt{(\sum a^{2})^{2}+2(\sum a\sqrt{bc})^{2}}$
We have to prove:  $(\sum a^{2})^{2}+2(\sum a\sqrt{bc})^{2}\ge 3(\sum{bc})^{2}\Leftrightarrow \sum a^{4}+4abc\sum\sqrt{a}\ge\sum b^{2}c^{2}+4abc\sum a\Leftrightarrow \sum S_{a}(b-c)^{2}\ge 0$
$S_{a}=(b+c)^{2}(\sqrt{b}+\sqrt{c})^{2}-4abc$
Assume that $a \ge b \ge c$
$S_{b}=(c+a)^{2}(\sqrt{a}+\sqrt{c})^{2}-4abc\ge 4ca[(\sqrt{a}+\sqrt{c})^{2}-b]\ge 0$
$S_{c}=(a+b)^{2}(\sqrt{a}+\sqrt{b})^{2}-4abc\ge 4ab[(\sqrt{a}+\sqrt{b})^{2}-c]\ge 0$
$S_{b}+S_{a}\ge 4c[b(\sqrt{b}+\sqrt{c})^{2}+a(\sqrt{a}+\sqrt{c})^{2}-2ab]=4c[a^{2}+b^{2}-2ab+bc+ca+2b\sqrt{bc}+2c\sqrt{ca}]\ge 0$


Does there exists a set $S$ with empty interior ($\mathring{S}=\phi$) such that $S$ has [b]non-[/b]zero Lesbegue-measure?

Of course there is- the empty set, or a single point, or...

Now, what did you really mean?

Perhaps whose complement has null interior.  But the rationals satisfy that condition.

what about $[0;1] - \mathbb{Q}$ ? or a modified Cantor set ?

An interior point $x_0$ of $S$ is a point for which there exists $\epsilon>0$ such that $(x_0-\epsilon,x_0+\epsilon)\subset S$. But in every such interval there lies a rational number. Hence, $[0,1]\setminus \mathbb{Q}$ has empty interior.

Am I wrong?

but what do you want ? $S$ is a set with empty interior and Lebesgue measure $1$.


calculus
function
analytic geometry
graphing lines
slope
integration
logarithms

You can find an explanation on MathWorld: [url=http://mathworld.wolfram.com/Mean-ValueTheorem.html]mathworld.wolfram.com/Mean-ValueTheorem.html[/url]

Hopefully your problem will become clear.

I still don't know how to solve it....sorry

The only theorem regarding functions that I know that remotely sounds like "Mean Value Theorem" is the Intermediate Value Theorem which is kind of intuition if you think about it  :? but doesn't really have much to do with this problem.

I'd think this is a calculus problem yep (and I don't know calculus so can't help lol)

[quote="probability1.01"]The Mean Value Theorem is calculus, so I don't think it should be in this forum.  There's also another so-called Mean Value Theorem about average values of functions I think, which is still calculus-ish.  The problem is not very clear though, so I don't know what they want.  Maybe there's still another mean value theorem I don't know about.  Really, they could be more specific with their names of theorems :D.[/quote]

They clearly want you to find the value of $c$ for which $(b-a)f'(c)=f(b)-f(a)$ with $c\in (a,b)$

[quote="Kle"]What x on the interval (1,4) satisfies Mean Value Theorem for f(x)=5-4/x ?

What is Mean Value Theorem? Please explain to me[/quote]

The mean value theroem is the average slope between the given interval...

Im going to move this

Since nobody posted a solution I guess I will.

$f'(x)=\frac{4}{x^2}=\frac{f(4)-f(1)}{3} = 1 \Rightarrow x=2$

All this means is the point in the interval at which the average rate of change (secant slope) equals the instantaneous rate of change (tangent slope).

The other mean value theorem probability1.01 was referrring to is $f_{\text{ave}} = \frac{1}{b-a}\int_a^b f(x) \ dx$, which gives the average value of the function ($f_{\text{ave}}$) over an interval from $a$ to $b$.

I was searching for a particular application of the MVT (mean-value theorem) myself when I ran across your post.

To put it simply, the MVT says that for any function differentiable on a closed interval there is a point in that interval where the slope of the line tangent to the curve equals the slope of the line containing the end points of the interval; i.e. where the tangent line and the secant line are parallel.

In other words, there is at least one point on every curve where the instantaneous rate of change at said point matches the average, or mean, rate of change across the interval containing the point.

For example if I said, my average speed while driving from my home to Boston was 60 mph then the MVT implies that at least once, perhaps more often, my speed was exactly 60 mph in my journey from my home to Boston, assuming of course that the route I took was continuous and differentiable between here and there.

The MVT provides a quick and dirty way to differentiate a continuous curve if its values are only known empirically.

In your problem, you gave f(x) = 5-4/x, in the interval [1,4].

The slope of the secant line = (f(4)-f(1))/(4-1) = (4-1)/(4-1) = 1.

The slope of the tangent line at point c = f'(c) = 4/x^2|(x=c) = 4/c^2.

The MVT --> f'(c) = (f(4)-f(1))/(4-1)

4/c^2 = 1  --> 4 = c^2  --> c = sqrt(4) = 2.

2 is contained in (1,4)  Q.E.D.

the maen value theorem is one of the best theorems in calculus.
it basically says if the average rate of change of a function on a given interval, such as $[a,b]$, is some number like $C$. then there exists atleast one point on the interval $[a,b]$ that the rate of change of function at that particular point is $C$.
to make sense out of it think of this: if you travel at an average speed of $40$ miles per hour for one hour, during that one hour there was at least one time where you were travleing exactly at $40$ miles per hour.

[quote="kunny"]For $0<a<b$, prove the following inequality.

\[1-\frac{a}{b}<\ln \frac{b}{a}<\frac{b}{a}-1\]

Source: 1980 Gunma University[/quote]
i've seen this before.

There are several ways to prove Kunny's inequality(ies). But, since this thread is originally about the mean-value theorem, we can use it to solve the problem and illustrate its applications.

We start with the right hand inequality $\ln \frac{b}{a}<\frac{b}{a} -1 $. We can prove it indirectly by proving the following (very useful) inequality:

[b]Lemma[/b]. Denote $\frac{b}{a}$ by $x$. Now, if $x \in R$, then $e^x\geq x +1$.
[b]Proof[/b]. Let $f(x)=e^x$. Then, $f'(x)=f(x)$. Now assume that $x \not= 0$. Then there exists, according to the mean-value theorem, an $y$ between $x$ and zero such that \[e^x-1=f(x)-f(0)=(x-0)f'(y)=xe^y\]Now, if $x>0$, we have $y>0$ and $e^y>1$, which implies $x \cdot e^y>x$. If $x<0$, then $y<0$ and $e^y<1$, which implies $x \cdot e^y>x$. If $x=0$ we obviously have equality, but this is irrelevant since $a$ and $b$ are disctinct. $\Box$

Now we can prove our original problem,  $\ln x < x - 1$. It follows from the Lemma in the following way: If $x>0$ then $x=e^{\ln x}>\ln x + 1$. $\Box$

Now we can proceed to the left inequality, which in terms of $x$ becomes $\ln x > 1-\frac{1}{x}$. Substitute $x$ by $\frac{1}{x}$ in our right logarithmic inequality. Then, $-\ln x = \ln 1 - \ln x =  \ln \frac{1}{x} < \frac{1}{x} - 1$, by some logarithmic laws. This proves our inequality. $\Box$

[quote="Jimmy"]The other mean value theorem probability1.01 was referrring to is $f_{\text{ave}} = \frac{1}{b-a}\int_a^b f(x) \ dx$, which gives the average value of the function ($f_{\text{ave}}$) over an interval from $a$ to $b$.[/quote]

Actually, Jimmy, "the other mean value theorem," so called is a special case of the mean-value theorem proper.

