inequalities
quadratics
absolute value
linear algebra

What is AM-HM-... whatever else there is?  I tried googling, but it didn't work.  Sorry if this was already answered or if I should post this elsewhere.

The Arithmetic Mean-Harmonic Mean Inequality, abbreviated AM-HM, says that the arithmetic mean of n real numbers is greater than or equal to the harmonic mean of those numbers.

In other words, if a1, a2,... ,an are n real numbers, then 1/n*(:Sigma: ak) :ge: n*(:Sigma: 1/ak)-1, with equality iff all the numbers are equal.

See if you can try and prove this.

Thanks.  btw, isn't there like another GM or something else in there?

There's the RMS-AM-GM-HM inequality chain.  Basically, it says that

RMS  :ge:  AM  :ge:  GM  :ge:  HM

Where RMS (root mean square; also called quadratic mean) is sqrt((a_1^2 + a_2^2 + ... + a_n^2)/n)
AM (arithmetic mean) is (a_1 + a_2 + ... + a_n)/n
GM (geometric mean) is (a_1*a_2*...*a_n)^(1/n)
HM (harmonic mean) is n/(1/a_1 + 1/a_2 + ... + 1/a_n)

By the way, I'm moving this from AMC to the Other Problem Solving Topics forum.

It's actually part of the Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality, which says that for n non-negative real numbers the AM :ge: GM :ge: HM, with equality iff all the numbers are equal.

Which is itself part of the more general Power Mean Inequality, which says that for n non-negative real numbers, a1, a2,... ,an, if Pr = [1/n*:Sigma: akr]1/r, (r any real number not equal to zero), then Pr :ge: Ps when r > s, with equality iff all the numbers are equal.  P0 is just defined to be the Geometric Mean.

Edited to fix for Ravi's comment.

By the way, don't we have to specify that the numbers are positive, or at least nonnegative?

I think Ravi's right.  Cauchy-Schwartz only holds for non-negative reals I think, and since AM-HM can be proven by Cauchy, I'd assume that it would no longer hold since Cauchy no longer holds for negatives.  Plus a counterexample: (-1 + 0.5)/2 = -0.25 < 2 = 2*1/(1/-1 + 1/0.5)

No, Cauchy-Schwarz holds for any real numbers.  The AM-GM-HM inequality needs to assume that the numbers are nonnegative.  Same with the more general Power Mean inequality.

Thanks for the clarification (for me, anyway).

Often, you'll see AM-GM (or power mean) written with absolute value signs to get rid of the uncertainty.  Also, this is more useful when you do things like linear algebra, when it pops up.

Note that in Cauchy, switching some of the signs will, if it has any change at all, make the smaller side smaller and not change the larger side.

About Cauchy and sign restrictions:

The original form of Cauchy
(x_1^2 + x_2^2 + ... x_n^2)(y_1^2 + y_2^2 + ... y_n^2) >= (x_1*y_1 + x_2*y_2 + ... x_n*y_n)^2 does indeed hold for all real values of the x's and y's.  However, if you use it to prove other results (eg AM-HM: let x_i^2 = a_i, y_i^2 = 1/a_i) as is commonly done, you have to remember than x_i^2 and y_i^2, being squares, can ONLY be positive integers.  This is why AM-HM does not hold for all reals even though it is a consequence of Cauchy.

Can someone please state what the Power Mean Inequality is?  I don't know what it is.  Thanks

[tex]\displaystyle\Biggl(\frac{\sum_{k=1}^n a_k^l}{n}\Biggr)^{1/l}\ge\Biggl(\frac{\sum_{k=1}^n a_k^m}{n}\Biggr)^{1/m}[/tex]

if [tex]l\ge m[/tex] and all the [tex]a_k[/tex]s are nonnegative.


HAPPY NEW YEAR!!!!

GO TO A NICE RESTAURANT AND PREPARE FOR THE FUTURE!!! AND BEYOND!


A well known problem : when $ \frac{a^2 \plus{} b^2}{1 \plus{} ab} \in \mathbb Z$, then it is a square.

But I can't find it in here? Some links, or sols or tips for searching?

