geometry
parallelogram
ratio
geometry solved

Let ABCD be a parallelogram. Join each vertice with the midpoints of the opposite sides. Determine the area of the formed octagon in terms of the area of the parallelogram.

It seems to be a generalization of problem 5, round 10,International Mathematical Talent Search.

We can turn the parallelogram into a square by applying an affine transformation. It will, of course, conserve everything we're interested in: ratios of lengths (and of areas).

We can reason easier on a square: Let $M,N,P,Q$ be the midpts of $AB,BC,CA,DA$ . Now let $X=AP\cap BQ,\ Y=BQ\cap CM,\ Z=CM\cap DN,\ T=DN\cap AP$. Let $l$ be the side of the square $XYZT$. We can see that if we take a side of the octagon out of $l$ we're left with two small pieces with length $x$ each, s.t. the side of the octagon is $x\sqrt 2$. The area of the octagon is the area left from $XYZT$ after taking out four small isosceles right triangles of sides $x,x,x\sqrt 2$. The ratio between the area of the octagon and the area of $XYZT$ must be $\frac {l^2-2x^2}{l^2}$. We have $l=(2+\sqrt 2)x\Rightarrow \frac {[octagon]}{[XYZT]}=2\sqrt 2-2$ (it's easy to compute). It's also easy to see that the area of $XYZT$ is one fifth of the area of $ABCD$, so the ratio $\frac {[octagon]}{[ABCD]}=\frac {2\sqrt 2-2}5$.

The advantage of the square is that we have right isosceles triangles.

Aaargh...that's what I was thinking about, but too slowly.... :D

Pierre.

What is an affine transformation?Why does it conserve the ratios?

That's the definition: a transformation which preserves collinearity and the ratio of distances.

Try [url=http://mathworld.wolfram.com/AffineTransformation.html]this[/url]. Maybe that's not satisfying, but you can always try to Google it :).


algorithm
number theory open
number theory

See the [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]Extended Euclidean Algorithm[/url].

[quote="Victory.US"]for positive integers $ m,n$ , Show that there exist interger $ x$ and $ y$ such that $ mx \plus{} ny \equal{} gcd(m,n)$   :maybe:[/quote]

Bezout's Theorem!


ok so you have to use 4 4's to make it equal 0-9. I got some of them but need help with for 1,3,4,5, and 6. Here's an example if it doesn't make sense to you:
ex. (4+4)-(4/4) =
(8)-(1) =
8-1 =7
 Help please


Let f,g be functions such that f:[2,$\infty$)-->R and g:A-->R , f(x)=x-2 and
g(x)=$\sqrt{4-3\sqrt{4-3\sqrt{10-3x}}}$.It's easy to prove that f is increasing and g is decreasing $\Rightarrow$ x=3 is the only solution

Could you tell me please what "infinite nesting" refer to ?

It means we can substitute $t$ by itself infinitely:

\begin{eqnarray*}t &=& \sqrt{4-3\sqrt{4-3\sqrt{4-3t}}}\\ &=& \sqrt{4-3\sqrt{4-3\sqrt{4-3\sqrt{4-3\sqrt{4-3\sqrt{4-3t}}}}}}\\ &=& \sqrt{4-3\sqrt{4-3\sqrt{4-3\sqrt{4-3\sqrt{4-3\sqrt{4-3\sqrt{4-3\sqrt{4-3\sqrt{4-3t}}}}}}}}}\\ &=&\dots\end{eqnarray*}

[quote="Farenhajt"]Or substitute $t=x-2$ to get

$t=\sqrt{4-3\sqrt{4-3\sqrt{4-3t}}}$

Now by infinite nesting we get $t=\sqrt{4-3t}\iff t=1\iff x=3$.[/quote]
It is not proved, that t=1 is unique solution.
Let $f(t)=\sqrt{4-3t}, g(t)=\frac{4-t^{2}}{3}$.
Equation $f(f(f(t)))=t$ equavalent to $g(g(g(t)))=t$ with $t>0,g(t)>0$ or $0<t<2$. We have $\frac{|g(t)-1|}{|t-1|}=\frac{|t+1|}{3}<1$, therefore this equation had unique solution.


algebra
polynomial
analytic geometry
Ring Theory
superior algebra
superior algebra theorems

Hello,

I was just wondering about this. :huh:

Let $Y$ be a projective variety, in $\mathbb{P}^{n}$ .  Let $S$be the ring of polynomials in the variables $x_{0},\ldots , x_{n}$. Let $I(Y)$ be the homogeneous ideal, generated by the homogeneous polynomials vanishing on $Y$.  Let $(Y)=S/I(Y)$be the homogeneous coordinate ring.

Now, how much is this "like the affine case"?  I mean, does every (homogeneous?)  maximal ideal of $(Y)$ correspond to a specific point on the projective variety $Y$?
I know this is true in the affine case: the points on $Y$ correspond with the maximal ideals in $A(Y)$, or thus with the maximal ideals containing $I(Y)$ in the total polynomial ring $A$.

Note that I am cautious when I write homogeneous, since I'm not even  sure whether or not $S(Y)$ is a graded ring.
 :ninja:

Does anyone know anything about this? :(
I'm quite confused? My guess would be that in $S$ , the maximal ideal generated by $x_{0},\ldots , x_{n}$  (which is a homogeneous ideal) does not correspond with any points, just the empty set?

[quote="owk"]Yes, there is a correspondence between homogeneous prime ideals not equal to the irrelevant ideal $S_{+}$ and irreducible closed subsets. In particular, ideals maximal with respect to these conditions correspond to points. This is all in Hartshorne ch. 1. owk[/quote]
 :oops:  I'm actually asking this for an exam I have within eight hours, and another on the next day so getting my hands on that book in time will be hard. :P
But do I understand you correctly: the points correspond to "the homogeneous prime ideals different from the irrelevant ideal, not contained in a bigger homogeneous prime ideal different from the irrelevant ideal"? :huh:

Yes, but I think this is too ugly to be useful. [url=http://modular.math.washington.edu/scans/252/hartshorne/Hartshorne-Algebraic_Geometry.pdf]Hartshorne (PDF, 155MB)[/url] owk


geometry
geometric transformation
reflection
LaTeX
summer program
Mathcamp
AMC

For the non-alumni and mere mortals,

Sit back, relax, and wait for an answer....

I only solved 7/10, so I will have to wait and see. I sent my solutions via fax, so hopefully they get there. Also, I hope it is not too late.

It was not too late (wow David that was fast! )

Anyone else?

Meeeeee~!

Just a random reflection on my application: I only solved 6 and a half (my quiz solutions were my first LaTeX document ever.. whee~) and unfortunately, my personal rec didn't turn out too... personal :( (my recommender showed it to me
afterwards) But I actually liked my essay (something that rarely ever happens)...  so hmm.

Yep, time to sit back and relax, and hopefully get in!

My brother's now polishing up his personal statement on the other computer... I'm feeling sort of accomplished since I'm usually the procrastinator. :P

I'm a procrastinator and I'm getting really crammed...
I persuaded myself into thinking that the deadline was by the end of April 27, instead of midnight April 26th, :? I still have six hours to go :)
I never seem to get my essays straight... I had two different drafts that I tore up. One ended like a boring three-page autobiography (sort of), Another just got really crazy. :( So now I'm working on my third draft... ugh. Hopefully this one maybe I'll like.
I'm polishing my quiz solutions simultaneously, I've solved 6 problems (not sure if they are all correct) and two incomplete ones.

haha i am frantically reminding my personal recommendation to get in her rec... she says she cant do it till 11:30 Pm so... yeah... i have 8/10 problems but i doubt 1 of them is completely correct... also i gotta polish up my essays and what not... today is gonna be rough ;)

6,7, and 8 are good scores folks. You guys shouldn't worry about getting in. ^_^ These problems - especially the last 4 - were not easy at all (at least in my opinion).

i have 6 full solutions and partials and/or insight to the others.
knowing me, some will be totally incorrect.

lets hope!

i am proly gonna submit my application around 11:45... when do you think a response will come back?

Well, I solved the first five fully, 90% for 7, and partial solutions for the rest.  I was really cramming last night and I sent them in this morning. I think I wrote a good essay though, so hopefully I'll get in.  (I only heard about Mathcamp like 2 wks before the deadline, ugh I wish I had more time for the quiz).  Yeah, I thought the last 4 were extremely difficult, too, but for some reason I didn't get #6 at all.  Maybe I was missing a "critical insight."

6 isn't too easy. It requires a bit of insight at you would probably gain after MathCamp. The last 4 are definitely all approachable.

The site is slooooow now.  Let me upload my solutionsss.

If everything I did was correct, I got 8 complete, 1 partial, 1 purely conjectures/deductions.

1-7, and 9 = complete
8 I couldn't prove only solutions
10 purely conjectures/deductions

Aw man, I just got a

[quote]
Something went awry!

Whoops! Something broke. It could just be a temporary thing, or maybe it's a serious bug.

Please log back in and try it again.

Please also send an email to application@mathcamp.org telling us what you were doing when the error occurred, so we can fix the problem.
[/quote]

Edit: x2

Submitted

We can't talk about solutions after the deadline, can we, since after today admissions are still rolling?

Oh, and #9b ... took me around a week.

[quote="sonar"]The site is slooooow now.[/quote]

I don't know [i]what's[/i] going on with the site right now, but it ain't good.  The site worked smoothly all day, but now there's been an out of memory error every 5 minutes for the past 45 minutes.  I'm sitting in my apartment watching the java logs, restarting the application whenever it dies...  so at least it's never out for more than a few seconds.

I've got our computer people trying to figure out what's going on....

[quote="ztbb"]It should be long enough to say what you want to say, and doesn't need to be any longer than that![/quote]
I wrote it 6 sentences. I think it's better than my other two drafts. :D

1,2,5,7,9 complete
3,6 no idea
4,8 posted my "findings" but no proof.
10  did it, but not so sure

[quote="bubala"]so have you gone to Mathcamp before and solve all the problems for fun?[/quote]

No, but I do know a handful of MathCamp alums and they all give it glowing recommendations. Personally, I got just about everything on 1-8. Unfortunately, AP studying is preventing me for working on 9 and 10. :(

wow... i have no chance...

i barely did 1
i did 2,3,4,5 dcompletely
6 partly
9 partly...

have NO CHANCE... whatsoever... and i am a USAMO quali too.... i feel somewhat embarassed... :?

haha i live in the east and i barely submitted mine in time (by my watch anyways) 11:56 lol... i can finally relax lol  :)

As you might tell from the time, I am currently trying to finish my answers before the deadline. So far I've got 1,2, mostly 3, 4, and half of 7. Hopefully I'll figure a couple more out in the next couple hours.

Wow. What time zone are you in? Now thats what I would call last minute -- literally!  Anyways good luck...and just keep thinking! It does seem like its all worth it!

[quote="xxreddevilzxx"]actually, i talked to the director... he said it was okay if we missed it by not much. I live in Pacific Time Zone, so i had to fax mine at 9:00PM here... I faxed at 9:06PM PT, and he accepted it.[/quote]

We aren't usually strict anyway -- obviously, because of the computer problems we will have to be very lenient this year.


There will be a math jam on this very subject next Wendsday.

Its a free thing that anyone who has an account can join, hosted on artofproblemsolving/mathlinks, by going to the classroom at the time of the scheduled math jam

Also, do a search for generatingfunctionology.  It's a book that is available for free over the internet.  Just about anything you would want to know about them are in this book (although I've only read the first couple pages....so I really can't tell you everything that is in there).

yes its a good book.  but i cant understand past the first like 10 pages :)

http://www.math.upenn.edu/~wilf/gfology.pdf

power series?

hmm i guess for me now, (IF i am not wrong) power series are those series like Maclaurin's or Taylor and so on. i've only learnt maclaurins so far :blush: .. Correct me if i am wrong

Maclaurin's:
f(x) = f(0) + xf'(0) + [(x^2)/2!]f''(0) + [(x^3)/3!]f'''(0) +... [(x^n)/(n!)]f-nth derivative(0)

e.g

sin x = 0 + x + 0 - (x^3)/6 + 0 + (x^5)/120 + .......
        = x - (x^3)/6 + (x^5)/120 + ......

so sin x is approximately x - (x^3)/6 assuming powers above 3 could be ignored.

hmm am i right?

Generating functions are, in themselves, above getting started and we are now getting into calculus.  This thread should be moved...to Other Problem Solving Topics perhaps.


As the subject suggested, I would like to ask: Prove or disprove: $ p^k\minus{}1$ is divisible by $ pk\plus{}1$ for all prime number $ p>2$ and natural number $ k>1$.

Obviously not, take $ p\equal{}5$ and $ k\equal{}3$.

I 'm sorry. I 've stated the problem wrongly:

Prove or disprove: $ p^k\minus{}1$ is [b]not divisible[/b] by $ pk\plus{}1$ for all prime number $ p>2$ and natural number $ k>1$.

Let $ a\equal{}pk\plus{}1$, then from $ a|p^k\minus{}1$ we get $ a|k^kp^k\minus{}k^k$ or $ a|k^k\minus{}(\minus{}1)^k$ and from $ a|k^k\minus{}(\minus{}1)^k, a\equal{}1\mod k$ we get $ a|(\frac{a\minus{}1}{k})^k\minus{}1$.
Therefore sufficiently consider divisors $ k^k\minus{}(\minus{}1)^k$
1. $ k\equal{}2\to p\equal{}1$,
2. $ k\equal{}3\to p\equal{}2$,
3. $ k\equal{}4\to p\equal{}1,4,21$ - not primes,
4. $ k\equal{}5\to p\equal{}1,104,625$ - not primes,
5. $ k\equal{}6\to p\equal{}1,5,7,36,50,222,1555$ had prime solutions $ p\equal{}5$ and $ p\equal{}7$.
It give at least to solutions $ p\equal{}5,k\equal{}6$ and $ p\equal{}7,k\equal{}6$.


Prove that $\frac{1}{2 \sqrt{2}n}<\{n \sqrt{2}\}\leq 1-\frac{2(3-2 \sqrt{2})}{n}$ for all natural $n$.

Check for $n \le 12$.