As you said, $f_{\text{ave}} = \frac{1}{b-a}\int_a^b f(x) \ dx$; yet,

$\int_a^b f(x) \ dx=F(b)-F(a)$ by the first fundamental theorem of calculus.

Thus $f_{\text{ave}}=\frac{F(b)-F(a)}{b-a}$, which is the mean-value theorem.

[quote="blahblahblah"]^

There's one more so-called mean value theorem that says if $f$ is continuous (and hence integrable) on $[a,b]$, there exists $c\in [a,b]$ such that

$(b-a)f(c)=\int^b_a f(t)dt$

the proof is not difficult - try and find it![/quote]

I can't prove it (yet), but is it saying that the average value of the function on a given interval exists on that interval, if the function is continuous?

Yes.

Big hint below:
[hide]
If $f$ is continuous on $[a,b]$, it attains its minimum $m$ and maximum $M$.  We have the inequalities $m(b-a)\leq\int^b_a f(t)dt\leq M(b-a)$.  Proceed.[/hide]

$m(b-a) \le \int_a^b f(t) \ dt \le M(b-a)$

$m(b-a) \le F(b)-F(a) \le M(b-a)$

$m \le \frac{F(b)-F(a)}{b-a} \le M$

$m \le f_{ave} \le M$

So since the funtion is continuous, it must have a value corresponding to all points between $m$ and $M$, thus including $f_{ave}$.

I don't like this, because it's not necessarily true that a continuous function has an antiderivative (although all continuous functions are integrable)

anyways, we have $m\leq \frac {1}{b-a}\int^b_a f(t)dt\leq M$, and by continuity, there exists $c\in [a,b]$ such that $f(c)=\frac {1}{b-a}\int^b_a f(t)dt$, the end.

[quote]I don't like this, because it's not necessarily true that a continuous function has an antiderivative (although all continuous functions are integrable)[/quote]

The antiderivative of a function is its indefinite integral by definition.

$ F(x) =\int\nolimits f(x)\mathrm{d}x$

If all continuous functions are integrable, then all continuous functions have antiderivatives.

I'm sorry, it seems that I am just spewing absurdities today.  Anyways, you're (obviously) correct, as the simplest antiderivative can just be given as

$F(x)=\int^x_a f(t)dt$


AMC
AIME
arithmetic sequence
geometric sequence

Hello everyone,

       Would you please tell me how to solve this problem?

In an increasing sequence of 4 positive integers, the 1st three terms form an arithmetic progression, the last 3 terms form a geometric progression, and the 1st and 4th terms differ by 30.  Find the sum of the 4 terms.

       Thank you.


integration
logarithms
calculus
calculus computations

Consider $\int{e^{x}f\prime(x)}\,dx$ to find an expression for:

$\int^{b}_{a}{e^{x}[f(x)+f\prime(x)]}\,dx$ $a,b\in\Re$ $b>a$

Hence, by letting $f(x)=x\ln{x}-x$ evaluate

$\int^{e}_{1}{e^{x}(1+x)\ln{x}}\,dx$

Yes it does...anyone like to attempt a solution?  :lol:


algebra
polynomial
combinatorics proposed
combinatorics

A [i]self-avoiding rook walk[/i] on a chessboard (a rectangular grid of unit squares) is a path traced by a sequence of moves parallel to an edge of the board from one unit square to another, such that each begins where the previous move ended and such that no move ever crosses a square that has previously been crossed, i.e., the rook's path is non-self-intersecting.

Let $ R(m, n)$ be the number of self-avoiding rook walks on an $ m \times n$ ($ m$ rows, $ n$ columns) chessboard which begin at the lower-left corner and end at the upper-left corner. For example, $ R(m, 1) \equal{} 1$ for all natural numbers $ m$; $ R(2, 2) \equal{} 2$; $ R(3, 2) \equal{} 4$; $ R(3, 3) \equal{} 11$. Find a formula for $ R(3, n)$ for each natural number $ n$.

Answer of Reynaldo Gil Pons:

Sorry for my English, it is not so good. I speak Spanish.

Let $ R(3,n) \equal{} y_n$ and let $ x_n$ be the number of self-avoiding rook walks on an 3x$ n$ ($ n$ columns) chessboard which begin at the lower-left corner and end at the upper-square of the lower-left corner.

That is, if the chessboard squares are denoted as follows:

  $ a_{11} \hspace{1cm} a_{12} \hspace{1cm} ...$
  $ a_{21}\hspace{1cm} a_{22} \hspace{1cm} ...$
  $ a_{31}\hspace{1cm} a_{32} \hspace{1cm} ...$

$ x_n$ and $ y_n$ are the number of walks from $ a_{31}$ to $ a_{21}$ and from $ a_{31}$ to $ a_{11}$, respectively.

Let’s establish a correspondence between each walk from $ a_{31}$ to $ a_{21}$ and an only one walk from $ a_{31}$ to $ a_{11}$. If a walk from $ a_{31}$ to $ a_{21}$ pass through $ a_{11}$, this walk necessarily end at $ a_{11}\minus{}a_{21}$. Then, we can remove this last step and a walk from $ a_{31}$ to $ a_{11}$ is obtained. On the contrary, if a walk from $ a_{31}$ to $ a_{21}$ does not pass through $ a_{11}$, we can add the step $ a_{21}\minus{}a_{11}$ at the end and a walk from $ a_{31}$ to $ a_{11}$ is obtained.

However, the inverse correspondence is not true (i.e., for each walk from $ a_{31}$ to $ a_{11}$ no only one walk from $ a_{31}$ to $ a_{21}$ exists). Notice that all walks beginning at $ a_{31}\minus{}a_{21}\minus{}a_{22}$ to $ a_{11}$ have not considered yet. All walks beginning at $ a_{31}\minus{}a_{21}\minus{}a_{22}$ to $ a_{11}$ necessarily end at $ a_{12}\minus{}a_{11}$. Hence, the number of these walks coincides with the number of walks from $ a_{22}$ to $ a_{12}$ on the chessboard with $ n$-1 columns (first column is disregarded). It is not difficult to see that the number of walks from $ a_{22}$ to $ a_{12}$ in this chessboard is $ x_{n \minus{} 1}$.

Therefore:

        $ y_n \equal{} x_n \plus{} x_{n \minus{} 1} \hspace{2cm} (1)$

On the other hand, the shortest walk from $ a_{31}$ to $ a_{21}$ is $ a_{31}\minus{}a_{21}$ and all other walks from $ a_{31}$ to $ a_{21}$ begin at $ a_{31}\minus{}a_{32}$.

Let’s establish a correspondence between each walk from $ a_{32}$ to $ a_{12}$ (on the chessboard of $ n$-1 columns) and an only one walk from $ a_{32}$ to $ a_{21}$ (on the chessboard of $ n$ columns). We can add the steps $ a_{12}\minus{}a_{11}\minus{}a_{21}$ at the end of each walk from $ a_{32}$ to $ a_{12}$ to obtain all walks from $ a_{32}$ to $ a_{21}$ that end at $ a_{11}\minus{}a_{21}$. It is easy to see that the number of walks from $ a_{32}$ to $ a_{12}$ (on the chessboard of $ n$-1 columns) is $ y_{n \minus{} 1}$, and hence, the number of walks from $ a_{31}$ to $ a_{21}$ that end at $ a_{11}\minus{}a_{21}$ is $ y_{n \minus{} 1}$.

Let’s also establish a correspondence between each walk from $ a_{32}$ to $ a_{22}$ (on the chessboard of $ n$-1 columns) and an only one walk from $ a_{32}$ to $ a_{21}$ (on the chessboard of $ n$ columns). We can add the step $ a_{22}\minus{}a_{21}$ at the end of each walk from $ a_{32}$ to $ a_{22}$ to obtain all walks from $ a_{32}$ to $ a_{21}$ that end at $ a_{22}\minus{}a_{21}$. It is easy to see that the number of walks from $ a_{32}$ to $ a_{22}$ (on the chessboard of $ n$-1 columns) is $ x_{n \minus{} 1}$, and hence, the number of walks from $ a_{31}$ to $ a_{21}$ that end at $ a_{22}\minus{}a_{21}$ is $ x_{n \minus{} 1}$.