A classic.  It's from 1988 IMO:
[url]https://artofproblemsolving.com/community/c6h57282[/url]


Positive real numbers, $a_1, .. ,a_6$ satisfy $a_1^2+..+a_6^2=2$. Think six squares that has side length of $a_i$ ($i=1,2,\ldots,6$). Show that the squares can be packed inside a square of length $2$, without overlapping.

i wrote a solution for this, but it got erased?

here it is again, but in short form, since i dont have time to write it all

--------

sort a_i decreasing

we have a < sqrt(2)

a < 2/3 and we are done, because we can fit in 4/3 by 2 rectangle for sure.

else, put a_1 in the corner, a_6 in opposite corner, it is easy to show that $a_1 + a_6 < 2$, so it fits.

Now you want to show a_2 and a_5 fit in a $2-a_1$ x $2-a_6$ rectangle, and so do a_3 and a_4.

We need two statements: $a_1 + a_2 \leq 2$, and $a_2 + a_5 + a_6 \leq 2$.  These are easily proved.

Hence, result.

It can be extended when $6$ replace by $n$.
With same strategy :!: .

WLOG that $a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5 \ge a_6.$ We need the following bounds.

[b] Bound 1. [/b] $a_1 + a_2 \le 2$.

[b] Proof of Bound. [/b] We have that $(a_1 + a_2)^2 \le 2(a_1^2 + a_2^2) \le 4$, and so the bound is proven.

[b] Bound 2. [/b] $a_1 + a_3 + a_5 \le 2.$

[b] Proof of Bound. [/b] We have that $a_3^2 + a_5^2 \le \frac{a_2^2 + a_3^2 + a_4^2 + a_5^2}{2} \le 1 - \frac{a_1^2}{2}.$ This means that $a_3 + a_5 \le \sqrt{2 - a_1^2}.$ Hence, we have that $a_1 + a_3 + a_5 \le a_1 + \sqrt{2 - a_1^2} \le 2.$

With this bounds, we are prepared to describe the algorithm. First, cut off a strip of width $a_1$ from the bottom of the square, then a strip of width $a_3$ from the remaining part, and finally a strip of size $a_5$ from the remaining part. This is doable by Bound $2.$ Now, into the first strip, place the $a_1 \times a_1$ and the $a_2 \times a_2$, which can fit because $a_1 + a_2 \le 2$ by Bound $1.$ Into the second strip, place the $a_3 \times a_3$ and the $a_4 \times a_4$, which fit because $a_3 + a_4 \le a_1 + a_2 \le 2.$ Finally in the third strip, place the $a_5 \times a_5$ and the $a_6 \times a_6$, which fit because $a_5 + a_6 \le a_1 + a_2 \le 2.$

We are done.

$\square$


function
trigonometry
Putnam
real analysis
real analysis solved

Find the differentiable functions $f(x),g(x),h(x)$ which satisfy the following conditions.

(ⅰ) $f(x)f'(x)+g(x)g'(x)+h(x)h'(x)=3\sin x\cos x$

(ⅱ) $f(x)g'(x)+g(x)h'(x)+h(x)f'(x)=2\cos x-\sin x\cos x$

(ⅲ) $f(x)h'(x)+g(x)f'(x)+h(x)g'(x)=-\sin x\cos x$

(ⅳ) $f(x)=g(x)+1$

(ⅴ) $f(0)+g(0)+h(0)=2$

Just an observation

(iv) gives $f'=g'$

(i), (ii), (iii) can be written like

$(1+g)g'+gg'+hh'=3\sin{x}\cos{x}$
$(1+g)g'+gh'+hg'=2\cos{x} - \sin{x}\cos{x}$
$h'+gg'+hg'+gh' = -\sin{x}\cos{x}$

Integrate the equations get term with $g,h$ use (v) for the constant

I'm sorry to have kept you waiting for a while, MathLinks everyone, especially, moderator [b]Moubinool[/b].

I will post my solution as follows.

Adding (ⅰ),(ⅱ) ,and (ⅲ),we have $(f+g+h)(f+g+h)'=\sin x+2\cos x$.
Set $F(x)=f(x)+g(x)+h(x)$, we obtain $F(x)\cdot F'(x)=\sin x+2\cos x$.
We rewrite  $[(F(x))^2]'=2\sin x+4\cos x$.
Integreating of both sides, we have  $(F(x))^2=\sin ^2 x+4\sin x+C$.
From (ⅴ) ,we have $F(0)=2$,yielding $C=2$,
we obtain $F(x)=\sin x+2$,i.e.$f(x)+g(x)+h(x)=\sin x+2\cdots\ [1]$,
then from (ⅳ) and [1], we have $2g(x)+h(x)=\sin x+1\cdots[2]$, differentiating of both sides, we obtain $2g'(x)+h'(x)=\cos x\cdots[3]$.
On the other hand, subtracting (ⅲ)　from　(ⅱ),$f(g'-h')+g(h'-f')+h(f'-g')=2\cos x$.
In using $f(x)-g(x)=1,f'(x)=g'(x)$,we obtain $g'(x)-h'(x)=2\cos x\cdots[4]$.
Thus from [3],we have $g'(x)=\cos x,h'(x)=-\cos x$.Substituting these for (ⅲ),
we have $f(x)(-\cos x)+[g(x)+h(x)]\cos x=-\sin x\cos x$,yielding $h(x)=1-\sin x$.