Let $m$ be the nearest integer to $n\sqrt 2$.  Then $\vert m-n\sqrt 2\vert . \vert m+n\sqrt 2 \vert = \vert m^2-2n^2 \vert \ge 1$.

If $m < n\sqrt 2$ then $\{n \sqrt 2 \} = \vert m-n\sqrt 2\vert \ge 1 / \vert m+n\sqrt 2 \vert  > 1/(2n\sqrt 2)$.
If $m > n\sqrt 2$ then $\{n \sqrt 2 \} = 1 - \vert m-n\sqrt 2\vert$ and $\vert m-n\sqrt 2\vert \ge 1 / \vert m+n\sqrt 2 \vert  > 1/(2n\sqrt 2+1)$.  Now $2n \sqrt 2 + 1 < (3+2\sqrt 2)n/2$ for $n > 1/(3/2-\sqrt 2) > 12$.  So $\{n \sqrt 2 \} < 1 - 2/((3+2\sqrt 2)n) = 1 - 2(3-2\sqrt 2)/n$.

Nice...

There is a trivial mistake. You should write $n > 1/(3/2-\sqrt 2) \Rightarrow n \geq 12$, instead of $n > 1/(3/2-\sqrt 2) > 12$, so effectively you only need to check for $n < 12$. This doesn't affect the proof... nevertheless, it's still nice...


LaTeX
number theory unsolved
number theory

If a,b are positive integers, a|(b^2+1), b|(a^2+1), b>a, we have (3b-a)|b^2+1

If $a,b$ are positive integers. Given that $a|b^2+1$ and $b|a^2+1$ show that $(3b-a)|b^2+1$.

[LaTeX-ed]

@hawk: i was thinking so to.

I am sure the similar(or exactly the same ) problem  has been posted many times.
we can gain that a,b satisfy
$a^2+b^2+1-3ab=0$

How do you get that relation, HawkTiger?
I get the fact that if a^2 + 1 = kb and b^2 + 1 = sa, then a | k^2 + 1 and k | a^2 + 1. Also we have that k < a because b > a and a^2 + 1 = kb. So we can decrease our solution pairs until we reach a = 1. And the only solution pair with a = 1 is (a,b) = (1,2). So from this solution we get all the other pairs by:

a_(n+1) = b_n
b_(n+1) = ((b_n)^2 + 1)/a_n

How can you prove that 3b - a = b_n+1?

Oh...I saw that it was posted as an earlier problem hehe. Never mind.


Do you mean $ a^2\plus{}b^2\plus{}c^2\plus{}abc\equal{}4$??

[quote="kihe_freety5"]a,b,c>0 a²+b²+c²+abc=4;
prove that :
                   3(a+b+c)≥abc+8[/quote]
$ a\equal{}2cosA;b\equal{}2cosB;c\equal{}2cosC$ with $ A;B;C$ be three angle in a triangle.
We need prove that:
$ 6(cosA\plus{}cosB\plus{}cosC) \geq 8(cosAcosBcosC\plus{}1)$

[quote="kihe_freety5"]a,b,c>0 a²+b²+c²+abc=4;
prove that :
                   3(a+b+c)≥abc+8[/quote]
This inequality is not always true.
And I found it is Jack Garfunkel Inequality.
My solution for it :)


algebra
polynomial
AMC
AMC 12
AIME
system of equations

Given $s_{0}= 3$, and $s_{1}= 4$, under the recurrence relation $s_{n}= 4s_{n-1}-3s_{n-2}$, find $s_{2008}$ .

$(\text{A}) \frac{12+2^{2005}}{2}$

$(\text{B}) \frac{3^{2008}+5}{2^{1004}}$

$(\text{C}) \frac{4^{2008}+7}{2^{2007}}$

$(\text{D}) \frac{3^{2008}+5}{2}$

$(\text{E}) \frac{2^{4016}-3^{2008}+5}{2}$

[hide="My Solution"]:
The characteristic polynomial of the recurrence relation is $s^{2}=4s-3$, or $s^{2}-4s+3 = 0$. This polynomial factors as $(s-3)(s-1)$, with roots $r_{1}= 3$ and $r_{2}= 1$.

By Binet's theorem, we can write the general form for the Nth term of a linear recurrence as $s_{n}=A(r_{1})^{n}+B(r_{2})^{n}$ with constants A and B.

Now we solve a system of equations using our starting values.
$3 = A(3)^{0}+B(1)^{0}\longrightarrow 3 = A+B$
$4 = A(3)^{1}+B(1)^{1}\longrightarrow 4 = 3A+B$
Obtaining values of A = 0.5, B = 2.5, we complete our closed form expression.
$s_{n}= 0.5(3)^{N}+2.5(1)^{N}\longrightarrow s_{n}= 0.5(3)^{N}+2.5$

Our value is then $\frac{3^{2008}+5}{2}$, or Choice D.
[/hide]

Any comments? I designed this problem to be ~23 level on the AMC12, or 4-5 level on the AIME.

Alternate solutions are welcome as well. Please let me know if my solution has errors, as I'm thinking about throwing together a Mock AIME for my school and I might do some variants on this problem. I don't want any errors, so let me know if something seems funky and I'll try to explain. Thanks! Matt Redmond.

Characteristic polynomials aren't really AMC level...

What level would you classify this as, then?

That's pretty clean, quick, and clear. Did you like the problem? (What difficulty should I put this as (roughly?))
I like your solution a lot.

thank you! but i saw this method on aops before, and i thought it was nice, too, so i learned/remembered it.

As an aspiring problem writer, is there anything I could change to make this problem better? I don't want to repeat too many old ideas.

[quote="DiscreetFourierTransform"]I don't want to repeat too many old ideas.[/quote]

Unfortunately, this is a pretty common idea.  You generally wouldn't find this question on a real test because it is too well-known and, by itself, does not pose enough of a thinking challenge.

I've never heard of characteristic polynomials.

Does it always work????

It's why Binet's formula for the Fibonacci numbers works. It always works for linear recursions.

Say we are given an arbitrary LINEAR recursion
$a_{n}= Xa_{n-1}+Ya_{n-2}+Za_{n-3}$ (there could be as many more terms as you want, but it's enough of an example with three on the RHS.)

We can then rewrite this as
$a_{n+3}= Xa_{n+2}+Ya_{n+1}+Za_{n+0}$

Now transform the subscripts into exponents. I'll not include a proof of why this works, but I believe that Mathworld and Wikipedia both have one.
The characteristic polynomial of the above recursion is then
$a^{3}= Xa^{2}+Ya^{1}+Za^{0}$
or $a^{3}-Xa^{2}-Ya-Z = 0$

By Binet's theorem, we can rewrite any linear recursion as a closed form expression
$a_{n}= A(r_{1})^{n}+B(r_{2})^{n}+C(r_{3})^{n}$... where various $r_{i}$ are the roots of the characteristic polynomial and A, B, and C are constants given by the initial values.
We then see that if we have a recursion with three terms on the RHS to begin with, our characteristic polynomial will have three roots, and thus we can set up a system of three equations.

We put
$a_{0}= A(r_{1})^{0}+B(r_{2})^{0}+C(r_{3})^{0}$
$a_{1}= A(r_{1})^{1}+B(r_{2})^{1}+C(r_{3})^{1}$
$a_{2}= A(r_{1})^{2}+B(r_{2})^{2}+C(r_{3})^{2}$
then we substitute in our original values and our roots of the characteristic polynomial, letting us set up a system where we can solve for constants A, B, and C.

Our final equation then takes the form
$a_{n}= A(r_{1})^{n}+B(r_{2})^{n}+C(r_{3})^{n}$
with known constants A, B, C and known roots $r_{1}, r_{2}, r_{3}$
From there, it's a matter of evaluating this closed form expression to find nth term values.

The book [url=http://www.math.upenn.edu/~wilf/DownldGF.html]Generatingfunctionology[/url] (available free at that link) provides in its opening chapter some of the theoretical background for why this should actually work.

Incidentally, if you ask me, the real problem with this problem is that it admits only two types of solution: method 1 is to happen to know how to solve questions like this and then is a direct application of the method.  Method 2 is to get lucky and notice a pattern in the first few terms of the sequence.  Neither of these is very appealing from a problem-solving point of view.

The solution posted by minsoens was different from the application of the Method, was quite "amc level", and worked elegantly. Perhaps there are other methods then either seeing a pattern or using a formula.

Fair enough, and shame on me for not reading the whole thread.  The same nice features of the answer allowed a simplification to a much more recognizable pattern.


Does there exist a function $ f: {\Bbb N} \to {\Bbb N}$ such that $ f(f(n \minus{} 1)) \equal{} f(n \plus{} 1) \minus{} f(n)$ for all $ n > 2$.

We have for $ n \ge 2$, $ f(n)$ is strictly increasing, so $ f(n) \ge n\minus{}1$ for all $ n \ge 2$

Now we also have $ f(f(n\minus{}1)) \le f(f(n\minus{}1))\plus{}f(f(n\minus{}2))\plus{}...\plus{}f(f(1))\equal{}f(n\plus{}1)\minus{}f(2)<f(n\plus{}1)$ for all $ n \ge 2$, so $ f(n\minus{}1) \le n$

So now we have $ n\minus{}1 \le f(n) \le n\plus{}1$ for all $ n \ge 2$,

$ n\minus{}3 \le f(f(n\minus{}1))\equal{}f(n\plus{}1)\minus{}f(n) \le 3$ for all $ n>3$, impossible. So there doesn't exist such function.

$ f(n \plus{} 1) \minus{} f(n) \equal{} f(f(n \minus{} 1))\ge 1$ $ \Rightarrow f(n)\ge n$.
Thus $ f(n\plus{}1)\equal{}f(n)\plus{}f(f(n\minus{}1))\ge f(n)\plus{}f(n\minus{}1)\ge 2n\minus{}1$. Notice that $ f(n)\equal{}2n\minus{}1$ obviously doesn't satisfy the condition. Therefore, there exist such $ c$ that $ f(c)\ge 2c$. Clearly for $ n>c$ we will obtain that $ f(n)\ge 2n$.

Let $ f(n)\ge 2^in$ ($ i\ge 1$). Then $ f(n \plus{} 1) \minus{} f(n) \equal{} f(f(n \minus{} 1))\ge 2^{i}f(n \minus{} 1)\ge 2^{2i}(n \minus{} 1)$ $ \Rightarrow f(n \plus{} 1)\ge f(n) \plus{} 2^{2i}(n \minus{} 1)\ge 2^in \plus{} 2^{2i}(n \minus{} 1)\ge 2^{i \plus{} 1}(n \plus{} 1)$
$ \Leftrightarrow 2^{i}(n \minus{} 1)\ge n \plus{} 2$, which is obvious for $ n > 3$ for example . Hence, $ f(n)\ge \lim_{i\rightarrow \infty}2^in \equal{} \infty$ for $ n > 3$, which is impossible, since $ f(n)$ is a constant.

not so hard
we can trivially get f is strictly increasing
so $f(n)\ge n$
since $f(n+1)=f(n)+f(f(n-1))$
hence $f(n-1)<n+1$,or $f(n)\le n+1$
so for every n,f(n)equals to either n or n+1.
hence $f(n+1)-f(n)=1 or 2$
yielding $f(n-1)=f^{-1}(f(n+1)-f(n))=1 or 2$
this contradicts with the increasingness of f.
QED


??
$4^{16} = 64^4 > 16^4 + 8$ and $3^{12} = 9^6 > 9^4 > 6^4$ thus the expression is positive...Did I miss something?

Pierre.


function
integration
calculus
calculus computations

Prove:
\[ \zeta (s) = s \int_{1}^{\infty} \frac{[x]}{x^{s+1}}dx \]
P.S. this might be trivial but I'm beginner in this field :oops:

The hint in the book says compute the integral from $1$ to $N$, and compare it with the $N$-th partial sum of $\frac{1}{i^s}$

$\displaystyle s \int_{1}^{N} \frac{[x]}{x^{s+1}}dx = s \left( \int_{1}^{2} \frac{1}{x^{s+1}}dx + \int_{2}^{3} \frac{2}{x^{s+1}}dx + ... + \int_{N-1}^{N} \frac{N-1}{x^{s+1}}dx \right)$

$\displaystyle = -s \left [ (\frac{1}{s 2^s} - \frac{1}{s}) +(\frac{2}{s 3^s} - \frac{2}{s 2^s}) + ... + (\frac{N-1}{s N^s} - \frac{N-1}{s (N-1)^s}) \right ]$

$\displaystyle = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + ... + \frac{1}{N^s} = \sum_{i=1}^{N}\frac{1}{i^s}$


A mixture of methanol and ethanol contains $41\%{\text{ }}O$
 (percentages of mass). Which is the molar ratio of the two alcohols in the mixture?

Thanks!

[hide="Hint"]Consider a sample of the mixture with 100 g of mass. What is the mass of oxygen in this sample? What is the combined mass of carbon and hydrogen? (Write the chemical formulas of methanol and ethanol). Write those two masses in terms of number of moles of methanol and ethanol and the molar masses of carbon, oxygen, and hydrogen.[/hide]

[hide="Answer"]$\frac{n_{methanol}}{n_{ethanol}}\approx 1$[/hide]

[hide="Solution"]

$m(O) = \frac{41}{100}(m(C_{2}H_{5}OH)+m(CH_{3}OH))$

Assume there is 1 mol of $C_{2}H_{5}OH$ and $n$ mol of $CH_{3}OH$

Then you get $16(1+n) = \frac{41}{100}(46+32n) \Rightarrow 288n=286$

So the molar ratio is basically 1:1. [/hide]


If $a,b,c \in R$ so that $ \sum \frac{1}{1+a^2}=2$ then $abc(a+b+c-abc) \leq \frac{5}{8}$.