Then, all walks from $ a_{31}$ to $ a_{21}$ (on the chessboard of $ n$ columns) have been considered: $ a_{31}\minus{}a_{21}$, the walks that end at $ a_{11}\minus{}a_{21}$ and the ones that end at $ a_{22}\minus{}a_{21}$.

Therefore:

      $ x_n \equal{} 1 \plus{} y_{n \minus{} 1} \plus{} x_{n \minus{} 1}$

This implies that:

     $ x_{n \plus{} 1} \equal{} 1 \plus{} y_n \plus{} x_n \hspace{2cm} (2)$

From (1) and (2):

 $ x_{n \plus{} 1} \minus{} 2 x_n \minus{} x_{n \minus{} 1} \minus{} 1 \equal{} 0 \hspace{2cm} (3)$

This implies that:

$ x_{n \plus{} 2} \minus{} 2 x_{n \plus{} 1} \minus{} x_n \minus{} 1 \equal{} 0 \hspace{2cm} (4)$

Subtracting (3) from (4):

$ x_{n \plus{} 2} \minus{} 3 x_{n \plus{} 1} \plus{} x_n \plus{} x_{n \minus{} 1} \equal{} 0$

Using the characteristic polynomial of the linear recurrence mentioned above, we can obtain $ x_n \equal{} \frac { \minus{} 2 \plus{} (1 \plus{} \sqrt {2})^{n \plus{} 1} \plus{} (1 \minus{} \sqrt {2})^{n \plus{} 1}}{4}$. Hence,
$ y_n \equal{} \frac {(2 \plus{} \sqrt {2})(1 \plus{} \sqrt {2})^n \plus{} (2 \minus{} \sqrt {2})(1 \minus{} \sqrt {2})^n}{4} \minus{} 1$ for all $ n$.


Let $a,\ b,\ c,\ d$ be non negative real numbers such that $2(ab+ac+ad+bc+bd+cd)+abc+abd+acd+bcd=16.$

Prove that $a+b+c+d\geq \frac{2}{3}(ab+ac+ad+bc+bd+cd).$

When does the equality holds?

See here http://www.mathlinks.ro/Forum/viewtopic.php?t=85836 . Same idea :)


Pipe A will fill a tank in 6 hours.  Pipe B will fill the same tank in
4 hours.  Pipe C will fill the tank in the same number of hours
that it will take Pipes A and B working together to fill the tank.
What fraction of the tank will be filled if all three of the pipes
work together for one hour?  Express your answer as a
common fraction.

[quote="yaofan"]pipe A does 1/6 of the job in an hour. pipe B does 1/4 of the job in an hour.
1/6+1/4=10/24 so it takes pipe C 12/5 hours to finish the job. you add all the fractions up.

1/6+1/4+5/12=5/6

so it takes 6/5 of an hour for all 3 pipes to finish the job.

please correct me if i'm wrong.[/quote]
Uh, yeah, you are incorrect.  You were correct all the way until the part on [hide="yaofan"]1/6+1/4+5/12=5/6

so it takes 6/5 of an hour for all 3 pipes to finish the job. [/hide]
The question did not ask for how long it takes for all 3 pipes to finish the job, it asked for what fraction of the tank will be filled when all the pipes work together for 1 hour.  Please see below.[quote="GameBot"]Pipe A will fill a tank in 6 hours.  Pipe B will fill the same tank in
4 hours.  Pipe C will fill the tank in the same number of hours
that it will take Pipes A and B working together to fill the tank.
What fraction of the tank will be filled if [b]all three of the pipes
work together for one hour[/b]?  Express your answer as a
common fraction.[/quote]
So the answer is just the sum of the fraction which turned out to be $\boxed{\frac{5}{6}}$.

Lol, I made the same mistake the first time I answered this problem.  I thought it asked for the amount of hours it will take to finish the job, not the fraction of the tank that will be filled if all three pipes work for one hour.  :blush:


analytic geometry
geometry solved
geometry

There are four paths (lines) in plane. No two parallel and no three collinear. On each path one person (let their names be $A$, $B$, $C$, $D$) is walking with the constant speed. Each pair (if they meet in the same point) drinks 2 beers. Prove that the total number of beers which were drunk can not be equal to $10$  :rotfl:

This is quite a known problem. The idea behind it: Work in 3-dimensional space, with time being the third dimension.

So consider our plane as the xy plane of a 3-dimensional coordinate system. Let $A\left(x_A\left(t\right);\;y_A\left(t\right);\;0\right)$, $B\left(x_B\left(t\right);\;y_B\left(t\right);\;0\right)$, $C\left(x_C\left(t\right);\;y_C\left(t\right);\;0\right)$, $D\left(x_D\left(t\right);\;y_D\left(t\right);\;0\right)$ be the coordinates of the points A, B, C, D in dependence of the time t. Now, consider in our 3-dimensional space the points A', B', C', D' defined by the coordinates $A^{\prime}\left(x_A\left(t\right);\;y_A\left(t\right);\;t\right)$, $B^{\prime}\left(x_B\left(t\right);\;y_B\left(t\right);\;t\right)$, $C^{\prime}\left(x_C\left(t\right);\;y_C\left(t\right);\;t\right)$, $D^{\prime}\left(x_D\left(t\right);\;y_D\left(t\right);\;t\right)$. As the time t goes, each of these points A', B', C', D' moves along a straight line (in fact, the first and second coordinates of these points behave linearly since they are the coordinates of the points A, B, C, D which move with constant speed along straight lines, and the third coordinates are t and thus also vary linearly; hence, all coordinates of the points A', B', C', D' behave linearly, and thus these points move with constant speed along straight lines). Let a', b', c', d' be these straight lines along which the points A', B', C', D' move.

Note that, because of the coordinate representations $A\left(x_A\left(t\right);\;y_A\left(t\right);\;0\right)$ and $A^{\prime}\left(x_A\left(t\right);\;y_A\left(t\right);\;t\right)$, the point A is the projection of the point A' on the xy plane. Thus, the line along which the point A moves is the projection of the line along which the point A' moves - i. e. of the line a' - on the xy plane. Similarly, the lines along which the points B, C, D move are the projections of the lines b', c', d' on the xy plane. Now, if two of the lines a', b', c', d' were parallel - say, a' and b' -, then their projections on the xy plane would be parallel as well, i. e. the lines along which the points A and B move would be parallel, but this is impossible by a condition of the problem. Hence, no two of the lines a', b', c', d' are parallel.

Now, the main observation [b](*)[/b]: Two of the four persons - say, A and B - meet in the same point (and thus drink 2 beers) if and only if the corresponding lines a' and b' intersect (remember, these are lines in space). In fact, because of the coordinate representations $A\left(x_A\left(t\right);\;y_A\left(t\right);\;0\right)$ and $B\left(x_B\left(t\right);\;y_B\left(t\right);\;0\right)$, the persons A and B meet in the same point if and only if there exists a moment t such that $x_A\left(t\right)=x_B\left(t\right)$, $y_A\left(t\right)=y_B\left(t\right)$ and 0 = 0. On the other hand, because of the coordinate representations $A^{\prime}\left(x_A\left(t\right);\;y_A\left(t\right);\;t\right)$ and $B^{\prime}\left(x_B\left(t\right);\;y_B\left(t\right);\;t\right)$, the lines a' and b' intersect if and only if there exist reals u and v such that $x_A\left(u\right)=x_B\left(v\right)$, $y_A\left(u\right)=y_B\left(v\right)$ and u = v. Trivially, this is one and the same condition, so the persons A and B meet in the same point if and only if the corresponding lines a' and b' intersect.