Thus from [2],the desired functions are $f(x)=\sin x+2,g(x)=\sin x,h(x)=1-\sin x$.

Thank you.

kunny

This is the first I have seen such system of differential equation, and I find it
very interesting, because it does not use a lot of theory on differential equation.

If you have other problem like this one please post it.

Thank you very much kunny for your nice solution.

Own http://www.mathlinks.ro/Forum/viewtopic.php?t=34186&search_id=898826459

2009 Putnam A2 http://www.mathlinks.ro/Forum/viewtopic.php?t=316667

very very thankful to you Kunny, for giving a very interesting solution. I am one of many who never got such problems.

but don't those solutions not satisfy iv?

sin(x)+2 is not equal to sin(x)+1

edit: i think the problem is when you set c=2. if F(0)=2 and F(0)^2=sin(0)^2+4sin(0)+c, then 2^2=c=4


Prove that if $ f$ is continuous at $ a$ and the $ \lim_{x \rightarrow a} f'(x)$ exists then so does $ f'(a)$ and they must be equal.

$ \lim_{x\to a} f'(x) \equal{} \lim_{x \to a} \lim_{h \to 0} \frac{f(x\plus{}h)\minus{}f(x)}{h}\equal{}$

letting $ x_n \to a$ as $ n \to \infty$

$ \lim_{x_n \to a} \lim_{h \to 0} \frac{f(x_n\plus{}h)\minus{}f(x_n)}{h}\equal{}$
$ \lim_{n \to \infty} \lim_{h \to 0} \frac{f(x_n\plus{}h)\minus{}f(x_n)}{h}\equal{}$
$ \lim_{h \to 0} \lim_{n \to \infty} \frac{f(x_n\plus{}h)\minus{}f(x_n)}{h}$
$ \lim_{h \to 0} \frac{f(a\plus{}h)\minus{}f(a)}{h}\equal{}f'(a)$

?? not sure if this is legal.

[quote="kenn4000"]$ \lim_{x\to a} f'(x) = \lim_{x \to a} \lim_{h \to 0} \frac {f(x + h) - f(x)}{h} =$

letting $ x_n \to a$ as $ n \to \infty$

$ \lim_{x_n \to a} \lim_{h \to 0} \frac {f(x_n + h) - f(x_n)}{h} =$
$ \lim_{n \to \infty} \lim_{h \to 0} \frac {f(x_n + h) - f(x_n)}{h} =$
$ \lim_{h \to 0} \lim_{n \to \infty} \frac {f(x_n + h) - f(x_n)}{h}$
$ \lim_{h \to 0} \frac {f(a + h) - f(a)}{h} = f'(a)$

?? not sure if this is legal.[/quote]  I think so  :D .. Any way ...
It's simpler with the mean value theorem : $ \frac {f(x)-f(a)}{x-a} = f'(u(x))$ where $ 0 < |u(x)-a|< |x-a|$.
Let $ L$ be the limit  ${ L=\lim_{x\rightarrow a}f'(x)}$ . We obtain :
$ (x\rightarrow a) \Longrightarrow (u(x)\rightarrow a) \Longrightarrow (f'(u(x))\rightarrow L) \Longrightarrow (\frac {f(x)-f(a)}{x-a}\rightarrow L)$
Manifestly $ f'(a)$ exists and $ L=f'(a)$
 :cool:

[quote="kenn4000"]$ \lim_{x\to a} f'(x) \equal{} \lim_{x \to a} \lim_{h \to 0} \frac {f(x \plus{} h) \minus{} f(x)}{h} \equal{}$

letting $ x_n \to a$ as $ n \to \infty$

$ \lim_{x_n \to a} \lim_{h \to 0} \frac {f(x_n \plus{} h) \minus{} f(x_n)}{h} \equal{}$
$ \lim_{n \to \infty} \lim_{h \to 0} \frac {f(x_n \plus{} h) \minus{} f(x_n)}{h} \equal{}$
$ \lim_{h \to 0} \lim_{n \to \infty} \frac {f(x_n \plus{} h) \minus{} f(x_n)}{h}$
$ \lim_{h \to 0} \frac {f(a \plus{} h) \minus{} f(a)}{h} \equal{} f'(a)$