Multiply both side of the given equation by (a 2 +1)(b 2 +1)(c 2 +1),we have
               a 2 b 2 +b 2 c 2 +c 2  a 2 +2a 2b  2 c 2 =1
            Let ab=x ,bc=y ,ac=z ,we have:
                x 2 +y 2  +z 2 +2xyz=1
            So x=cos A ,y=cos B ,z=cosC ,with A,B,C are angles of an acute triangle
     The inequality we have to prove is equivalent to
   CosAcosB+CosBcosC+cosCcosA-CosAcosBcosC \leq 5/8 (*)
 But
          CosACosBcosC=(Sin 2 A+Sin 2 B+Sin 2 C)/2 -1
So (*) is equivalent to
      2(CosAcosB+CosBcosC+cosCcosA)-(3- cos 2 A- cos 2 B- cos 2 C) +2  <=5/4
     2(cosAcosB+cosBcosC+cosCcosA)+(cos 2 A+cos 2 B+cos 2C)  \leq 9/4
  <-->  cosA+cosB+cosC \leq 3/2 ,which is a well-known inequality

You should check your solution. Because a,b,c are real numbers,you can't
have that trigonometric replacement. :D

1/(1+a 2 )=x 1/(1+b 2 )=y 1/(1+c 2 )=z
x=2u/(u+v+w)
y=2v/(u+v+w)
z=2w/(u+v+w)
a 2 =(v+w-u)/2u
b 2 =(w+u-v)/2v
c 2 =(u+v-w)/2w
u=q+r
v=r+p
w=p+q
the ineq is:
a 2 bc+ab 2 c+abc 2  \leq 5/8+a 2 b 2 c 2  so it's enough to prove for a,b,c>0
a= \sqrt p/(q+r)
b= \sqrt q/(r+p)
c= \sqrt r/(p+q)
 \sum 1/(1+a 2 )=2
1=a 2 b 2 +b 2 c 2 +c 2 a 2 +2(abc) 2
So the ineq becomes:
2(a 2 bc+b 2 ca+c 2 ab) \leq 5/4+2(abc) 2
2(a 2 bc+b 2 ca+c 2 ab) \leq 5/4+1-a 2 b 2 -b 2 c 2 -c 2 a 2
(ab+bc+ca) 2  \leq 9/4
ab+bc+ca \leq 3/2
 \sum  \sqrt pq/(p+r)(q+r)  \leq 3/2
2 \sqrt pq/(p+r)(q+r) \leq p/(p+r) + q/(q+r)
The same with the others and adding we get the result


Show that if $ m, n$ and $ r$ are positive integers and: $ 1 + m + n\sqrt {3} = (2 + \sqrt {3})^{2r - 1}$ then $ m$ is a perfect square.

$ (2\plus{}\sqrt 3)^a\equal{}m\plus{}1\plus{}n\sqrt 3, (2\minus{}\sqrt 3)^a\equal{}m\plus{}1\minus{}n\sqrt 3\to 1\equal{}(m\plus{}1)^2\minus{}3n^2$.
Therefore $ m(m\plus{}2)\equal{}3n^2$ and if $ (6,m)\equal{}1$, then $ 3\not m, (m,m\plus{}2)\equal{}1$ . so $ m$ is square,
$ m\equal{}2^a\minus{}1\mod 3\equal{}1\mod 3$ because a is odd and $ m\plus{}1\equal{}0\mod 2$ because a is odd. It give result.


vector
function
integration
calculus
derivative
calculus computations

Consider the vector field  F = 2xi - 4yj + (2z - 3)k.

Find the potential function for f .

I am kind of stuck and don't know how to solve this....

The purpose is to find a scalar field V such that $ \vec{F} \equal{} \nabla V$. If you cannot think of a such a V, then suppose $ \vec{F}$ is a force field and let's calculate the work done by this field between to positions A and B:

$ W \equal{} \int_A^B \vec{F} \cdot d \vec{r} \equal{} \int_A^B [2x\,dx \minus{} 4y \,dy \plus{} (2z\minus{}3) \,dz]$.

I think you can now proceed from here by yourself.

[quote="Carcul"]The purpose is to find a scalar field V such that $ \vec{F} \equal{} \nabla V$. If you cannot think of a such a V, then suppose $ \vec{F}$ is a force field and let's calculate the work done by this field between to positions A and B:

$ W \equal{} \int_A^B \vec{F} \cdot d \vec{r} \equal{} \int_A^B [2x\,dx \minus{} 4y \,dy \plus{} (2z \minus{} 3) \,dz]$.

I think you can now proceed from here by yourself.[/quote]

hmm..but what is A and B here?

Could made it  an indefinite integral, and the constant would depend on the choice of the origin or reference of the potential.

There is a systematic way to do this. We're trying to find a function $ v$ such that $ \nabla v\equal{}\vec{F}.$ which is to say:
\[ \frac{\partial v}{\partial x}\equal{}2x;\ \frac{\partial v}{\partial y}\equal{}\minus{}4y;\ \frac{\partial v}{\partial z}\equal{}2z\minus{}3\]
The procedure is to take it one variable at a time, finding the antiderivative with respect to that variable. The "constant of integration" won't be a constant but rather a yet unknown function of the variables that haven't been used yet. In between each step, we find the partial derivative of our new expression for $ v$ with respect to the next variable.

$ v\equal{}\int 2x\,dx\equal{}x^2\plus{}g(y,z).$ (That's what I mean by an unknown function.)

$ \frac{\partial v}{\partial y}\equal{}0\plus{}\frac{\partial g}{\partial y}\equal{}\minus{}4y,$ so

$ \frac{\partial g}{\partial y}\equal{}\minus{}4y$

$ g\equal{}\int \minus{}4y\,dy\equal{}\minus{}2y^2\plus{}h(z),$ so $ v\equal{}x^2\minus{}2y^2\plus{}h(z).$

$ \frac{\partial v}{\partial z}\equal{}0\plus{}0\plus{}h'(z)\equal{}2z\minus{}3,$ so

$ h(z)\equal{}\int (2z\minus{}3)\,dz\equal{}z^2\minus{}3z\plus{}C.$

Hence we have $ v\equal{}x^2\minus{}2y^2\plus{}z^2\minus{}3z\plus{}C.$

In this case, the problem was so easy we didn't need this method: the variables separate fully and we can see that we can just do three unconnected integrations. In general, however, we do need what I've outlined above.

The method is self-correcting in the sense that if you attempt to do this for a vector field for which $ \text{curl}\,\vec{F}\equal{}\nabla\times\vec{F}\ne\vec{0},$ then you will at some point arrive at a false equation and be unable to proceed.

[quote="Kent Merryfield"]There is a systematic way to do this. We're trying to find a function $ v$ such that $ \nabla v \equal{} \vec{F}.$ which is to say:
\[ \frac {\partial v}{\partial x} \equal{} 2x;\ \frac {\partial v}{\partial y} \equal{} \minus{} 4y;\ \frac {\partial v}{\partial z} \equal{} 2z \minus{} 3
\]
The procedure is to take it one variable at a time, finding the antiderivative with respect to that variable. The "constant of integration" won't be a constant but rather a yet unknown function of the variables that haven't been used yet. In between each step, we find the partial derivative of our new expression for $ v$ with respect to the next variable.

$ v \equal{} \int 2x\,dx \equal{} x^2 \plus{} g(y,z).$ (That's what I mean by an unknown function.)

$ \frac {\partial v}{\partial y} \equal{} 0 \plus{} \frac {\partial g}{\partial y} \equal{} \minus{} 4y,$ so

$ \frac {\partial g}{\partial y} \equal{} \minus{} 4y$

$ g \equal{} \int \minus{} 4y\,dy \equal{} \minus{} 2y^2 \plus{} h(z),$ so $ v \equal{} x^2 \minus{} 2y^2 \plus{} h(z).$

$ \frac {\partial v}{\partial z} \equal{} 0 \plus{} 0 \plus{} h'(z) \equal{} 2z \minus{} 3,$ so

$ h(z) \equal{} \int (2z \minus{} 3)\,dz \equal{} z^2 \minus{} 3z \plus{} C.$

Hence we have $ v \equal{} x^2 \minus{} 2y^2 \plus{} z^2 \minus{} 3z \plus{} C.$

In this case, the problem was so easy we didn't need this method: the variables separate fully and we can see that we can just do three unconnected integrations. In general, however, we do need what I've outlined above.

The method is self-correcting in the sense that if you attempt to do this for a vector field for which $ \text{curl}\,\vec{F} \equal{} \nabla\times\vec{F}\ne\vec{0},$ then you will at some point arrive at a false equation and be unable to proceed.[/quote]

thanks for the answers.. I'd appreciate it.. if I am asked to:

Use the Fundamental Theorem of Line Integrals to find the exact value of  the line integral for C the line from (1, 1, 1) to (4, 5, -8). What should I do? can I utilize the potential function I get above and plug in numbers and get the difference?

If $ \vec{F}\equal{}\nabla v,$ and $ C$ is a curve that starts at point $ p_0$ and ends at point $ p_1,$ then
\[ \int_C\vec{F}\cdot d\vec{r}\equal{}\int_C\nabla v\cdot d\vec{r}\equal{}v(p_1)\minus{}v(p_0).\]
What did you think was meant by "Fundamental Theorem of Line Integrals" anyway? This is it. Since you've found $ v,$ all you have to do is evaluate it at each end. And the answer is independent of path - it doesn't matter that that this is the line segment between those points; it could be any (rectifiable) path with these endpoints. The result would be the same.

[quote="Kent Merryfield"]If $ \vec{F} \equal{} \nabla v,$ and $ C$ is a curve that starts at point $ p_0$ and ends at point $ p_1,$ then
\[ \int_C\vec{F}\cdot d\vec{r} \equal{} \int_C\nabla v\cdot d\vec{r} \equal{} v(p_1) \minus{} v(p_0).
\]
What did you think was meant by "Fundamental Theorem of Line Integrals" anyway? This is it. Since you've found $ v,$ all you have to do is evaluate it at each end. And the answer is independent of path - it doesn't matter that that this is the line segment between those points; it could be any (rectifiable) path with these endpoints. The result would be the
same.[/quote]

well I tried to plug in numbers and take the difference and I got is:

=(16 - 50 + 64 + 24) - (1-2+1-3)
=54-3
=51

and I input this into webassign and gives me a wrong result

[quote="Kent Merryfield"][quote]=(16 - 50 + 64 + 24) - (1-2+1-3)
=54-3
=51 [/quote]
The arithmetic mistake is right in front of you. Find it.[/quote]

sorry I think it's 54-2 = 52.. but still that's not the answer

I'm still waiting for you to do some correct arithmetic. Hasn't happened yet.

You've got a double-negative.  It's $ 54\minus{}(\minus{}3)\equal{}54\plus{}3\equal{}57$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="Jan"]Fix a prime $ p$, and suppose that a subgroup $ H \subseteq G$ has the property that $ C_G(x) \subseteq H$ for every element $ x \in H$ having order $ p$.
Show that $ p$ cannot divide both $ \left|H\right|$ and $ \left|G : H\right|$.[/quote]

Let P be a Sylow p-subgroup of H, a subgroup of the finite group G.  If [G:H] is divisible by p, then [G:P] is divisible by p, and so [G:N_G(P)] is divisible by p, and so P is contained as a normal subgroup in some larger p=group Q.  Then P intersect Z(Q) is non-trivial (if P is non-trivial), so there is some x in P ≤ H, x of order p, such that Q is contained in the centralizer of x.  Hence if |H| and [G:H] are both divisible by p, then H contains an element of order p centralized by elements outside of H.


Each of the interior angles of a heptagon (a seven-sided polygon) is obtuse and the number
of degrees in each angle is a multiple of $ 9$. No two angles are equal. Find in degrees the
sum of the two largest angles in the heptagon.

[hide]
The interior angles of a heptagon = 180 (n-2) = 900 degree
All angles are multiples of nine so let the angles be labled 9a,9b,9c,9d,9e,9f,9g where a<b<c..<g
therefore 9(a+b+c...+g) = 900
since all angles are greater than 90, a>10
since a is also the smallest a =11
Also, b=12 c=13 d=14 e=15
and 9(f+g)= 900 - 9(11+12+...15) = 315

Answer: 315
[/hide]


Prove that there are infinitely many composite numbers $ n$ such that $ n$ divides $ 3^{n\minus{}1}\minus{}2^{n\minus{}1}$.

A good problem . Have some ways to solve this problem .
Here is my solution .
Take $ n = 3^k - 2^k$
We have $ n|3^{n - 1} - 2^{n - 1} %Error. "Leftrighatarow" is a bad command.
k|3^k - 2^k - 1$
We prove that have infinite k such that $ k|3^k - 2^k - 1$
Take $ k = 2^m$
We will prove that $ 2^m|3^{2^m} - 2^{2^m} - 1$
$ \Leftrightarrow 2^{2^m}|3^{2^m} - 1$
But we knew that $ v_{2}(3^{2^m} - 1) = m + 1$
So $ 2^m|3^{2^m} - 1$
It mean that $ k = 2^m$ satisfy condition .
Easy to check that $ n = 3^{2^m} - 2^{2^m}$ where $ m > 1$ is a composite number.
Problem was proved .


http://yashar.biz/mathcounts/mathcounts_files/2001/2001%20State%20Target.pdf

click on the link and go to problem 3.

[hide]and in case you don't know... the form of a circle is:
(x-h)^2+(y-k)^2=r^2
where r=radius
and (h,k) is the midpoint...
so the midpoint would be (0,0) with a radius of 2.[/hide]

[quote="Ihatepie"][hide]
graphing out x^2+y^2≤4 while keeping it in the rectangle, you get a quarter circle with radius 2. so The area of the quarter circle is 2^2pi/4 or pi. Since the area of the rectangle is 32, the probability is pi/32.[/hide][/quote]

thanks i get it now


inequalities
inequalities proposed
algebra

Assume that $ c\equal{}\min\{a,b,c\}.$ Then, we have
$ \frac{a^2}{(b\minus{}c)^2} \ge \frac{a^2}{b^2},\frac{b^2}{(a\minus{}c)^2} \ge \frac{b^2}{a^2},$
and
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \minus{}\frac{a^2\plus{}b^2}{ab} \equal{}\frac{c[abc\minus{}(a\plus{}b)(a^2\plus{}b^2)]}{ab(ab\plus{}bc\plus{}ca)} \le 0.$
From these inequalities, we obtain the following remarks:

For (a), it suffices to prove that
$ \frac{a^2}{b^2}\plus{}\frac{b^2}{a^2} \ge \frac{a^2\plus{}b^2}{ab},$ which is true.