Now, assume that the total number of beers drunk is equal to 10. Since any 2 beers correspond to two persons meeting in one point, this means that there are exactly 5 pairs of persons which meet in one point. WLOG, assume that the persons A and B meet in one point, the persons A and C meet in one point, the persons A and D meet in one point, the persons B and C meet in one point, and the persons B and D meet in one point, but the persons C and D don't meet in one point. According to the observation [b](*)[/b], this can be rewritten as follows: The lines a' and b' intersect, the lines a' and c' intersect, the lines a' and d' intersect, the lines b' and c' intersect, the lines b' and d' intersect, but the lines c' and d' don't intersect.

But is this possible? In fact, since the lines a' and b' intersect, they lie in a common plane. The line c', intersecting both of these lines, must also lie in this plane. Similarly, the line d' lies in this plane. Thus, all four lines a', b', c', d' lie in one plane. Hence, the lines c' and d' are either parallel or intersect. Since they cannot be parallel and cannot intersect either, we obtain a contradiction. Thus, our assumption was wrong, and the total number of beers drunk cannot be equal to 10.

  darij


superior algebra
superior algebra unsolved

Give an example of a ring $ R$ and its ideal $ \mathfrak{a}$ such that $ \mathfrak{a}$ is a maximal in the set of ideals but it is not a maximal left ideal.

Take the ring of $ n\times n$-matrices over some field and the zero ideal. It is maximal (the ring is simple) but there are nontrivial left ideals.


trigonometry
analytic geometry
graphing lines
slope

Find the mass M of the hanging block in the figure which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and pulleys are light

I feel this is a problem from H. C Verma, isn't it?

ya this is from HC verma
.i am getting a wrong answer and the answer i am getting   is slightly different f from what is given in hc .

[quote="Srinath"]
well here is the figure if anyone wants it.

[/quote]

Anyone wants it? How do you expect one to solve it without the diagram?

Draw the FBD, write down your equations.

These should look something like this-

\begin{eqnarray}
Mg-T = Ma \\
T = (M'+m)a \\
ma\cos\theta = mg\sin\theta
\end{eqnarray}

That should help you through!

What happens when $ \theta \equal{} {\pi \over 4}\;\;$ ?

$ a \equal{} g ?$

What is your take on this problem, Immanuel?

If  a = g, from your equation (1) T =0 ?

Add $ (1)$ and $ (2)$.

$ Mg \equal{} (M\plus{}M'\plus{}m)a \\ \\ \\
a \equal{} \frac{Mg}{M\plus{}M'\plus{}m} \\ \\ \\
\cot\theta \equal{} \frac{g}{a} \equal{} \frac{M\plus{}M'\plus{}m}{M} \\ \\ \\
M(\cot\theta\minus{}1) \equal{} M'\plus{}m \\ \\ \\
\boxed{M\equal{} \frac{M'\plus{}1}{(\cot\theta\minus{}1)}}$

Which happens to be the right answer (I have the book and the answers are at the back but sadly, no solutions)

The books answers may be wrong...

And Rajiv, i dont see why you can write that (3) equation. If we are solving the problem according to the pic, i.e. the slope itself is moving.. then we cant write that, since the block on the slope is not in an inertial frame since the slope is accelerating, and you know that Newtons laws are valid only in inertial frames...

if we take the frame of reference as the triangular block which is non inertial then newton laws can be applied if pseudo force is included and rajiv has included the pseudo force $ ma$ on the small block
so rajiv is correct isnt it ? i dont see why should he wrong yeahnigga.pls explain your reason why he is wrong.
forgive if i am wrong

Well i dont see the pseudo-force in his equation...

(3) is right through wrong lines, even Rajiv not being God (comes from a kind of saying:"God writes right through wrong lines).

Either in the inertial frame the normal force $ N$ the triangular block exert on the small block, has an vertical component that equilibrates his weight , $ N\cos\theta \equal{} mg$ and the horizontal component produce the acceleration , $ N \sin \theta \equal{} ma$.

So, $ {a \over g} \equal{} \tan \theta$.

Either in the triangular block's frame, where the small block is at rest,  the vertical component from the normal force still balances the weight, and the horizontal one balances the inertial force.

Anyway, it could be made much much simple: All the system (internal forces, like T, cancels out) of mass M+M'+m gains the acceleration due Ms weight.

So the answer , say in the book, say in Rajivs post is Ok.

The point I mentioned before in another form is about $ \theta$
being greater or equal to 45°. Here M becomes infinite or  negative ??!!
btw what means pic?

The equation $ (3)$ is valid only for $ \theta \in \left(0, \frac {\pi}{4}\right)$ since $ a < g$. If $ \theta \geq \frac {\pi}{4},$ then $ m$ cannot be in equilibrium w.r.t the slope, whatsoever. For a certain $ m$ the system remains at rest or accelerates in the other direction. Find the range of $ m$. Remember, $ \theta \in \left(0, \frac {\pi}{4}\right)$.

[quote="I"]
[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/1/3/c/13c996e2738536cc815b5da231c5b54b2b550c7a.gif[/img]

[/quote]

A small correction is required here.

For equilibrium of $ m$ $ a_{req} = g \tan\theta$.

Now the three equations:

\begin{eqnarray}Mg-T = Ma\\
T = (M'+m)a\\
ma_{req}\cos\theta = mg\sin\theta\end{eqnarray}

Now consider only the slope $ M'$ and the $ m$ on it. If $ M'$ is accelerated with $ a_{req}$, force up the plane is $ ma_{req}\cos\theta$.
For equilibrium,
$ ma_{req}\cos\theta = mg\sin\theta \\ \\
a_{req}= g\tan\theta$

Recall that $ \tan\theta > 1 \forall \theta \in \left[\frac{\pi}{4},\frac{\pi}{2}\right)$ which means $ a_{req}>g \forall \theta \in \left[\frac{\pi}{4},\frac{\pi}{2}\right)$.

Look at equation $ (1): \;T = M(g-a)$. If $ a_{req} \geq g, \; T\leq0$ which is impossible!

In my opinion, my initial analysis of the problem is correct.


[quote="cuopbientcd"]If $f(n)=n(an+b)$ and $(a,p)=(b,p)=1$ then $\sum_{1}^{p-1}(f(n)|p)=-(a|p)$[/quote]
What do you mean by $f(n)|p$?

[quote="Hawk Tiger"]What do you mean by $f(n)|p$?[/quote]
Not $f(n)|p$ but $(f(n)|p)$ or $(\frac{f(n)}{p})$. This is Legendre's simbol.

But why the problem is hot?  Hometask? :maybe:

Well, I think it would be not bad thing in any ways to give not solution but a
[hide="hint"]Let $x'$ be inverse for $x$, i.e. $xx'\equiv 1 (mod p)$. Then $(x|p)=(x'|p)$. Combine this with $\sum_{x=0}^{p-1}(x|p)=0$[/hide]


The probability function of a certain discrete random satisfies the following relationship:
$P(0) = P(1)$ and $P(k+1) = \frac{1}{k}\cdot P(k)$ for some postive integer $k$.  Find $P(0)$.