For (b), since $ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)} \equal{}\frac{a^2\plus{}b^2\plus{}c^2}{2(ab\plus{}bc\plus{}ca)}\plus{}1 \le \frac{a^2\plus{}b^2}{2ab}\plus{}1,$ it suffices to prove that
$ \frac{a^2}{b^2}\plus{}\frac{b^2}{a^2} \ge \frac{a^2\plus{}b^2}{2ab}\plus{}1,$ which is true.

For (c), it suffices to prove that
$ \frac{a^2}{b^2}\plus{}\frac{b^2}{a^2}\plus{}2 \ge \left( \frac{a^2\plus{}b^2}{ab}\right)^2,$ which leads to an equality.

[quote="Vasc"]

What about this

$ (d) \ \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2 \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge 4$ ?[/quote]
I think my way still works, but it is a little more complicated.

Without loss of generality, we may assume that $ c$ is the smallest number among $ a,b,c.$ We will show that
\[ \frac{a^{2}}{(b\minus{}c)^{2}}\plus{}\frac{b^{2}}{(a\minus{}c)^{2}}\plus{}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ca}\geq \frac{a^{2}}{b^{2}}\plus{}\frac{b^{2}}{a^{2}}\plus{}\frac{a^{2}\plus{}b^{2}}{ab}.\]
Indeed, this inequality is equivalent to
\[ \frac{a^{2}}{(b\minus{}c)^{2}}\minus{}\frac{a^{2}}{b^{2}}\plus{}\frac{b^{2}}{(a\minus{}c)^{2}}\minus{}\frac{b^{2}}{a^{2}}\geq \frac{a^{2}\plus{}b^{2}}{ab}\minus{}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ca},\]
or
\[ \frac{ca^{2}(2b\minus{}c)}{b^{2}(b\minus{}c)^{2}}\plus{}\frac{cb^{2}(2a\minus{}c)}{a^{2}(a\minus{}c)^{2}}\geq \frac{c[(a\plus{}b)(a^{2}\plus{}b^{2})\minus{}abc]}{ab(ab\plus{}bc\plus{}ca)}.\]
Since
\[ \frac{2b\minus{}c}{(b\minus{}c)^{2}}\geq \frac{2b\minus{}2c}{(b\minus{}c)^{2}}\equal{}\frac{2}{b\minus{}c}\geq \frac{2}{b},\quad\frac{2a\minus{}c}{(a\minus{}c)^{2}}\geq \frac{2}{a},\]
and
\[ \frac{(a\plus{}b)(a^{2}\plus{}b^{2})\minus{}abc}{ab(ab\plus{}bc\plus{}ca)}\leq \frac{(a\plus{}b)(a^{2}\plus{}b^{2})}{a^{2}b^{2}},\]
it suffices to prove that
\[ \frac{2ca^{2}}{b^{3}}\plus{}\frac{2cb^{2}}{a^{3}}\geq \frac{c(a\plus{}b)(a^{2}\plus{}b^{2})}{a^{2}b^{2}},\]
which is true because
\[ \frac{2a^{2}}{b^{3}}\plus{}\frac{2b^{2}}{a^{3}}\equal{}\frac{2a^{4}}{a^{2}b^{3}}\plus{}\frac{2b^{4}}{a^{3}b^{2}}\geq \frac{2(a^{2}\plus{}b^{2})^{2}}{a^{2}b^{2}(a\plus{}b)}\geq \frac{(a\plus{}b)(a^{2}\plus{}b^{2})}{a^{2}b^{2}}.\]
Now, according to the above inequality, one can reduce the problem into proving that
\[ \frac{a^{2}}{b^{2}}\plus{}\frac{b^{2}}{a^{2}}\plus{}\frac{a^{2}\plus{}b^{2}}{ab}\geq 4,\]
which is true.

[quote="Vasc"]Nice, Can_hang. :lol:
What about this?

$ (e) \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2 \plus{} \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}\ge 6$ ?[/quote]

In my result, we even have

$ (f)\ \ \  \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2 \plus{} \frac{9}{4} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge \frac{13}{2}.$

[quote="can_hang2007"][quote="Vasc"]Nice, Can_hang. :lol:
What about this?

$ (e) \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2 \plus{} \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}\ge 6$ ?[/quote]

In my result, we even have

$ (f)\ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2 \plus{} \frac {9}{4} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge \frac {13}{2}.$[/quote]

Indeed, the inequality below holds for any distinct nonpositive real numbers $ a,b,c$ and $ \minus{}6\le k\le \frac {13}{2}$:

$ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2 \plus{} \frac {(k\minus{}2)(a^2 \plus{} b^2 \plus{} c^2)}{2(ab \plus{} bc \plus{} ca)}\ge k$.

But I think your above method does not work for $ k>4$.  :?:

[quote="Vasc"][quote="can_hang2007"][quote="Vasc"]Nice, Can_hang. :lol:
What about this?

$ (e) \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2 \plus{} \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}\ge 6$ ?[/quote]

In my result, we even have

$ (f)\ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2 \plus{} \frac {9}{4} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge \frac {13}{2}.$[/quote]

Indeed, the inequality below holds for any distinct nonpositive real numbers $ a,b,c$ and $ \minus{} 6\le k\le \frac {13}{2}$:

$ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2 \plus{} \frac {(k \minus{} 2)(a^2 \plus{} b^2 \plus{} c^2)}{2(ab \plus{} bc \plus{} ca)}\ge k$.

But I think your above method does not work for $ k > 4$.  :?:[/quote]
I also think so, Vasc. My way does not work for the case $ k>4$ :(. But in this case, we can prove the inequality as follows:

First, we see that for $ a^2\plus{}b^2\plus{}c^2 \ge 2(ab\plus{}bc\plus{}ca),$ the inequality is true according to
$ \sum \frac{a^2}{(b\minus{}c)^2} \ge 2,$
so it suffices to consider the case $ 2(ab\plus{}bc\plus{}ca)>a^2\plus{}b^2\plus{}c^2.$ Now, note that
$ \sum \frac{a^2}{(b\minus{}c)^2} \ge 2\plus{}\frac{9}{4} \frac{2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2}{a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca}.$
(There is a nice identity here)
Therefore, it suffices to show that
$ \frac{9}{4} \frac{2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2}{a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca} \plus{}\frac{(k\minus{}2)(a^2\plus{}b^2\plus{}c^2)}{2(ab\plus{}bc\plus{}ca)} \ge k\minus{}2,$
wich is equivalent to
$ \frac{9}{4} \frac{2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2}{a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca} \ge \frac{k\minus{}2}{2}\frac{2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2}{ab\plus{}bc\plus{}ca}.$
This is true because $ \frac{9}{4} \ge \frac{k\minus{}2}{2}>0,$ $ 2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2>0,$ and
$ ab\plus{}bc\plus{}ca>a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca.$

[quote="can_hang2007"] Now, note that
$ \sum \frac {a^2}{(b \minus{} c)^2} \ge 2 \plus{} \frac {9}{4} \frac {2(ab \plus{} bc \plus{} ca) \minus{} a^2 \minus{} b^2 \minus{} c^2}{a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca}.$
(There is a nice identity here)[/quote]
This is a very special method.  :wink:

$ (d) \ \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2\plus{}\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge 4$ ?
[img]http://s6.sinaimg.cn/middle/0018zOAxgy71GdIepXn75&690[/img]

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

$ (a) \ \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2\ge \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$;

[/quote]

We have
\[LHS-RHS=\frac{1}{2(ab+bc+ca)}\sum{\frac{c(a+b)(a^2+ab-2ac+b^2-2bc+c^2)^2}{(-c+a)^2(b-c)^2}}
+\frac{1}{2(ab+bc+ca)}\sum{\frac{ab(a-b)^2(a+b-c)^2}{(-c+a)^2(b-c)^2}}\ge{0}\]

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then
$ (b) \ \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2\ge \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}$.
[/quote]

We have
\[LHS-RHS=\frac{1}{8(ab+bc+ca)}\sum{\frac{c(3a+3b+c)(a^2+ab-2ac+b^2-2bc+c^2)^2}{(a-c)^2(b-c)^2}}
+\frac{5}{8(ab+bc+ca)}\sum{\frac{ab(a-b)^2(a+b-c)^2}{(a-c)^2(b-c)^2}}\ge{0}\]

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then
Moreover,
$ (c) \ \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2\plus{}2\ge \left(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^2$.[/quote]

We have
\[LHS-RHS=\frac{abc}{(ab+ac+bc)^2}[\sum{\frac{c(b+2c)(a+2c)}{(a-b)^2}}+4(a+b+c)]\ge{0}\]

[quote=Vasc][quote="can_hang2007"]Assume that $ c \equal{} \min\{a,b,c\}.$ Then, we have
$ \frac {a^2}{(b \minus{} c)^2} \ge \frac {a^2}{b^2},\frac {b^2}{(a \minus{} c)^2} \ge \frac {b^2}{a^2},$
and
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \minus{} \frac {a^2 \plus{} b^2}{ab} \equal{} \frac {c[abc \minus{} (a \plus{} b)(a^2 \plus{} b^2)]}{ab(ab \plus{} bc \plus{} ca)} \le 0.$
From these inequalities, we obtain the following remarks:

For (a), it suffices to prove that
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{a^2} \ge \frac {a^2 \plus{} b^2}{ab},$ which is true.

For (b), since $ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2(ab \plus{} bc \plus{} ca)} \plus{} 1 \le \frac {a^2 \plus{} b^2}{2ab} \plus{} 1,$ it suffices to prove that
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{a^2} \ge \frac {a^2 \plus{} b^2}{2ab} \plus{} 1,$ which is true.

For (c), it suffices to prove that
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{a^2} \plus{} 2 \ge \left( \frac {a^2 \plus{} b^2}{ab}\right)^2,$ which leads to an equality.[/quote]
Very nice, Can_hang. :lol:

What about this

$ (d) \ \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2\plus{}\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge 4$ ?[/quote]

[img]http://s4.sinaimg.cn/middle/006ptkjAzy7627c3HNN03&690[/img]

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

$ (a) \ \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2\ge \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$;

$ (b) \ \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2\ge \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}$.

Moreover,

$ (c) \ \ \ \ \left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2\plus{}2\ge \left(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^2$.[/quote]
For pairwise distinct nonnegative reals $a,b,c$, prove that
$$\left(\frac a{b \minus{} c}\right)^2 \plus{} \left(\frac b{c \minus{} a}\right)^2 \plus{} \left(\frac c{a \minus{} b}\right)^2>2$$
[url=https://artofproblemsolving.com/community/c6h1414671p7967479]Canadian Mathematical Olympiad 2017[/url]


When the repeating decimal $ 0.\overline{12}$ is expressed as a common fraction in lowest terms, what is the sum of its numerator and denominator?

Let $ x\equal{}0.\overline{12}$.

Then $ 100x\equal{}12.\overline{12}$.

If we subtract the first equation from the second, we get $ 99x\equal{}12$.  Thus, $ x\equal{}\frac{12}{99}$.  Simplifying, we get $ x\equal{}\frac{4}{33}$.  Adding those yields $ \boxed{37}$.

AIME15 just proved that $ 0.\overline{ab} \equal{} \frac{ab}{99}$.

This works for everything:

If you have $ x$ digits repeated just like the above, in fractional form, it is the number repeated (in $ x$ digits) over $ x$ $ 9$'s.


[quote="xRobynx"]What are the next two terms in this sequence : 594, 487, 566, 493, 310, 447? Please help me. let me know how it was worked out or how to work it out. Your help would be much appreciated. Ta x  :D[/quote]

 :wacko: This thing has a pattern? It's so crazy. I don't see the pattern. Sorry.

What do you need it for?  Often, context is necessary to answer something of this sort.

actually I checked that too but i didn't post :)

just google for "sequences" or "integer sequences"

[quote="lightrhee"][quote="Danbert"]Odd...  No matches in the Online Encylopedia of Integer Sequences...[/quote]
Wow! :o  Such thing exists! I didn't know that. Where is it?[/quote]

[url=http://www.learnstuff.com/integer-sequence-resources/]Direct Link[/url]

 :lol:

[quote="Danbert"][quote="lightrhee"][quote="Danbert"]Odd...  No matches in the Online Encylopedia of Integer Sequences...[/quote]
Wow! :o  Such thing exists! I didn't know that. Where is it?[/quote]

[url=http://www.research.att.com/~njas/sequences/]Direct Link[/url]

 :lol:[/quote]
This is so cool! :o My sequence from puzzle/brainteaser marathon is also on it.

where did you get this sequence? internet? book?

[quote="Danbert"][quote="lightrhee"][quote="Danbert"]Odd...  No matches in the Online Encylopedia of Integer Sequences...[/quote]
Wow! :o  Such thing exists! I didn't know that. Where is it?[/quote]

[url=http://www.research.att.com/~njas/sequences/]Direct Link[/url]

 :lol:[/quote]

  Awesome! Now my posts will increase from 500- 5000!!!!!!!!!!!


I think this is proven sometimes on the left hand side thingy.
Their prove is pretty good.

[quote="shadysaysurspammed"]P.t:$(1+2+3+...+n)^2=1^3+2^3+....+n^3$

No Induction allowed.[/quote]

They are see it in here:[b]toantuoitho.nxbgd.com.vn[/b]
This is a web in Vietnam! and it  have problem


i am just curious. yes, i mean chapter, not states. our chapter needed a 38
EDIT: something happened to the poll. just count 31-34 as 34 or less.

There is no poll.

I wish I was in your chapter, then I could go to countdown. Anyways, ours was like a 43 if I remember correctly.....or a 42.

In our chapter, I think 41 was the cutoff, although you would have to beat some other 41's in tiebreaks. I believe scores were 46, 45, 45, 44, 43, and some 42/41's.

i had a 32. and i was in second BEFORE the countdown round. :rotfl:

[quote="Yongyi781"]My coach wouldn't show me the results, but all I know is the first place person in my chapter got a 43...

... and that first place person was me! :first:[/quote]

Just goes to show that this chapter test was not a good indicator, since Yongyi was 3rd in the state, but would not have made countdown in another chapter in the same state!!  That's nuts.  Kids in my chapter must just be really good at not making silly mistakes.  Interestingly this means our nats team is made up of 4 kids that either finished 1st or tied for first at Chapter.....so maybe it wasn't that bad an indicator :lol:

I got 46.
2nd place 42
I estimate about mid 30's.

well at my state, The top 4 were

1. 1st at his chapter
2. (me) 1st at my chapter
3. 1st at his chapter
4. 1st at his chapter

So I think that it is a pretty good indicator of how well you are gonna do at states.

Unless I counted something wrong on the stats sheet, the cutoff at my chapter was a 30.
1st place was a 40.

I think the cutoff is around 25-30 ?

yeah i guess so


$ a_1, a_2, ..., a_n$ are positive numbers such that their sum is one. Find the minimum of:
$ a_1/(1 \plus{} a_2 \plus{} ... \plus{} n) \plus{} a_2/(1 \plus{} a_1 \plus{} a_3 \plus{}...\plus{}a_n) \plus{}... \plus{} a_n/(1 \plus{} a_1 \plus{} ... \plus{} a_{n \minus{} 1})$(and please prove it!).

[quote="bengal"]$ a_1, a_2, ..., a_n$ are positive numbers such that their sum is one. Find the minimum of:
$ a_1/(1 \plus{} a_2 \plus{} ... \plus{} n) \plus{} a_2/(1 \plus{} a_1 \plus{} a_3 \plus{} ... \plus{} a_n) \plus{} ... \plus{} a_n/(1 \plus{} a_1 \plus{} ... \plus{} a_{n \minus{} 1})$(and please prove it!).[/quote]

[hide="Solution"]
Assuming you meant to have $ 1\plus{}a_2\plus{}\cdots\plus{}a_n$ in the denominator of the first term,
Let $ S\equal{}\dfrac{a_1}{1\plus{}a_2\plus{}\cdots\plus{}a_n}\plus{}\dfrac{a_2}{1\plus{}a_1\plus{}\cdots\plus{}a_n}\plus{}\cdots \plus{}\dfrac{a_n}{1\plus{}a_1\plus{}\cdots \plus{}a_{n\minus{}1}}$.
We have that $ a_1\plus{}a_2\plus{}\cdots\plus{}a_n\equal{}1$,
Thus we can rewrite the original expression as,
\[ S\equal{}\sum_{i\equal{}1}^{n}{\dfrac{a_i}{2\minus{}a_i}}\]
We can then add one to each term then subtract $ n$ to get,
\[ S\equal{}\minus{}n\plus{}\sum_{i\equal{}1}^{n}{\dfrac{2}{2\minus{}a_1}}\]
Take out a factor of $ 2$ from the sum,
\[ S\equal{}\minus{}n\plus{}2\left(\sum_{i\equal{}1}^{n}{\dfrac{1}{2\minus{}a_1}}\right)\]
Use Cauchy-Schwarz to show that,
\[ (2n\minus{}1)\left(\sum_{i\equal{}1}^{n}{\dfrac{1}{2\minus{}a_1}}\right)\ge n^2\implies \sum_{i\equal{}1}^{n}{\dfrac{1}{2\minus{}a_1}}\ge \dfrac{n^2}{2n\minus{}1}\]
Hence,
\[ S\equal{}\minus{}n\plus{}2\left(\sum_{i\equal{}1}^{n}{\dfrac{1}{2\minus{}a_1}}\right)\ge \minus{}n\plus{}2\left(\dfrac{n^2}{2n\minus{}1}\right)\equal{}\dfrac{2n^2}{2n\minus{}1}\minus{}n\equal{}\boxed{\dfrac{n}{2n\minus{}1}}\]
And that's our answer.
Equality occurs when $ a_1\equal{}a_2\equal{}\cdots\equal{}a_n\equal{}\frac{1}{n}$
[/hide]

Let S be the value we want.
We use the Cauchy schuarz inequality and we get:
(S)(n-1)>=n^2
S=n^2/n-1

What I did was:

$ (2(a_1 \plus{} \cdots \plus{} a_{n}) \minus{} (a_{1}^2 \plus{} \cdots \plus{} a_{n}^2))\frac{a_{1}}{2 \minus{} {a_1}}\plus{}\frac{a_{2}}{2 \minus{} a_2}\plus{}\cdots\plus{}\frac{a_{n}}{2 \minus{} a_{n}} \geq (a_1 \plus{} \cdots \plus{} a_n)^2$
$ J \equal{} a_{1}^2 \plus{} \cdots \plus{}a_{n}^2$
$ (2 \minus{} J)S \geq 1$
$ 2 \minus{} 1/S \geq J \geq 1/n$.
Rearranging gives the desired result.


$ (n \in N*)$  $ f(x) \equal{} \sqrt{\frac{x^{2n\plus{}1} }{x\minus{}1}}\minus{}2x$
find $ \lim_{x\rightarrow \plus{}\infty}f(x)$

let  $ g(x)\equal{}\sqrt{x\plus{}\sqrt{x\plus{}\sqrt{x}}}\minus{}\sqrt{x}$
find  $ \lim_{x\rightarrow \plus{}\infty}g(x)$

[hide="Number 2"]$ \lim \left(\sqrt{x \plus{} \sqrt{x \plus{} \sqrt{x}}} \minus{} \sqrt{x}\right) \equal{} \lim \frac{\left(\sqrt{x \plus{} \sqrt{x \plus{} \sqrt{x}}} \minus{} \sqrt{x}\right)\left(\sqrt{x \plus{} \sqrt{x \plus{} \sqrt{x}}} \plus{} \sqrt{x}\right)}{\left(\sqrt{x \plus{} \sqrt{x \plus{} \sqrt{x}}} \plus{} \sqrt{x}\right)}$

$ \equal{} \lim \frac{x \plus{} \sqrt{x \plus{} \sqrt{x}} \minus{} x}{\sqrt{x \plus{} \sqrt{x \plus{} \sqrt{x}}} \plus{} \sqrt{x}} \equal{} \lim \frac{\sqrt{x}\sqrt{1 \plus{} \frac{1}{\sqrt{x}}}}{\sqrt{x}\left(1 \plus{} \sqrt{1 \plus{} \frac{1}{\sqrt{x}}\sqrt{1 \plus{} \frac{1}{\sqrt{x}}}}\right)}$

$ \equal{} \frac{\sqrt{1 \plus{} 0}}{1 \plus{} \sqrt{1 \plus{} 0}} \equal{} \frac{1}{2}$[/hide]

[hide="Number 1"]$ \lim \left(\sqrt{\frac{x^{2n\plus{}1}}{x\minus{}1}} \minus{} 2x\right) \equal{} \lim \left(\sqrt{\frac{x^{2n}}{1 \minus{} \frac{1}{x}}} \minus{} 2x\right) \equal{} \lim \left[x^n \left(1 \minus{} \frac{1}{x}\right)^{\minus{}1/2} \minus{} 2x\right]$

$ \equal{} \lim\,\, x \cdot \left[x^{n\minus{}1}\left(1 \plus{} \frac{1}{2x}\right) \minus{} 2\right] \equal{} \infty$[/hide]


percent
search
combinatorics unsolved
combinatorics

A group of $ 100$ students numbered $ 1$ through $ 100$ are playing the following game. The judge writes the numbers $ 1$, $ 2$, $ \ldots$, $ 100$ on $ 100$ cards, places them on the table in an arbitrary order and turns them over. The students $ 1$ to $ 100$ enter the room one by one, and each of them flips $ 50$ of the cards. If among the cards flipped by student $ j$ there is card $ j$, he gains one point. The flipped cards are then turned over again. The students cannot communicate during the game nor can they see the cards flipped by other students. The group wins the game if each student gains a point. Is there a strategy giving the group more than $ 1$ percent of chance to win?

This was [url=http://www.mathlinks.ro/viewtopic.php?search_id=975696286&t=175877]already posted[/url], more than [url=http://www.mathlinks.ro/viewtopic.php?search_id=975696286&t=191186]once[/url] in fact.


calculus
integration
absolute value
real analysis
real analysis unsolved

Elvaluate, \[ \int_{\minus{}\infty}^{\infty}\dfrac{1}{a^3}\exp\bigg(\minus{}\dfrac{y^2}{a^2}\bigg)dy\] where $ a\equal{}\sqrt{1\minus{}\dfrac{x^2}{y^2}}$

Are we to assume that $ x$ is fixed? How should we understand the square root when the expression under it is negative? Finally, what to do with a terrible (exponential) singularity as $ y\to x\minus{}$. Or, perhaps, you just forgot the absolute value in the definition of $ a$? :?


tamame tavabe f,g,h:R-->R ra biabid ke:

[b]f( h(g(x))+y )+g( z+f(y) )=h(y)+x+g( y+f(z) )[/b]

find all the functions $ f,g,h: \mathbb{R}\rightarrow \mathbb{R}$ s.t:
$ f(h(g(x))+y)+g(z+f(y))=h(y)+x+g(y+f(z))$


algebra polynomial function trigonometry factorial algorithm Vieta

can we turn a non polynomial into a polynomial ?i mean can we change an equation or an identity that is not polynomial to a polynomial ?

$f(x)=e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots+\frac{x^{n}}{n!}+\cdots ; ~n=0,1,\ldots$

So yes, I suppose if you consider a series a polynomial.

Expanding on what skimnc said: look up Taylor polynomials. Any differentiable function can be estimated to some extent as a polynomial, and in some cases you can prove that a continuing Taylor series (like the one above for e, and also the ones for sine and cosine) converge to the actual function with an infinite number of terms.

As a brief intro. For a differentiable function F(x), the nth order Taylor approximation centered at the value x=a is:

$F(x)\approx{F(a)+F'(a)(x-a)+\frac{F''(a)}{2!}(x-a)^{2}+\frac{F'''(a)}{3!}(x-a)^{3}}\dots$

$=\sum_{k=0}^{n}{\frac{F^{\text{kth deriv}}(a)}{k!}(x-a)^{k}}$

[quote="binomial_4eva"]can we change an equation or an identity that is not polynomial to a polynomial ?[/quote]

I'm not really sure what you mean.  Are you talking about identities such as

$\sin^{2}x+\cos^{2}x = 1$

?  In what meaningful way could you hope to change this into a "polynomial" identity?  (Aside from the Taylor series expansions, which don't really say much.)

i mean can we turn this $\frac{(n-1)!}{(n-b)!}+\frac{(n-2)!}{(n-b)!}+\frac{(n-3)!}{(n-b)!}+...+\frac{(n-i)!}{(n-b)!}+...+1$ i dont know what you call this, and identity or an equation or expression, but can we turn that to a polynomial ?
or a series, like the one skimnc said for $e^{x}$

An [i]identity[/i] and an [i]equation[/i] require that you have two separate [i]expressions[/i] which you identify or equate.  What you have there is an expression.

This is an example of "why it's important to actually let people know what you're talking about" -- expressing the thing you've given us as a polynomial will use special properties of the functions involved; it won't be some general process for magically making things into polynomials (and I'm not sure why you would expect it to be).

Finally, this post will actually require you to take out a piece of paper and pencil, write things down, and think about them.

Let's just deal with a single term of your expression.  We're going to want to treat $b$ as a fixed constant (let's say, a positive integer bigger than 1) and $n$ as the variable.  Then we have, for example $\frac{(n-1)!}{(n-b)!}$.  What does this [i]mean[/i]?  Unpack the factorial notation and you should see that many things (including the entire denominator) cancel in a very nice way.  What's left is a product of some number of terms, each involving our variable $n$.  Now, here's something to be careful of: $n!$ is a product of terms involving a variable $n$, but as $n$ grows, the number of terms in the product increases -- this ensures that we aren't dealing directly with a polynomial.  Do we have this problem in our case?

When you've finished with one term, hopefully you can see that the same method can be applied to all the terms; since the sum of polynomials is a polynomial, we're done.  You should also be able to see what the degree of the polynomial is.  Unfortunately, in general, this polynomial is not easy to write down, i.e. it doesn't have a simple, general form.  For a particular choice of $b$, we are now able to figure out what the polynomial is and express it in a conventional fashion, but it hasn't given us a way of expressing it in general without going through this process.  (We have an [i]algorithm[/i], but not a [i]formula[/i].)  Looking at the result for the first few values of $b$, I don't notice any really nice patterns.

JBL thanks i get wat u mean, its hard i know to find a poly for this, but i seriously need a polynomial for this.

[quote]We're going to want to treat $b$ as a fixed constant (let's say, a positive integer bigger than 1) and $n$ as the variable.[/quote]
let me add something , $n$ and $b$ are both fixed constant, only the $i$ changes, $i$ is the variable. and ya lets just consider this term, $\frac{(n-i)!}{(n-b)!}$

[quote="binomial_4eva"]JBL thanks i get wat u mean, its hard i know to find a poly for this, but i seriously need a polynomial for this.[/quote]

Why?  It would be easier if you just told us what you were trying to do or what problem you're trying to solve.

[quote="binomial_4eva"]only the $i$ changes, $i$ is the variable. [/quote]

But increasing $i$ just adds terms - it doesn't change the terms before it.  It's a little like asking for a polynomial representation of $\sum_{k=1}^{i}k \cdot k!$.  There is a simplified form, yes, but it's not a polynomial in $i$.

You should go re-read my post.  See the line where I wrote [quote="JBL"]this post will actually require you to take out a piece of paper and pencil, write things down, and think about them.[/quote] These are the important steps you have neglected to do.  Sit down and actually write down and think about what I wrote -- right now, you just sound pathetic.  I've given you a complete recipe for what to do, [i]but you actually need to do it[/i].

[quote="binomial_4eva"]let me add something , $n$ and $b$ are both fixed constant, only the $i$ changes, $i$ is the variable.[/quote]

No.  $i$ is a "dummy variable" -- it will not appear in the final expression.  The expression you are looking for is a polynomial in the variable $n$ with degree that depends on $b$.

I explained to you what to do for $\frac{(n-1)!}{(n-b)!}$, but this explanation wasn't specific to $n-1$ -- as I said, you need to do exactly the same thing for each other term (for each different value of $i$).