[hide="Hint 1"]
Remember that the probabilities given by the probability function must sum to 1[/hide]

[hide="Hint 2"]
Notic$e$ anything about this sum?[/hide]

You should get a different result.
[hide="Solution"]
$P(1) = P(0)$
$P(2) = 1\cdot P(1) = P(0)$
$P(3) = \frac{1}{2}\cdot P(0)$
$P(4) = \frac{1}{6}\cdot P(0)$
$P(5) = \frac{1}{24}\cdot P(0)$
Recall that $\sum_{k=0}^{\infty}P(k) = 1$ to satisfy a probability function.
We have $P(0)+\sum_{k=1}^{\infty}\frac{P(0)}{(k-1)!}= P(0)(1+e)$
Since the sum must equal 1 to be a probability function it follows that $P(0) = \frac{1}{1+e}$.
   [/hide]

[quote="illcrowflu"]We have $P(0)+\sum_{k=1}^{\infty}\frac{P(0)}{(k-1)!}= P(0)(1+e)$[/quote]

You are right. :blush: , but you should justify this equation. :D

We have $P(0)+P(0)\sum_{k=1}^{\infty}\frac{1^{k}}{(k-1)!}= P(0)+P(0)\sum_{k=0}^{\infty}\frac{1^{k}}{k!}$.  Here we make use of the identity $e^{t}= \sum_{k=0}^{\infty}\frac{t^{k}}{k!}$.  Our expression becomes $P(0)+P(0)e = P(0)(1+e).$


[quote="Vasc"]If $a,b,c$ are nonnegative numbers, then

$\sum \sqrt{(a^{2}+b^{2})(a^{2}+c^{2})}\leq (a+b+c)^{2}$.[/quote]

So :) , let's follow the flow of my thoughts when I was solving this problem :) .

First I tried to find the points where the equality holds. Of course the first thing to check is the case when $a=b=c$, but it obviously doesn't work. So I procceded to the next common case when one of the numbers is equal to $0$. Putting  $a=0$ I got the following inequality
\[bc+\sqrt{b^{2}+c^{2}}(b+c)\leq (b+c)^{2}\]
<=>
\[bc\leq (b+c)(b+c-\sqrt{b^{2}+c^{2}})=(b+c)\frac{(b+c)^{2}-(b^{2}+c^{2})}{b+c+\sqrt{b^{2}+c^{2}}}=(b+c)\frac{2bc}{b+c+\sqrt{b^{2}+c^{2}}}\]
<=>
\[\sqrt{b^{2}+c^{2}}\leq b+c\]
<=>
\[0\leq 2bc\] :D

So, I asumed that equality case holds when $ab+bc+ac=0$(that is two of the numbers are $0$)

Then I noticed that I had proved an inequality with the expretion $\sqrt{x^{2}+y^{2}}$; I thought that it can help to prove the initial inequality.

So what was the inequality

\[\sqrt{b^{2}+c^{2}}\leq b+c-\frac{bc}{b+c}=a+b+c-\frac{ab+bc+ac}{b+c}\]
Then to make my work with expretions easier I assumed that $a+b+c=1$
so we have that
\[\sqrt{b^{2}+c^{2}}\leq 1-\frac{ab+bc+ac}{b+c}\]

and putting this inequality for all the expretions of $\sqrt{x^{2}+y^{2}}$ kind in the initial inequality and opening the paranthesies, I got that I needed to prove that

\[3-2(ab+bc+ac)(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c})+\frac{2(ab+bc+ac)^{2}}{(a+b)(b+c)(a+c)}\leq 1\]
<=>
\[3-2\frac{ab+bc+ac}{(a+b)(b+c)(a+c)}\leq 1\]
<=>
\[(1-a)(1-b)(1-c)\leq ab+bc+ac\]
<=>
\[ab+bc+ac-abc\leq ab+bc+ac\]
<=>
\[0\leq abc\]  :D

I hope I haven't boried you :blush: .

[quote="Vasc"]If $a,b,c$ are nonnegative numbers, then

$\sum \sqrt{(a^{2}+b^{2})(a^{2}+c^{2})}\leq (a+b+c)^{2}$.[/quote]
Nice, VasC. Here is my solutions:
WLOG, we may assume $c=\min\{a,b,c\}$, then
\[a^{2}+b^{2}\le u^{2}+v^{2},a^{2}+c^{2}\le u^{2},b^{2}+c^{2}\le v^{2}\]
Where $u=a+\frac{c}{2},v=b+\frac{c}{2}$.
So we have to prove
\[(u+v)^{2}\ge (u+v)\sqrt{u^{2}+v^{2}}+uv \]
But it is easy. :wink:
We are done. :)

Nice, Tigran and Toanhocmuon! :)
My solution is identical to the second.


Let b be a positive integer. How can we show that the period length of the base b expansion of 1/m is m-1 iff m is prime and b is a primitive root modulo m?


trigonometry
inequalities
geometry proposed
geometry

$ RHS$
It is easy to see $ \tan\frac{A}{2}\equal{}\frac{r}{s\minus{}a}$ and because $ sr\equal{}S\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$
Therefore $ (s\minus{}a)\tan^{2}\frac{A}{2}\plus{}(s\minus{}b)\tan^{2}\frac{B}{2}\plus{}(s\minus{}c)\tan^{2}\frac{C}{2}$
$ \equal{}\frac{\sum_{cyc}(s\minus{}a)(s\minus{}b)}{s}$

$ \frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{2(a\plus{}b\plus{}c)} \geq \frac{\sum_{cyc}(s\minus{}a)(s\minus{}b)}{s}$
$ <\equal{}> a^{2}\plus{}b^{2}\plus{}c^{2} \geq s^{2}\plus{}(ab\plus{}bc\plus{}ca)$
$ <\equal{}> 0 \geq 2ab\plus{}2bc\plus{}2ca\minus{}a^{2}\minus{}b^{2}\minus{}c^{2}$
By the way, $ \sqrt{a}\plus{}\sqrt{b})^{2}>a\plus{}b>c\equal{}(\sqrt{c})^2$
therefore $ \sqrt{a},\sqrt{b},\sqrt{c}$ are also three length of triangle.
Therefore last inequality holds.

$ LHS$
$ a\plus{}b\plus{}c\equal{}4s$
$ \equal{}>$ ETS  $ 9R^{2} \geq a^{2}\plus{}b^{2}\plus{}c^{2}$
$ well \minus{} known S\equal{}\frac{abc}{4R}$
$ \therefore$ ETS $ 9(abc)^{2} \geq (a^{2}\plus{}b^{2}\plus{}c^{2})(a\plus{}b\plus{}c)\prod_{cyc}(a\plus{}b\minus{}c)$
equivalent to $ sum_{cyc}a^{6}\plus{}3a^{2}b^{2}c^{2}\geq \sum_{sym}a^{4}b^{2}$
it's schur form. Then we are done.

[quote="Heebeen, Yang"]$ RHS$
It is easy to see $ \tan\frac {A}{2} \equal{} \frac {r}{s \minus{} a}$ and because $ sr \equal{} S \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$
Therefore $ (s \minus{} a)\tan^{2}\frac {A}{2} \plus{} (s \minus{} b)\tan^{2}\frac {B}{2} \plus{} (s \minus{} c)\tan^{2}\frac {C}{2}$
$ \equal{} \frac {\sum_{cyc}(s \minus{} a)(s \minus{} b)}{s}$

$ \frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{2(a \plus{} b \plus{} c)} \geq \frac {\sum_{cyc}(s \minus{} a)(s \minus{} b)}{s}$
$ < \equal{} > a^{2} \plus{} b^{2} \plus{} c^{2} \geq s^{2} \plus{} (ab \plus{} bc \plus{} ca)$
$ < \equal{} > 0 \geq 2ab \plus{} 2bc \plus{} 2ca \minus{} a^{2} \minus{} b^{2} \minus{} c^{2}$
By the way, $ \sqrt {a} \plus{} \sqrt {b})^{2} > a \plus{} b > c \equal{} (\sqrt {c})^2$
therefore $ \sqrt {a},\sqrt {b},\sqrt {c}$ are also three length of triangle.
Therefore last inequality holds.