Up until you wrote [quote="binomial_4eva"]$(n-1)(n-2)\cdots\!(n-i_{b})= n^{b}-3n+2$[/quote] you were doing fine.  (Although I'm not sure what this "$i_{b}$" thing is -- it should just be $b-1$.)  The part I quoted is just wrong.  Of course, you already have it as a polynomial (as I noted before, a polynomial in the variable $n$) -- it just happens that your polynomial is in factored form.  It's easy to say what the coefficients are, since we have the roots -- they are the appropriate symmetric sums of a set of consecutive positive integers with a power of $-1$ tossed in (this is what Vieta's says) -- but there aren't nice closed forms for these expressions in general.  Of course, you now have all the tools to write out your polynomial for any [i]fixed[/i] value of $b$.

One other error: in the general term of your sum in the last equation you've written, you have $(n-i)(n-(i+1))$ -- it should be (using your notation) $(n-i)(n-(i+1))\cdots(n-i_{b})$.

thanks JBL, $i_{b}$ was $b-1$ but neway i edited it.
$(n-1)(n-2)\cdots\!(n-(b-1))=n^{b}-(1+2+\cdots\!+(b-1))n^{b-1}+(2+3+\cdots\!+(b-1)b)n^{b-2}+\cdots\!+(-1)^{b}(b-1)!$
what will these look like if we expand out? $(n-2)(n-3)\cdots\!(n-(b-1))$ and $(n-3)(n-4)\cdots\!(n-(b-1))$ and $(n-4)(n-5)\cdots\!(n-(b-1))$
what is the formula for symmetric sums of a set of consecutive positive integers?

You still haven't explained why you need a polynomial.


Let $ x,y,z$ be non-negative real numbers. Prove that
\[ (8x^{2}+yz)(8y^{2}+zx)(8z^{2}+xy)\leq (x+y+z)^{6}.\]

[quote="pvthuan"]Let $ x,y,z$ be non-negative real numbers. Prove that
\[ (8x^{2}+yz)(8y^{2}+zx)(8z^{2}+xy)\leq (x+y+z)^{6}. \]
[/quote]
Expanding kills it by Schur and Muirhead.

yeah, but solution will be simpler if you use substitution: $ p=a+b+c,q=ab+bc+ca,r=abc$ and assume that $ p=a+b+c=3$ ...  :roll:

[quote="Vasc"]I found this inequality in my archive, with the solution

$ E(x,y,z \le E(x,a,a)\le E(b,b,b)$,

where E(x,y,z) is the left side, $ a=\frac{y+z}{2}$ and $ b=\frac{x+y+z}{3}$, on the assumption $ x\le y\le z$.
Also, I found the related inequalities:
\[ (2x^{2}+7yz)(2y^{2}+7zx)(2z^{2}+7xy)\geq 27(xy+yz+zx)^{3}\]
and
\[ (5x^{2}+yz)(5y^{2}+zx)(5z^{2}+xy)\geq 8(xy+yz+zx)^{3}. \]
[/quote]
they are very nice problems, vasc and pvthuan  :wink:
I think all of them can be solve by a technique. I used to elementary expressions to solve them.
I will post one of them, after that, you can solve remain inequalities easily.
We put $ p=a+b+c,q=ab+bc+ca,r=abc$. We can assume that $ q=ab+bc+ca=3$
By expanding expression, we have
\[ (2x^{2}+7yz)(2y^{2}+7zx)(2z^{2}+7xy)\geq 27(xy+yz+zx)^{3}\]

\[ \iff 729r^{2}+27+98r(p^{3}-12p)+42pr\ge 0 \]

\[ \iff 729r^{2}+27-882r^{2}+126r^{2}\ge 0 \]

\[ \iff 1\ge r \]
which is clearly true, because:
\[ p=3\ge 3\sqrt[3]{r^{2}}\to r\le 1 \]
On the other hand, by Schur's inequality:
\[ p^{3}-12p\ge-9r \]

\[ pr\ge 3r^{2}\]
We are done

[quote="zaizai-hoang"][quote="Vasc"].....Also, I found the related inequalities:
\[ (2x^{2}\plus{}7yz)(2y^{2}\plus{}7zx)(2z^{2}\plus{}7xy)\geq 27(xy\plus{}yz\plus{}zx)^{3}\]
and
\[ (5x^{2}\plus{}yz)(5y^{2}\plus{}zx)(5z^{2}\plus{}xy)\geq 8(xy\plus{}yz\plus{}zx)^{3}.\]
[/quote]
they are very nice problems, vasc and pvthuan  :wink:
I think all of them can be solve by a technique. I used to elementary expressions to solve them.
I will post one of them, after that, you can solve remain inequalities easily.
We put $ p \equal{} a\plus{}b\plus{}c,q \equal{} ab\plus{}bc\plus{}ca,r \equal{} abc$. We can assume that $ q \equal{} ab\plus{}bc\plus{}ca \equal{} 3$
By expanding expression, we have
\[ (2x^{2}\plus{}7yz)(2y^{2}\plus{}7zx)(2z^{2}\plus{}7xy)\geq 27(xy\plus{}yz\plus{}zx)^{3}\]

\[ \iff 729r^{2}\plus{}27\plus{}98r(p^{3}\minus{}12p)\plus{}42pr\ge 0\]

\[ \iff 729r^{2}\plus{}27\minus{}882r^{2}\plus{}126r^{2}\ge 0\]

\[ \iff 1\ge r\]
which is clearly true, because:
\[ p\equal{}3\ge 3\sqrt[3]{r^{2}}\to r\le 1\]
On the other hand, by Schur's inequality:
\[ p^{3}\minus{}12p\ge\minus{}9r\]

\[ pr\ge 3r^{2}\]
We are done[/quote]

Dear ZaiZai, for the second inequality of Vasc

 \[ (5x^{2}\plus{}yz)(5y^{2}\plus{}zx)(5z^{2}\plus{}xy)\geq 8(xy\plus{}yz\plus{}zx)^{3}.\]

I found using your method

$ 216r^{2}\plus{}459\minus{}270pr\plus{}5p^{3}r\geq 0$

with $ p\equal{}a\plus{}b\plus{}c$, $ q\equal{}ab\plus{}bc\plus{}ca\equal{}3$ and $ r\equal{}abc$

I tried to complete it using SHUR applied to $ ab,bc,ca$ and $ a,b,c$, that is

$ 27\plus{}9r^{2}\geq 12pr$

$ p^{3}\plus{}9r\geq 12p$

but I failed :blush: Can you, please, help me to complete it?

Thank you very much.


calculus
integration
real analysis
calculus computations

I need help on this Calculus problem for my Calc AB class. I just started work problems today so I am not sure what to do.

A cable that weighs 2 lb/ft is used to lift 800 lb of coal up a mineshaft 500 ft deep.
-Show how to approximate the required work by a Reimann Sum
-Then, express the work as an integral and evaluate it.

Thanks.

Suppose you lift the bucket from height $x$ above the bottom of the shaft to height $x+\Delta x.$ Work equals force times distance. The distance is $\Delta x.$ What is the force? The force would be the weight of the bucket plus the weight of the cable. The bucket weighs 800 lbs. The hanging length of the cable is changing but when the bucket height is $x,$ we have $500-x$ feet of cable weighing $1000-2x$ pounds. Thus, at least for a moment, the weight (and hence the force) is $800+1000-2x=1800-2x$ pounds. Multiply by the distance, and the force times distance is approximately (approximately because the force is changing) $(1800-2x)\,\Delta x.$

Sum over all the increments of distance over sum partition and we get that

Work $\approx\sum(1800-2x)\,\Delta x$

That's a Riemann sum; taking the limit at the distance increments (the mesh of the partition) go to zero, we get that

Work $=\int_{0}^{500}(1800-2x)\,dx.$

This is not the only valid way to set it up; we could also look at a section of cable and ask how far that section has to go. In that method, the coal is handled separately.


quadratics
number theory proposed
number theory

Show that the number of $2$-tuples $(n,n+1)$, twin residues, with both $n$ and $n+1$ being quadratic residues modulo $p$ is exactly

\[\frac{p - {-1 \overwithdelims () p}}{4} - 1.\]

The leading ideas can be found here:

[url]http://www.mathlinks.ro/viewtopic.php?t=4891[/url]

Yes I saw that thread and decided to post this in hope of gaining some new proofs.

My proof goes like this, we consider the sum of $1 + \left(\frac{n}{p}\right) + \left(\frac{n+1}{p}\right) + \left(\frac{n(n+1)}{p}\right)$. Then obviously that is $4$ times what we are searching for and the last term (or sum) can be evaluated easily by noting that $\left(\frac{n(n+1)}{p}\right) = \left(\frac{n^{-1} + 1}{p}\right)$.

(Okay, after a closer read even this was more or less what grobber wrote earlier)

This problem was already discussed and solved. It is in the unsolved section and its name is: at most p+1/4 consecutive cuadratic residues.......I dont remember who solved it, but he used all the hints given in the book (the presceding problems proposed there)


Ascertain :

$1\blacktriangleright\min_{x\in \mathrm R}\ (x+1)(x+2)(x+3)(x+4)\ .$

$2\blacktriangleright \{\lambda\in\mathrm R\ |\ \mathrm{the\ equation\ }(x+1)(x+2)(x+3)(x+4)=\lambda\ \mathrm{has\ at\ least\ one\ real\ root}\}\ .$

[quote="Virgil Nicula"]Ascertain :

$1\blacktriangleright\min_{x\in \mathrm R}\ (x+1)(x+2)(x+3)(x+4)\ .$

[hide="Answer"]min=-1\ $\left(x=\frac{-5\pm \sqrt{5}}{2}\right).$[/hide]

$2\blacktriangleright \{\lambda\in\mathrm R\ |\ \mathrm{the\ equation\ }(x+1)(x+2)(x+3)(x+4)=\lambda\ \mathrm{has\ at\ least\ one\ real\ root}\}\ .$

[color=red]Good Practice[/color]

The number of real roots are folllows.

[hide="Answer"]
$\lambda <-1\ \cdots 0.$
$\lambda =-1\ \cdots 1.$
$-1<\lambda \leq \frac{9}{16}\ \cdots 2.$
$\lambda >\frac{9}{16}\ \cdots 1$[/hide]


I'm sure it isn't 4, but... is the answer rational?

Maybe I'm just overestimating the problem.

[hide]$(a+b+c)^2=a^2+b^2+c^2 + 2(ab +ac+bc)  \implies (ab +ac+bc)=-\frac{1}{2}$

$(a+b+c)^3=a^3 + b^3 + c^3 + 3a(ab + ac) + 3b(ab + bc) + 3c(ac+bc) + 6abc$, consequently:

$(a+b+c)^3=a^3 + b^3 + c^3 + 3a(-\frac{1}{2} - bc) + 3b(-\frac{1}{2} - ac) + 3c(-\frac{1}{2} - ab) + 6abc$

$ \implies (a+b+c)^3=a^3 + b^3 + c^3 - 3\frac{1}{2}(a+b+c) - 3abc \implies abc=\frac{1}{6}$

now let's find $a^4 + b^4 + c^4$:

$3 \cdot 1=(a^3 + b^3 + c^3)(a+b+c)=a^4 + b^4 + c^4 + a^2(ab + ac) + b^2(ab+bc) + c^2(ac + bc)$ factoring analogously:

$3=a^4 + b^4 + c^4 -\frac{1}{2}(a^2+b^2+c^2) - (a + b + c)(abc)$

$\implies a^4 + b^4 + c^4=3 + 1 + \frac{1}{6}=\frac{25}{6}$[/hide]

[color=green][size=75]mod edit: use \cdot for multiplication, not "." and use \implies, not :arrow: ;)

oh, and hide your solutions :P [/size][/color]

Oh crap,i found it till abc=1/6 and then got lost............................
I cant understand whaatya did after that newmember,factorisin analougously? :?

[quote="solafidefarms"]Could ya'll please not use c***? It's not a nice word. Maybe in the title instead of c****y  use ambagious, meaning difficult, complicated? It's a much more impressive word then c***.[/quote]

Request granted :D

But still, it's A-M-B-I-G-U-O-U-S, not A-M-B-A-G-I-O-U-S. Sorry about pointing out the subtlety, but it would be more impressive if it were spelled correctly...... ;) :D

It would be more impressive if you had a dictionary... http://www.google.com/search?hl=en&lr=&q=define%3Aambagious

Never argue with a National Word Power Champion. ;)

:P

Oh, I was thinking of "ambiguous". Sorry :blush:

Now I have a new addition to my vocabulary!!! :w00t:

But back to the original problem: Is there a shorter way?

""Could ya'll please not use c***? It's not a nice word. Maybe in the title instead of c****y use ambagious, meaning difficult, complicated? It's a much more impressive word then c***.""

 lol , is crappy  sorry c*****  that bad? sorry i didnt know. :D  no probs in changin the title at all

Riddler,in ur sol,in one part,u took the value of abc to be 1/6,u didnt prove it ;) Same method here,riddler,but then i got abc diiferent than hta the new member got.i initially got abc=1/6,but on recalculatin it came different,any other method for calculatin abc=?

[quote="4everwise"][quote="Chess64"]It would be more impressive if you had a dictionary... http://www.google.com/search?hl=en&lr=&q=define%3Aambagious

Never argue with a National Word Power Champion.
 ;) [/quote]

Chess64, that's great!  :D[/quote]

Er, actually, if you didn't know, solafidefarms is the National Word Power Champion :P lol


$ a,b,c > 0$  and  $ x\geq\frac {a \plus{} b \plus{} c}{3\sqrt {3}} \minus{} 1$

prove that
\[ \frac {(b \plus{} cx)^2}{a} \plus{} \frac {(c \plus{} ax)^2}{b} \plus{} \frac {(a \plus{} bx)^2}{c}\geq{abc}
\]

Cauchy-Schwartz engel-form tells us L.H.S.$ \ge$

$ \frac {((1 \plus{} x)(a \plus{} b \plus{} c))^2}{a \plus{} b \plus{} c} \equal{} (1 \plus{} x)^2(a \plus{} b \plus{} c)\ge (\frac {a \plus{} b \plus{} c}{3\sqrt {3}})^2(a \plus{} b \plus{} c) \equal{} (\frac {a \plus{} b \plus{} c}{3})^3\ge abc$


[quote="shaaam"]if $ a,b \geq 0$, from Cauchy-Schwarz, we know that:

$ \frac {1}{a^2 \plus{} ab} \plus{} \frac {1}{b^2 \plus{} ab} \geq \frac {4}{(a \plus{} b)^2}$

We have the stronger inequality:

$ \frac {1}{a\sqrt {2(a^2 \plus{} b^2)}} \plus{} \frac {1}{b\sqrt {2(a^2 \plus{} b^2)}} \geq \frac {4}{(a \plus{} b)^2}$

Prove it.