$ LHS$
$ a \plus{} b \plus{} c \equal{} 4s$
$ \equal{} >$ ETS  $ 9R^{2} \geq a^{2} \plus{} b^{2} \plus{} c^{2}$
$ well \minus{} known S \equal{} \frac {abc}{4R}$
$ \therefore$ ETS $ 9(abc)^{2} \geq (a^{2} \plus{} b^{2} \plus{} c^{2})(a \plus{} b \plus{} c)\prod_{cyc}(a \plus{} b \minus{} c)$
equivalent to $ sum_{cyc}a^{6} \plus{} 3a^{2}b^{2}c^{2}\geq \sum_{sym}a^{4}b^{2}$
it's schur form. Then we are done.[/quote]

 :)


inequalities
real analysis
real analysis theorems

Here s a simple looking result, with a very interesting proof.
[b]Result: [/b] if $ u_{n}$ be a monotonic increasing sequence, such that for each $ n, u_{n} \le k$, where $ k$ is a fixed real number.
then the claim is that $ u_{n}$ converges to a limit less than or equal to $ k$.

[n]Proof:[/b] Let us consider dividing the set $ \mathbb{Q}$ of rationals into two parts: $ S_{1}$ and $ S_{2}$, $ S_{1}$ consisting of those rationals $ a$ such that $ u_{n} \ge a$ for some $ n$. Similarly, $ S_{2}$ be the set of those rationals $ b$ such that $ u_{n} < b$ for some $ n$.
Now firstly all $ a$'s are less than all $ b$'s owing to the monotonicity of our sequence.
Moreover all rationals are completely classified under this categorization.

This classification defines for us a real number $ A$.
Now notice that $ u_{n} \le A$ for each $ n$; for if $ u_{n} >A$, then $ u_{n}$ would be greater than each rational between $ A$ and $ u_{n}$, such a rational however is absorbed in to our set $ S_{2}$.
Now we chose a rational number $ r$ satisfying $ A\minus{} \epsilon \le r < A$ for a positive $ \epsilon$.

Clearly $ r$ belongs to the set $ S_{1}$.
This means that there is some $ u_{m}$ with $ u_{m} >r$ and consequently this inequality holds for each $ n>m$.

From the above two statements, we have $ |u_{n} \minus{} A| < \epsilon$ for each $ n \ge m$.
Hence it follows that $ u_{n} \to A$ as $ n \to \infty$.
We assert that $ A \le k$, for if not we would get a blatant contradiction by using the fact that $ u_{n} \le k$ for each $ n$.
This completes the proof.

We can obtain a corresponding theorem for monotonically decreasing sequence by merely defining $ v_{n} \equal{} \minus{} u_{n}$, and so on and so forth.

For eg we can use this idea to find the limit of the sequence $ x_{n} \equal{} n(x^{n} \minus{} 1)$.

The number $ k$ is the supremum of the sequence.  What you are showing is that a bounded monotone sequence converges.  Can you prove the analogous result for a monotone decreasing sequence?

[quote]Can you prove the analogous result for a monotone decreasing sequence?[/quote]
yes sir as i ve mentioned, we can do so by defining $ v_{n} \equal{} \minus{}u_{n}$, which is obviously a decreasing sequence.
In its statement, the words [i]greater than/less than[/i] in the statement for increasing sequences would have to be replaced by the words [i]less than/greater than[/i] respectively.


inequalities
trigonometry
floor function
limit
calculus
derivative
calculus computations

im assuming you mean less than or equal to? since x=0 is equal

$ f(x)\equal{}x\plus{}e^{x^2}\minus{}e^x\equal{}x\plus{}e^x[e^{x(x\minus{}1)}\minus{}1]$
$ f'(x) \equal{} 1 \plus{}2xe^{x^2}\minus{}e^x\equal{}1\plus{}e^x[2xe^{x(x\minus{}1)}\minus{}1]\equal{}1\plus{}g(x)$

so $ x\ge 1 \to e^{x(x\minus{}1)}\ge 1 \to f(x) \ge 0$

other cases seem less trivial =(

I am working on it but the case kenn4000 showed does seem trivial since $ x \ge 1$ means $ e^x \le e^{x^2 }$ by sheer monotonicity.

I will come back after work to note progress.

EDIT: Ok, I have solved it but I will wait for someone else while I take a shower so I can make sure  :maybe: .

I have it proved now but I will give some hints.

It is of course sufficient to prove such an inequality for positive x.

The simplest way is to try to 'approximate' exponentials by polynomials. It is not at all easy to figure out how to do though.  :lol:

Sorry everyone. I'm going back to school so I won't have time till Saturday or probably Sunday. The proof is really quite elementary though. If I have time I'll squeeze it in.

$ e^x \equal{} \sum_{k\equal{}0}^\infty \frac{x^k}{k!}$

$ e^{x^2} \equal{} \sum_{k\equal{}0}^\infty \frac{x^{2k}}{k!}$

how is it sufficient for positive x..?

Here is my thinking:

The objective is to show $ \sum\limits_{k \equal{} 2}^\infty {\frac {{x^k }}{{k!}}} \le \sum\limits_{k \equal{} 1}^\infty {\frac {{x^{2k} }}{{k!}}}$

but to do this you need to show: $ \sum\limits_{k \equal{} 2}^\infty {\frac {{|x|^k }}{{k!}}} \le \sum\limits_{k \equal{} 1}^\infty {\frac {{x^{2k} }}{{k!}}}$, where $ \sum\limits_{k \equal{} 2}^\infty  {\frac{{x^k }}{{k!}}}  \le \sum\limits_{k \equal{} 2}^\infty  {\frac{{|x|^k }}{{k!}}}$

Please tell me if this is correct.  :ninja:

im not sure i believe the reasoning for the absolute values =(

lets take the idea and run with it

$ \sum_{k \equal{} 1}^\infty \frac {x^{2k} }{k!}\minus{}\sum_{k \equal{} 2}^\infty \frac {x^k }{k!}\equal{} \sum_{k \equal{} 2}^\infty \left[\frac{\cos^2 \left(\frac{k\pi}{2}\right)}{\lfloor\frac{k}{2}\rfloor !}\minus{}\frac {1 }{k!} \right]x^k\equal{}$

$ \sum_{k \equal{} 2}^\infty \left[\frac{k\cdots (\lfloor\frac{k}{2}\rfloor\plus{}1) \cos^2 \left(\frac{k\pi}{2}\right)\minus{}1}{k!} \right]x^k \ge 0$ ?

We have,
$ f(x) \equal{} e^{x^2} \plus{} x \minus{} e^{x} \equal{} \left(e^{x} \minus{} \frac {1}{2}\right)^{2} \plus{} \left(\sqrt {x} \minus{} \frac {1}{2}\right)^{2} \plus{} \left(\sqrt {x} \minus{} \frac {1}{2}\right)$.

Edited: My e^{x^2} is confused with (e^x)^2

this inequality doesn't hold either.

Take $ x \equal{} \minus{} 100$

$ e^{ \minus{} 100} < \minus{} 100 \plus{} (e^{ \minus{} 100})^{2}$

$ \left(\frac {1}{e}\right)^{100} < \minus{} 100 \plus{} \left(\frac {1}{e}\right)^{10 000}$

but we know that for $ |r| < 1$ that $ r^{n}$ is very small for large $ n$, so we might as well be saying;

$ \epsilon_{1} < \minus{} 100 \plus{} \epsilon_{2}$

sure enough, if you take $ \lim_{x \rightarrow \minus{} \infty}$ you get $ 0 < \minus{} \infty$, contradiction.

Edit: Also confused $ (e^{x})^{2}$ with $ e^{x^{2}}$

Well it's crucial to my proof so let's make sure it's correct so I know if I have extra work to do on my flight.

We are asked to show: (a) $ \sum\limits_{k \equal{} 2}^\infty {\frac {{x^k }}{{k!}}} \le \sum\limits_{k \equal{} 1}^\infty {\frac {{x^{2k} }}{{k!}}}$

Note what happens when we replace x, with |x|. We get $ \sum\limits_{k \equal{} 2}^\infty {\frac {{|x|^k }}{{k!}}} \le \sum\limits_{k \equal{} 1}^\infty {\frac {{x^{2k} }}{{k!}}}$. Therefore if the original inequality is true, then this inequality must be true as well.