 :)[/quote]
It is equivalent to
$ (a\plus{}b)^3 \ge 4ab\sqrt{2(a^2\plus{}b^2)}$
On the other hand: $ 4ab\sqrt{2(a^2\plus{}b^2)} \equal{}4\sqrt{ab}.\sqrt{a^2\plus{}b^2}.\sqrt{2ab} \le 4.\frac{a\plus{}b}{2}.\frac{(a\plus{}b)^2}{2} \equal{}(a\plus{}b)^3$
Then, we have Q.E.D   :wink:  :)

[quote="shaaam"]if $ a,b \geq 0$, from Cauchy-Schwarz, we know that:

$ \frac {1}{a^2 \plus{} ab} \plus{} \frac {1}{b^2 \plus{} ab} \geq \frac {4}{(a \plus{} b)^2}$

We have the stronger inequality:

$ \frac {1}{a\sqrt {2(a^2 \plus{} b^2)}} \plus{} \frac {1}{b\sqrt {2(a^2 \plus{} b^2)}} \geq \frac {4}{(a \plus{} b)^2}$

Prove it.

 :)[/quote]
The following inequality is true too.
\[ \frac {1}{a\sqrt [3] {4(a^3 \plus{} b^3)}} \plus{} \frac {1}{b\sqrt [3]{4(a^3 \plus{} b^3)}} \geq \frac {4}{(a \plus{} b)^2}
\]
for positive $ a$ and $ b.$

Maybe the following similar question is interesting.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}3.$
Prove that if the inequality
\[ \frac{1}{a^2\plus{}2bc}\plus{}\frac{1}{b^2\plus{}2ac}\plus{}\frac{1}{c^2\plus{}2ab}\geq\left(\frac{a^2\plus{}b^2\plus{}c^2}{3}\right)^k\]
is true then, $ k\leq0.$

[quote=jasperE3][quote=arqady]Maybe the following similar question is interesting.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}3.$
Prove that if the inequality
\[ \frac{1}{a^2\plus{}2bc}\plus{}\frac{1}{b^2\plus{}2ac}\plus{}\frac{1}{c^2\plus{}2ab}\geq\left(\frac{a^2\plus{}b^2\plus{}c^2}{3}\right)^k\]
is true then, $ k\leq0.$[/quote]

I don't think it's true. We can have $a=1,b=0.5,c=1.5$ and $k=0.05$.[/quote]
Id est, $0.05\leq0$ :maybe:

[quote=arqady]
The following inequality is true too.
\[ \frac {1}{a\sqrt [3] {4(a^3 \plus{} b^3)}} \plus{} \frac {1}{b\sqrt [3]{4(a^3 \plus{} b^3)}} \geq \frac {4}{(a \plus{} b)^2}
\]
for positive $ a$ and $ b.$[/quote]
Splendor. Set WLOG $a^3+b^3=2.$ So if $a+b:=2s,$ then $s\in\left(\frac1{\sqrt[3]4},1\right]$ and $ab=\frac{4s^3-1}{3s}.$
We need to prove
$$\frac s{ab}\geq\frac1{s^2}\iff3s^4+1\geq4s^3\iff s^4+s^4+s^4+1\geq4\sqrt[4]{s^4\cdot s^4\cdot s^4\cdot1}=4s^3.$$


inequalities
AMC
USA(J)MO
USAMO
number theory
prime factorization

Let $ a$ and $ b$ be positive integers such that $ a|b^2,b^2|a^3,a^3|b^4,b^4|a^5,...$ Prove that $ a\equal{}b$.

[hide="my sketch proof (try the problem yourself first!)"]

If k is odd and $ a^k|b^{k\plus{}1},a^{k\plus{}2}|b^{k\plus{}3}$ then $ a^k*a^{k\plus{}2}\equal{}a^{2k\plus{}2}|b^{2k\plus{}4}$
$ \implies a^{k\plus{}1}|b^{k\plus{}2}$.

Similarly if m is even and $ b^m|a^{m\plus{}1},b^{m\plus{}2}|a^{m\plus{}3}$ then we have $ b^{m\plus{}1}|a^{m\plus{}2}$

So the problem statement is equivalent to $ a|b^2, a^2|b^3,a^3|b^4, ...$
We shall prove that $ a|b$ by contradiction:
Assume a does not divide b, then there's a prime factor $ p$ in $ a$ such that its power $ A$ in the prime factorization is higher than that of b, denoted by $ B$. Then from $ a|b^2$ we required $ 2B\ge A$, and from $ a^2|b^3$ we require $ 3B\ge2A\iff 2B\ge A\plus{}(A\minus{}B)$. In the same manner we require $ 2B\ge A\plus{}2(A\minus{}B)$ and $ 2B\ge A\plus{}x(A\minus{}B)$ where $ A\minus{}B>0$ Therefore we can choose a sufficiently large $ x$ such that the inequality does not hold. Contradiction.

Likewise we can prove $ b|a$ in the same manner as above.

So is this proof rigorous enough? Are we allowed to omit the parts of proof that are very similar and just write the proof is similar in a proof-writing contest (ex. usamo)? Also is there an easier way to prove $ a|b$ and $ b|a$ as I feel my proof is quite long-winded...[/hide]

Instead of saying "in the same matter," you can just say "from $ a^k | b^{k\plus{}1}$ we require..."

Anyway, there's a [hide="shorter proof."] $ x | y$ implies that $ x \le y$.  [/hide]

[hide="Blah blah blah"]
The conditions $ a|b^2,a^3|b^4,\dots,a^{2k \minus{} 1}|b^{2k}$ imply that $ \frac {b^{2k}}{a^{2k \minus{} 1}} \equal{} b\left(\frac ba\right)^{2k \minus{} 1}\in \mathbb Z$ for all $ k$, therefore $ a|b$.

The conditions $ b^2|a^3,b^4|a^5,\dots,b^{2k}|a^{2k \plus{} 1}$ imply that $ \frac {a^{2k \plus{} 1}}{b^{2k}} \equal{} a\left(\frac ab\right)^{2k}\in \mathbb Z$ for all $ k$, therefore $ b|a$.

If $ a|b$ and $ b|a$, we have $ a \equal{} b$.
[/hide]

Yes $ x|y$ implies $ x\leq y$ but I can not see from here why $ x\equal{}y$  :blush:

[quote="enndb0x"]Yes $ x|y$ implies $ x\leq y$ but I can not see from here why $ x \equal{} y$  :blush:[/quote]
[hide="Solution"] $ a^k | b^{k\plus{}1}$ implies that $ a \le b^{1 \plus{} \frac{1}{k} }$ for every positive integer $ k$.  For sufficiently large $ k$, this implies $ a \le b$.  Similarly, $ b^k | a^{k\plus{}1}$ for sufficiently large $ k$ implies $ b \le a$. [/hide]

Lets look at the valuation mod $ p$ (is there a better way to phrase this...I've just picked up the term here). Say $ p^x\parallel{}a$ and $ p^y\parallel{}b$

So the terms $ a|b^2$, $ a^3|b^4$, ... imply that $ (2k\minus{}1)x\le (2k)y$ so $ \frac{x}{y}\le 1\plus{}\frac{1}{2k\minus{}1}$ for all $ k\in\mathbb{N}$  (1).

If $ \frac{x}{y}>1$, then choose the first natural number $ k_0$ such that $ 2k_0\minus{}1>\frac{1}{\frac{x}{y}\minus{}1}>0$. Rearranging gives a contradiction to (1).

Therefore $ \frac{x}{y}\le 1$. Using a similar argument on the $ b^2|a^3$, $ b^4|a^5$ terms shows that $ \frac{y}{x}\le 1$. These two relations imply that $ x\equal{}y$.

If either $ a$ or $ b$ have no prime factors, easily $ a\equal{}b\equal{}1$. If $ a$ or $ b$ do have prime factors, they have the same amount of each prime factor so $ a\equal{}b$ are equal (note the only unit in $ \mathbb{N}$ is $ 1$...).


ABCD is a square of side 10 units. Arc of circles with centres at ABCD are drawn in.

find the area of the central region bounded by the four arcs.

Let the vertices of the shape be labeled $ P,Q,R,S$. The shape BPC can be found by doing algebra on the arcs centered at $ A$ and $ D$.

This is also equal to the area of CQD, and since CPQ is just a  square minus a fourth of a circle, we can determine through subtraction the area of CPQ.

This also equals to the area of ARS, and since we can easily figure out CPSARQ, since it it just 2 quarter circles minus a square, we can then figure out the area of the region bounded by the four arcs through subtraction.

[asy]pair A=(0,0), B=(1,0), C=(1,1), D=(0,1);
path P=arc(A,1,0,90), Q=arc(B,1,90,180), R=arc(C,1,180,270), S=arc(D,1,270,360);
path E=intersectionpoint(P,S);
fill(buildcycle(A--B,P,S),lightred);
fill(buildcycle(S,E--A),lightgreen);
fill(A--E--D--cycle,lightblue);
draw(A--B--C--D--cycle,linewidth(1));
draw(P^^Q^^R^^S,linewidth(1));
draw(A--E^^E--D,linewidth(1));
label("A",A,SW);
label("B",B,SE);
label("C",C,NE);
label("D",D,NW);[/asy]

Red + Green : $ 10^2\pi\times\frac{30}{360}\equal{}\frac{25}{3}\pi$.

Blue + Green : $ 10^2\pi\times\frac{60}{360} \equal{} \frac{50}{3}\pi$.

Blue : $ 10^2\frac{\sqrt{3}}{4} \equal{} 25\sqrt{3}$.

Green : (Blue + Green) - Blue = $ \frac{50}{3}\pi \minus{} 25\sqrt{3}$.

Red : (Red + Green) - Green = $ \frac{25}{3}\pi \minus{} \left(\frac{50}{3}\pi \minus{} 25\sqrt{3}\right) \equal{} 25\sqrt{3} \minus{} \frac{25}{3}\pi$.

Find Area : Square - 4 Red = $ 10^2 \minus{} 4\left(25\sqrt{3} \minus{} \frac{25}{3}\pi\right) \equal{} \boxed{100\left(1\minus{}\sqrt{3}\plus{}\frac{\pi}{3}\right)}$.


What is piety?

You may want to read Euthyphro:
http://classics.mit.edu/Plato/euthyfro.html

Piety is the ability to remain in one belief.

Does anyone else want to propose a definition, or should we work off the one above?

piety, from merriam webster:

[b]1:[/b] the quality or state of being pious: as [b]a:[/b] fidelity to natural obligations (as to parents) [b]b:[/b] dutifulness in religion : devoutness
[b]2:[/b] an act inspired by piety
[b]3:[/b] a conventional belief or standard

OK I really don't understand any if those definitions. Care to translate, anyone?

Being consistent to the beliefs that are considered moral despite it being difficult, unpopular or uncommon (with others in similar positions).

Note that persistence is the same as stubbornness, the only difference is if the speaker (author) agrees with the goal being pursued with the property.

Similarly, piety is typically not used much today, but it would only be applied to beliefs that were commonly viewed by the society as moral or proper.


Suppose radius of largest circle, large subcircle = R, small subcircle = r.

R=1/2.
By pythagorean theorem:
(1-r)^2+1/4=(1/2+r)^2
r^2-2r+5/4=1/4+r+r^2
3r=1
so r=1/3, R=1/2, ratio of areas = 4/9.

However if they're talking about the largest circle O, our answer is 1/9.


In the game NetHack, dipping a long sword in a fountain has a probability $ p$ of turning it into Excalibur.  It also has a probability $ q$ of drying up the fountain.  If there are $ n$ fountains on your level, what's the probability of making Excalibur before using up your fountain supply?  Assume for simplicity that you continue to dip until one of these conditions are satisfied.

[hide]
This would be 1 minus the probability that you use up all of your fountains, which is $ q^{n}$.

Thus the answer is $ \boxed{1-q^{n}}$.[/hide]

Actually, the probability of using up a fountain, given that you try to use the fountain until something happens, is not $ q$, unless you assume that p and q are the only possibilities, that is: $ p+q=1$, which was not given. I might give some explanation of this if I get time later but for the moment I hope it's understandable without an explanation to say that the probability of using up a fountain is actually $ \frac{q}{p+q}$. The probability of getting excalibur is still 1 minus the probability of using up all the fountains, and thus the probability of getting excalibur is:

$ 1-(\frac{q}{p+q})^{n}$

Oh, I didn't see that, sorry.

While we're on this topic, are $ p$ and $ q$ mutually exclusive? Because that's another thing I assumed.

Sorry, I should have specified the problem more precisely.

Each fountain is dipped into until you either get Excalibur or the fountain dries up.  On each dip, there is a probability $ p$ of getting Excalibur, and a probability $ q$ of the fountain drying up.  These events are mutually exclusive, and $ p+q<1$.


#9 has already been asked in "Some Mandelbrot Questions" in the intermediate section.

is that necessarily a bad thing :twisted:

i think #10 is a school level mathcounts sprint question.

grr... [hide]i missed it cause i forgot to put back the square sign after i factored it[/hide]... hehehe...

i finally sent mine in, lolo, but i still took little time, i just didnt feel like doing them

and you had LOTS of people distracting you... including me... sorry if i hindered your score (which probably i didn't)

lolo, its okay if you did, but i dont care as long as i know i zam able to do it

yah... mistakes can hurt u like... woah!!!... 9 points gone right there...

yeah, but we all know you were able to do them

LoL... but we should pm this... just in case someone might...

15. Technically, by inclusion of "and" in that number, you made it 1900.80. Please correct.

huh? Shouldn't that be "one thousand nine hundred and eighty hundredth" or "one thousand nine hundred and eight tenth"?

The number he wants is written one thousand nine hundred eighty.

aalindsey100 wrote:Can one of you help me on #9 Ive tried to do it a couple times but it just doesnt click

highlight to read:[hide] Do you know the expansion x^3+3x^2+3x+1=(x+1)^3? Can you go from there?[/hide]

Yes thanks I forgot totally about doing that I think I can figure it out from there.