First observation:
(b) $ \sum\limits_{k \equal{} 2}^\infty {\frac {{x^k }}{{k!}}} \le \sum\limits_{k \equal{} 2}^\infty {\frac {{|x|^k }}{{k!}}}$ because the terms on the right hand side are greater than the corresponding terms on the left hand side.

Correct?

Now assume we are able to prove the inequality (c) $ \sum\limits_{k \equal{} 2}^\infty {\frac {{|x|^k }}{{k!}}} \le \sum\limits_{k \equal{} 1}^\infty {\frac {{x^{2k} }}{{k!}}}$. This would be equivalent to proving (a) for positive values of x.

Correct?

We can now deduce that if (a) is true then $ \sum\limits_{k \equal{} 2}^\infty {\frac {{x^k }}{{k!}}} \le \sum\limits_{k \equal{} 2}^\infty {\frac {{|x|^k }}{{k!}}} \le \sum\limits_{k \equal{} 1}^\infty {\frac {{x^{2k} }}{{k!}}}$ is true and if $ \sum\limits_{k \equal{} 2}^\infty {\frac {{|x|^k }}{{k!}}} \le \sum\limits_{k \equal{} 1}^\infty {\frac {{x^{2k} }}{{k!}}}$ is true then $ \sum\limits_{k \equal{} 2}^\infty {\frac {{x^k }}{{k!}}} \le \sum\limits_{k \equal{} 1}^\infty {\frac {{x^{2k} }}{{k!}}}$ is true by (b), (c), and transitivity.

Hence it is sufficient to carry out a proof for positive values.

guys above. its $ e^{x^2}$ not $ (e^x)^2$

[quote="kenn4000"]im not sure i believe the reasoning for the absolute values =(

lets take the idea and run with it

$ \sum_{k \equal{} 1}^\infty \frac {x^{2k} }{k!} \minus{} \sum_{k \equal{} 2}^\infty \frac {x^k }{k!} \equal{} \sum_{k \equal{} 2}^\infty \left[\frac {\cos^2 \left(\frac {k\pi}{2}\right)}{\lfloor\frac {k}{2}\rfloor !} \minus{} \frac {1 }{k!} \right]x^k \equal{}$

$ \sum_{k \equal{} 2}^\infty \left[\frac {k\cdots (\lfloor\frac {k}{2}\rfloor \plus{} 1) \cos^2 \left(\frac {k\pi}{2}\right) \minus{} 1}{k!} \right]x^k \ge 0$ ?[/quote]

That's not the way I proceeded. It looks confusing, actually. Once you are done showing that it is sufficient to prove the inequality for positive values, you no longer need to consider the exponentials in series form. Once someone confirms my claim that it is sufficient to prove the inequality for positive values I will go to the next step.

Damned consecutive post time limit..

P.S. I will be happy to post in a more timely manner if someone will help me with my posts on the Legendre transform and the trigonometric inequality. :p I've seen way more difficult problems answered in "Computations & tutorials".

This solution was posted by someone whom I consulted for an alternative approach. It is NOT the solution I came up with and I take no credit for it, although it is awfully similar in just about every respect:

If $ f(x) \equal{} e^{x^2 }  \plus{} x \minus{} e^x$, a second derivative test shows that 0 is a local minimum of f. We can infer from an observation that f'' > 0 that f has no other zeros.

it seems like solving for these critical points is too elusive... each derivatives positivity kind of relies on what we are trying to prove right?

would like to see the proof of f''>0

$ f''(a) \equal{} 2e^{a^2}(2a^2 \minus{} a \plus{} 1) \minus{} 1 \equal{} 2e^{a^2}\left[ 2\left(a \minus{} \frac {1}{4}\right)^2 \plus{} \frac {7}{8}\right] \minus{} 1 > \frac {7}{4}e^{a^2} \minus{} 1 > 0$

Since it's not my solution I wouldn't want to appear to be taking credit by elaborating.

I did carry out essentially the same calculations in my own solution, the true difference being that his approach was motivated by a derivative test, whereas mine was motivated by examining the upward convexity of $ f(x) \equal{} e^x  \minus{} e^{x^2 }$on ]0,1[.
From looking at $ f''$ and the fact that $ e^x  < (2 \plus{} 4x^2 ) < (2 \plus{} 4x^2 )e^{x^2 }$on ]0,1[ everything follows.

Time to go to bed.. :sleep2:


inequalities
quadratics
algebra
linear equation
inequalities unsolved

Suppose $A$, $B$, $C$ are real numbers such that the relation

$A(x-y)(x-z) + B(y-z)(y-x) + C(z-x)(z-y) \ge 0$

holds for all real numbers $x$, $y$, $z$. State the condition(s) that $A$, $B$, $C$ must satisfy.

The answer is $A \ge 0$, $B \ge 0$, $C \ge 0$ and $A^2+B^2+C^2 \le 2(AB+BC+CA)$.

The first condition is easy to establish. By setting $x$, $y$, $z$ equal to $0$ two at a time, we can infer that $A$, $B$ and $C$ are nonnegative.

But I got stuck in the second part.

Can anyone solve this?

My idea is that we may try to treat this as a quadratic equation in $x$, $y$ and $z$ respectively. Then for the graph to lie no lower than the $x$-axis ($y$-axis, $z$-axis), we must first have $A > 0$ ($B > 0$, $C > 0$ respectively). ('$\ge 0$' will not do since the case of equality reduces to a linear equation which may not be greater than or equal to $0$ for all real values of $x$, $y$, $z$.)

In my opinion, the second condition will follow from the use of the discriminant. For the inequality to hold, it is necessary that $b^2-4ac \le 0$. This is the scary part which leads to very untidy algebra. And I have been unable to make any headway with this. My sketch suggests that we may need to use the discriminant at least twice if it is to be used at all.

So does anyone have a better idea?

WLOG : $x-y=u, y-z = v,  x-z = u+v $
Our condition transforms:
$Au^2+Cv^2\geq uv(B-A-C)$
$Au^2+Cv^2\geq 2|uv|\sqrt {AC}\geq uv(B-A-C)$

because we can find such u,v that equality will hold. We obtain

if $A+C < B$ then $(\sqrt {A} +\sqrt{C})^2\geq (\sqrt {B})^2 $
if $A+C > B$ then $(\sqrt {B})^2\geq (\sqrt {A} -\sqrt{C})^2 $

which gives us condition that $\sqrt{A},\sqrt{B},\sqrt{C}$ must be sides of a triangle

$(\sqrt{A}+\sqrt{B}+\sqrt{C})(\sqrt{A}+\sqrt{B}-\sqrt{C})(\sqrt{A}-\sqrt{B}+\sqrt{C})(-\sqrt{A}+\sqrt{B}+\sqrt{C})\geq 0$
Which gives us $a^2+b^2+c^2\leq 2(ab+bc+ac)$

Can we still use discriminant anyway?

By the way, can we really assume the second condition posted by Prowler? Why so?

Dear Prowler, I don't understand your proof, sorry.

Why is this

[quote]$Au^2+Cv^2\geq 2|uv|\sqrt {AC}\geq uv(B-A-C)$[/quote]

true?

and how to deduce from this one

[quote]if $A+C < B$ then $(\sqrt {A} +\sqrt{C})^2\geq (\sqrt {B})^2 $
if $A+C > B$ then $(\sqrt {B})^2\geq (\sqrt {A} -\sqrt{C})^2 $[/quote]

Thank you very much.

Dear manlio !

Supose that $A+C < B$ then you can find such $u,v > 0$ such that

$Au^2+Cv^2= 2uv\sqrt {AC}$
For example:  $ u = t\sqrt{C/A}$ $ v = t\sqrt{A/C}$
By condition of the problem we need
$Au^2+Cv^2 \geq uv(B-A-C)$ for all $u,v$.
So $2t^2\sqrt{AC}\geq t^2(B-A-C)$
$(\sqrt {A} +\sqrt{C})^2\geq (\sqrt {B})^2 $
Another case analogously, just take $u>0,v <0 $

Thank you very much. Very nice proof. :)

I think I have a normal solution.
$ A(x-y)(x-z) + B(y-z)(y-x) + C(z-x)(z-y) \ge 0  \Longleftrightarrow  Ax^2-(Ay+Az+By-By+Cz-Cy)x+(Ayz+By^2-Byz+Cz^2-Cyz)  \Longleftrightarrow  \Delta \leq 0  \Longleftrightarrow ((A+B-C)^2-4AB)y^2+?y+? \geq 0$
so we have $A^2+B^2+C^2 \le 2(AB+BC+CA)$

I have completed Zhaobin computations to have

$(\sum A^2 -2 \sum AB )(z-y)^2 \leq 0$

so the proof is complete


An equilateral polygon with 6 sides is inscribed in a circle of radius 20 cm. What is the area of the polygon (in cm2)?

[hide="Hint 1"] A regular hexagon is composed of six equilateral triangles. [/hide] [hide="Hint 2"] The side length of the hexagon is $20$ (because the above-mentioned triangles are equilateral). [/hide] [hide="Hint 3"] The area of an equilateral triangle with side length $s$ is $\frac{ s \cdot s \frac{ \sqrt{3}}{2}}{2}= \frac{s^{2}\sqrt{3}}{4}$. [/hide]

[hide]It implies that the sidelength of the hexagon is 20, so we can apply the regular hexagon area formula to get that the area is $600\sqrt{3}$.[/hide]


Μόλις το είδα στο Ουγγρικό  περιοδικό Κεμαλ

 [b]C. 962.[/b]

M is the orthocentre of the isosceles triangle ABC of base AC. Given that AC=BM, find the angles of the triangle.

[b][i] Μπάμπης[/i][/b]

[hide]$ O$ περίκεντρο, $ N$ μέσο $ AC$. Τότε $ ON\equal{}BM/2\equal{}AC/2\equal{}AN\iff \angle{AON}\equal{}45\iff \angle{B}\equal{}45$[/hide]


[hide="Obvious Solution"]
$ P(x)\equal{}x$

To check, simply plug into equation,
$ P(x^{100}\plus{}y^{100})\equal{}x^{100}\plus{}y^{100}\equal{}(P(x))^{100}\plus{}(P(y))^{100}$[/hide]

can you write more about you solution ?thanks

MJ GEO, You have posted 5 same titles posts.

You must change these titles.

Japanese Communities Modertator

kunny :wink:

[quote="mismatchtea"][hide="Obvious Solution"]
$ P(x) \equal{} x$

To check, simply plug into equation,
$ P(x^{100} \plus{} y^{100}) \equal{} x^{100} \plus{} y^{100} \equal{} (P(x))^{100} \plus{} (P(y))^{100}$[/hide][/quote]

This is not Sulution but Intuition.


LaTeX
function
floor function
ceiling function

First, he can make 25/5 = 5 cigars, and then he will smoke them. After that, he can take the 5 remaining cigar butts to make a new cigar and smoke that one. Then he only has 1 cigar butt left and can't make any new cigars, so the answer is 5 + 1 = [b]6[/b].

[quote][hide]25+5+1=31[/hide][/quote]
The first 25 he finds are cigar butts, not cigars.

[quote][hide]$\frac{25}{5}=5$
[/hide][/quote]
After he smokes those 5 cigars, he has 5 cigar butts left over.

[hide]We could also just subtract 4 as many times as possible from 25, without reaching 0.  This can be done 6 times.[/hide]

[quote="nebula42"]Interestingly, this problem is equivalent to finding the number of zeros in $25!$ (but that's only because every [b]5[/b] cigar butts becomes one cigar).

By the way, how do you use LaTeX to write the symbols for the floor and ceiling functions?[/quote]

\lfloor=$\lfloor$ , \rfloor=$\rfloor$, \lceil=$\lceil$ , \rceil=$\rceil$

$\lfloor \frac{25}{4} \rfloor=6$

because he has 25 butts,so he can makes 5 cigars,but then he'll makes another 5 butts,so he'll makes 5 plus 1=6

[quote="Arvind_sn"]Hobo Billy Bob likes to smoke cigars. From 5 cigar butts, Billy Bob can make one cigar. He finds 25 cigar butts in an ashtray. How many cigars will he smoke, assuming that he smokes every one he makes?[/quote]

$\frac{25}{5}= 5+1 = \fbox{6}$
I'm probably wrong...there must be a catch :rotfl:  :rotfl:

[hide="tricky part"]After he smokes 5 cigars, there are 5 cigar butts, so he makes one more[/hide]

[quote="Arvind_sn"]Hobo Billy Bob likes to smoke cigars. From 5 cigar butts, Billy Bob can make one cigar. He finds 25 cigar butts in an ashtray. How many cigars will he smoke, assuming that he smokes every one he makes?[/quote]

[hide]25/5=5, so 1 left over 5+1=6[/hide]


Given two segments with lengths $a$ and $b$, how to construct a segment with length $ab$?

careful with this one...you have to add a condition to make the segment unique

Hey, that's kind of neat, Altheman is right, the problem is unsolvable as posed.  If his hint isn't enough, have a couple more.

[hide="hint 1"] Consider using power of a point on various circles with a chords of length (a+b).  What type of circle/chord would we need to create another chord of length ab?[/hide]

[hide="hint 2"]  Let's say we measure segment $a$ and find it $\frac{1}{4}$ as long as segment $b.$  Suppose segment $a$ has length 1/2.  Then segment $b$ has length $2,$ and a segment of $ab$ would have length 1, which is between $a$ and $b.$  Now, suppose that segment $a$ has length 2.  Then segment $b$ has length 8, and a segment of length $ab$ would have length 16, making it [i]longer[/i] than either segment $a$ or $b.$  See what we need to add now? [/hide]

I'm just going to bump this because it's a nice problem and there has been no solution posted...

The easiest solution (I think) is to form an a*b rectangle, then change it into a rectangle of equal area except with side lenght 1. Then the other side has to be $ab$. This is quite a common construction, do i need to describe it?

PS you also need to be given the units your working with.


I've been considering taking the intermediate number theory seminar, although I noticed that the description says that it has no problem sets. Why is this so, and what drawbacks does this cause for the effectiveness of the course?

Yeah but there are like 23480324 message board problems each week. (Actually about 20-30.)


[color=darkblue][u][b]Problem 1:[/b][/u]
Find formula of sequence $ (u_n)$ known that: $ u_1\equal{}\minus{}1, u_2\equal{}1,u_3\equal{}\minus{}2,u_4\equal{}\minus{}1,u_5\equal{}0$

[u][b]Problem 2:[/b][/u]
Find formula of sequence $ (v_n)$ known that: $ v_1\equal{}0, v_2\equal{}\minus{}1,v_3\equal{}0,v_4\equal{}0,v_5\equal{}1$[/color]


Hi people, got a problem:

A piece of DNA (8000 bases) is cut with the restriction enzyme BglI and then with HpaII and finally with the mixture of both. The fragments are separated from each other , then a Southern Blot is done (again with these enzymes and with the mixture of them).
You can see the results in the two images I´ve attached to my post. (Image 1: restriction with BglI and HpaII results; image2: Southern Blot result)

My task is now to find out the length of the DNA fragments in the Southern Blot with the mixture of the enzymes. (The fragments thus are not shown on the image; there´s a blank column where they belong.)

Could anybody help and tell me how to proceed ? Thanks !
Please be quick it´s urgent ! Need the answer as quickly as possible !

Hey people it´s over now. Got the answer in another forum.
Anybody may delete this thread now.
Bye !


Can anyone who got 12 right tell me how to do it?
Thank you

check the Contests section

or the MathJam transcript

please try searching next time before making a new post