[quote="TripleM"]To me, 1980 is read one thousand nine hundred [b]and[/b] eighty. I've never seen leaving out the and, it doesn't sound right..
1900.80 would be read one thousand nine hundred point eight, or one thousand nine hundred and eighty hundredths or whatever. Maybe its american thingy tho.[/quote]
People generally say "and," but it technically isn't correct.

Could someone please post solutions to the one with the Fibonacci numbers and the other one: what are the rightmost three digits of 5^1993?

you know that any power of five has hte last two digits of 25...then, you can see that any even power of five ends in a 625 and an odd ends in a 125...since 1993 is odd, your answer is 125...at least that's what I got...

yeah, thats what I did too and he said it was wrong...im almost positive he said that was wrong

Also what I did, but I don't know if it was marked correct or incorrect.

anyone know how to do the fibonacci one? i wasn't really sure what he was asking...

it was asking which number from 0-9 shows up last as a unit digit

1,1,2,3,5,8,13,21,34,5,9,4,3,7,0,7,7,4,1,5,6

i decided to just use the units digits in the end. and as you can see every digit is then used. with 6 as the last one.

the answer is 125 for the 5^1993 one, i got it right...

Can anyone post just the answers at least? I'm not sure why I got a couple of them wrong though I forgot which...

pal---strange....Rep, where are you? Why did you count mine wrong?  :shock:  :twisted:


trigonometry Gauss calculus integration articles geometry parameterization

[b]2.[/b]  A spatially uniform magnetic field $ \overrightarrow{B}$ exists in the circular region S and this field is decreasing in magnitude with time at a constant rate.(See the fig)
The wooden ring $ C_1$ and the conducting rind $ C_2$are concentric with the magnetic field. The magnetic field is perpendicular to the plane of figure. Then,
(A)there is no induced electric field in $ C_1$
(B)there is an induced electric filed in $ C_1$ and it's magnitude is greater than the magnitude of induced electric field in $ C_2$
(C)there is an induced electric field in $ C_2$ and it's magnitude is greater than the magnitude of induced electric field in $ C_1$
(D)there is no induced electric field in $ C_2$.

[b]8.[/b]Consider a circle of radius R. A point charge lies at a distance a from it's centre and on it's axis such that $ R \equal{} a \sqrt3$. If the electric flux passing through the circle is $ \phi$ then the magnitude of the point charge is..
(A)$ \sqrt3 \epsilon_0 \phi$
(B)$ 2\epsilon_0 \phi$
(C)$ \frac {4\epsilon_0\phi}{\sqrt3}$
(D)$ 4\epsilon_0 \phi$

[b]20.[/b]Two positron($ e^ \plus{}$) and two protons(p) are kept on four corners of a square of side $ a$ as shown in fig. The mass of proton is much larger than mass of positron. Let $ q$ denote the charge on proton as well as positron. Then the kinetic energies of one of the positrons and one of the protons respectively after a[i] very long time[/i] will be..
(A)$ \frac {q^2}{4 \pi \epsilon_{0} a} (1 \plus{} \frac1{2 \sqrt2})$ , $ \frac {q^2}{4 \pi \epsilon_{0} a} (1 \plus{} \frac1{2 \sqrt2})$
(B)$ \frac {q^2}{2 \pi \epsilon_{0} a}$ , $ \frac {q^2}{4\sqrt2 \pi \epsilon_{0} a}$
(C)$ \frac {q^2}{4 \pi \epsilon_{0} a}$ , $ \frac {q^2}{4 \pi \epsilon_{0} a}$
(D)$ \frac {q^2}{4 \pi \epsilon_{0} a} (1 \plus{} \frac {1}{4 \sqrt2})$ , $ \frac {q^2}{8\sqrt2 \pi \epsilon_{0} a}$

[b]21.[/b]An electrostatic field line leaves at angle $ \alpha$ from a point charge $ q_1$ and connects with point charge $ \minus{} q_2$ at angle $ \beta$.
Then the relation ship between $ \alpha$ and $ \beta$ is
(A)$ q_1 \sin^{2}{\alpha} \equal{} q_2 \sin^2\beta$
(B)$ q_1 \tan{\alpha} \equal{} q_2 \tan{\beta}$
(C)$ q_1 \sin^{2}{\frac {\alpha}{2}} \equal{} q_2 \sin^{2}{\frac {\beta}{2}}$
(D)$ q_1 \cos{\alpha} \equal{} q_2 \cos{\beta}$

[b]25.[/b]A long straight wire carrying current $ I_1$ in $ \plus{} z$ direction. The $ x \minus{} y$ plane contains a closed cicular loop carrying current $ I_2$ and not encircling the straight wire. The force on the loop will be
(A)$ \frac {\mu_0 I_1 I_2}{2 \pi}$
(B)$ \frac {\mu_0 I_1 I_2}{4 \pi}$
(C)$ zero$
(D)depends on the distance of the centre if the loop from the wire.

[hide=" #2-Whats wrong here :?: "]
For the ring $ E_i \times 2\pi r_i \equal{} \minus{} \frac {d \phi}{dt} \equal{} \minus{} \pi r_i^2 \frac {dB}{dt}$
That gives me,  $ E_i \equal{} k \times r_i$  (k is some constant}
So my answer is C.
But that isn't correct acc. to the key :(
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[hide="#8"]- I've no idea how to proceed. I know somehow I've to use gauss theorem(may be) - $ \oint \overrightarrow{E} \bullet \overrightarrow{dS} \equal{} \frac {q}{\epsilon_0}$
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[hide="#20"]
My idea is - after a [i]long time[/i] the particles will be at infinity due to the repulsive forces. So all of the initial electrostatic potential energy is converted into kinetic energy. But how's that distributed btw protons and positrons? Any ideas?
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#21- Absolutely no idea. How to relate them?

[hide="#25"]
By intuition I feel that force depends on the distance between the centre of the loop and the straight wire.
For doing the actual calculation, I am clueless how to go on with the integration{to find the force}.
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[hide="Mistake in 2"]The flux is not $ \pi r_i ^2 B$, but instead $ \pi r_S ^2 B$, where $ r_S$ is the radius of the shaded circle $ S$.[/hide]

[hide="Hint for 8"]Since flux is a surface integral, what is meant by the flux through a loop -- which [i]surface[/i] do you calculate the flux over? Gauss's Law tells us it doesn't matter in a source-free region, as long as the boundary of that surface is the desired circle (determine for yourself why this must be true). So, you're free to choose the simplest surface.

Figure out what that surface should be in this case. [hide="Further hint"]The field whose flux you're calculating has only radial dependency, by symmetry.[/hide]

Once you have determined the desired surface, [url=http://en.wikipedia.org/wiki/Spherical_cap]this article[/url] contains a surface area formula that you might find very helpful if you don't already know it.[/hide]

[hide="Hint for 20"]The electrostatic potential energy goes to 0, so the kinetic energies of the four particles at time infinity must add up to the total electrostatic potential energy at time 0. This eliminates two of the answer choices.[/hide]

[hide="Hint for 21"]You can answer this question without calculation. Only one of the answers is remotely reasonable. Try drawing the field lines for $ q_1 \equal{} \text{large}, \; q_2 \equal{} \text{small}$ and eliminate by trying out different angles.

[b]Starter:[/b] $ \alpha \equal{} 0$ should automatically imply $ \beta \equal{} 0$, so choice (D) is clearly false.[/hide]

[hide="Hint for 25"]Let the current of the straight infinite wire flow along the z-axis.
Let the circle of current be parametrized as $ \left( A \plus{} R \cos \theta , \; R \sin \theta \right)$, where $ R < A$.
Let $ \vec{F} (\theta)$ be the force on the current component with parameter $ \theta$ as a result of $ \vec{B}$.

Calculate $ \vec{F}( \theta) \plus{} \vec{F} ( \minus{} \theta)$.[/hide]

[quote="savy2020"][b]20.[/b]Two positron($ e^ \plus{}$) and two protons(p) are kept on four corners of a square of side $ a$ as shown in fig. The mass of proton is much larger than mass of positron. Let $ q$ denote the charge on proton as well as positron. Then the kinetic energies of one of the positrons and one of the protons respectively after a[i] very long time[/i] will be..
(A)$ \frac {q^2}{4 \pi \epsilon_{0} a} (1 \plus{} \frac1{2 \sqrt2})$ , $ \frac {q^2}{4 \pi \epsilon_{0} a} (1 \plus{} \frac1{2 \sqrt2})$
(B)$ \frac {q^2}{2 \pi \epsilon_{0} a}$ , $ \frac {q^2}{4\sqrt2 \pi \epsilon_{0} a}$
(C)$ \frac {q^2}{4 \pi \epsilon_{0} a}$ , $ \frac {q^2}{4 \pi \epsilon_{0} a}$
(D)$ \frac {q^2}{4 \pi \epsilon_{0} a} (1 \plus{} \frac {1}{4 \sqrt2})$ , $ \frac {q^2}{8\sqrt2 \pi \epsilon_{0} a}$ [/quote]
[i]CORRECTION:[/i]
(D)$ \frac {q^2}{2 \pi \epsilon_{0} a} (1 \plus{} \frac {1}{4 \sqrt2})$ , $ \frac {q^2}{8\sqrt2 \pi \epsilon_{0} a}$

[hide="#2"]
Thanks I feel really bad for making such a silly mistake :(
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[hide="Solution for #8- Pls chk it"]
thanks again :)
Pls chk my solution.
The spherical cap constructed over the circle with radius $ R_s\equal{}\sqrt{R^2\plus{}a^2} \equal{} 2a$ subtends an angle of $ \Omega\equal{} \frac{A}{r^2}\equal{}{2\pi(2a)(a)}{{2a}^2}\equal{}\pi$ steradian where A=area of the spherical cap. So flux through this is $ \frac14 (\frac{q}{\epsilon_0})\equal{} \phi$ as the total angle at the centre is $ 4\pi$ steradian
Thank you very much for that surface area formula. I've searched for it's derivation and found [url=http://mathworld.wolfram.com/Zone.html]here[/url]

I think it could also be done as $ E \times A \plus{} \phi \equal{} 0$ $ \implies\phi \equal{} \frac{q}{4\pi \epsilon_0 (2a)^2} \times 4 \pi a^2$
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[hide="#20"]
The total electrostatic potential energy at t=0 is $ 4 ( \frac {q^2}{4 \pi \epsilon_{0} a} ) \plus{} 2 ( \frac {q^2}{4 \pi \epsilon_{0} \sqrt{2}a})$
Energy of a $ p$ and a $ e^\plus{}$ is $ E\equal{}2\frac {q^2}{4 \pi \epsilon_{0} a}(1\plus{} \frac{1}{2 \sqrt2})$
Option A and D agree with this. Now how to decide btw them? How do we justify whether the KEs will be equal or not. I feel they'll be equal and hence option A.
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[hide=" #21"]
Isn't there any mathematical method to solve?
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[hide="#25 solved"]
I get $ \overrightarrow{F} (\theta)\equal{} \minus{} \overrightarrow{F} (\minus{}\theta)$. So option C $ F_{net}\equal{}0$.  There will be torque though. Right?
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[quote="Diehard"]Sorry for reviving but I think you really overdid #8. The flux is just the area of the Gaussian surface multiplied by the Electric field strength since the position of the point charge is irrelevant. The flux is $(4\pi R^{2})(\frac{Q}{4\pi\epsilon_{0}R^{2}})$.[/quote]

No. He (ref. TZF) is correct. Whatever you wrote is the flux in the space and not through the given circle (As asked in the problem). Rather the flux through that circle will be given by $\phi=\frac{Q}{\epsilon_{0}}\cdot \frac{\Omega}{4\pi}$, Where $\Omega$ is the solid angle subtended by the point charge at the circle.

Note: $\Omega=2\Pi(1-\cos \theta)$, where $\theta$ is the half angle of the cone as [url=http://en.wikipedia.org/wiki/Solid_angle]here[/url]


How can I compute the number of subsets with 20 elements of a set with 52 elements using a C program?  :blush:

I know that the number is $C_{52}^{20}$ but..  how can I compute such a large number?  (I have made a post somewhere on this forum.. because I thought I was wrong, i though the number was not $C_{52}^{20}$)

thank you for very much!  :blush:

OK, you want to find the number $\left(\!\begin{array}{c}52\\20\end{array}\!\right)$. You can either calculate it directly from

$\left(\!\begin{array}{c}n\\k\end{array}\!\right)=\frac{n!}{k!(n-k)!}$

or from

$\left(\!\begin{array}{c}n\\k\end{array}\!\right)=\left(\!\begin{array}{c}n-1\\k-1\end{array}\!\right)+\left(\!\begin{array}{c}n-1\\k\end{array}\!\right)$

but any way, you'll get a number greater than int - fortunately, it is less than long long int (64 bit integer) so I suppose you could use the second formula to calculate it recursively.

(It is 125994627894135.)


on an online forum usernames can be 1-6 letters and can contain letters numbers or underscores. however, a username  cannot begin or end with an underscore. how many possible usernames are there?

[hide]If it is 1 character long, the number of possibilities is 36 (all 37 -1 for the underscore)
For it to be 2 characters long, we take $37^{2}= 1369$, and subtract 73 to get 1296.
For 3 characters: $37^{3}= 50653$, then subtract 2*36*37 to get 47989.
4 characters: $37^{4}= 1874161$, subtract 2*37*37*36 to get 1775593.
5 characters: $37^{5}= 69343957$, subtract 2*($37^{3}$)*36 to get 65696941.
6 characters: $37^{6}= 2565726409$, subtract 2*$37^{4}$*36 to get 2430786817.
Totaling this, we get $\boxed{2498308672}$[/hide]
hmm probably wrong oh well

can they contain anything else like # that for example?

magixter's looks correct.
[hide]
One symbol long $36$
Two symbols $36 \cdot 36$
Three symbols $36 \cdot 37 \cdot 36$
Four symbols $36 \cdot 37 \cdot 37 \cdot 36$
Five symbols $36 \cdot 37 \cdot 37 \cdot 37 \cdot 36$
Six symbols $36 \cdot 37 \cdot 37 \cdot 37 \cdot 37 \cdot 36$
Add them together.
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