?? not sure if this is legal.[/quote]

I don't see what good it would do to use $ x_n \to a$ as $ n \to \infty$.
Why can't we just say
$ \lim_{x_n \to a} \lim_{h \to 0} \frac {f(x_n \plus{} h) \minus{} f(x_n)}{h} \equal{}$
$ \lim_{h \to 0} \lim_{x_n \to a} \frac {f(x_n \plus{} h) \minus{} f(x_n)}{h} \equal{}$
$ \lim_{h \to 0} \frac {f(a \plus{} h) \minus{} f(a)}{h} \equal{}$
$ f'(a)$.

I think we should justify the step in switching the two limits.

In essence, we have to prove that f'(x) is also continuous at a.

As currently constituted, kenn4000's argument is not a proof. The interchange has to be justified, and I don't see an elementary way to do it.

What can we get from this discussion? The mean value theorem is better. If you want to prove anything about $ f$ starting from knowledge about $ f'$, it will almost certainly require either the MVT or an equivalent argument.


The probability of event A is 50% and the probability of event B is 75%. If the two events are not mutually exclusive, what is the range of probabilities that both event A and event B occur?

If A has a probability of 50 % and B of 75% it means taht on average from 100 tries A will occur in 50 cases, and B in 75. Together they can occur in minimum 25 cases, and maximum 50 cases, so the range should be 25%-50%.

[hide]
since they are not mutually exclusive, the least would be their difference, which is 25%, and the most would be the smaller amount, which is 50%
[/hide]

EDIT fdumps got it.


Question 2

If $ p$ is a prime, prove that:

$ p\,|\,a^{p}+(p-1)!a$ and $ p\,|\,(p-1)!a^{p}+a$

for any integer $ a$.

[quote="sludgethrower"]Question 1

Prove that there exist infinitely many composite numbers $ n$ for which $ a^{n-1}\equiv a\,(mod\,n)$.[/quote]

Do you mean $ a^{n-1}\equiv 1 \bmod n$?  These are known as the Carmichael numbers, or the Fermat pseudoprimes.  I'm not sure about your problem.

Edit:  Actually, I suppose your problem statement is right because the infinitude of the Carmichael numbers wasn't proven until 1994.

My question is correct, the thread is entitled "Fermat's Theorem" because from this, the result is a direct consequence.

p.s I thought that a composite integer $ n$ is a pseudoprime if and only if $ 2^{n}\equiv 2\,(mod\, n)$

[quote="sludgethrower"]My question is correct, the thread is entitled "Fermat's Theorem" because from this, the result is a direct consequence.[/quote]

What?  How does Fermat's Little Theorem follow from a question about composite moduli?

[quote="sludgethrower"]p.s I thought that a composite integer $ n$ is a pseudoprime if and only if $ 2^{n}\equiv 2\,(mod\, n)$[/quote]

A composite integer is a Carmichael number, or an absolute Fermat pseudoprime, if $ a^{n-1}\equiv 1 \bmod n$ for [b]all[/b] $ a$ relatively prime to $ n$.

t0rajir0u...I meant that the problem statement follows from Fermat's theorem. Thanks for clarifying about the pseudoprime thing. :)

Well, the way you've stated the problem is [hide="somewhat trivial"] Let $ n = 2p$ where $ p$ is prime.  Then $ \varphi(n) = p-1 | n-2 = 2p-2$. [/hide]


[quote="RED-and-black"]This is my first post here!
Show that if $x^2-xy+y^2\geq 1$ then we have:
$x^{2m}+y^{2m}\geq \frac{2}{3^m}$[/quote]

What's $m$ ??

also $x,y >0$  ?

 :)

Sorry about that, m is a positive integer and x,y>0


ratio
limit
search
calculus
calculus computations

Find the interval of $ x$ such that the series $ \sum_{n\equal{}1}^{\infty}\frac{x^{n}}{2n\minus{}1}$ converges

That's wrong.  At $ x \equal{}\minus{}1$ we have a convergent alternating series, so the interval is $ [\minus{}1, 1)$.

Yes, yes, you are obviously right. But my answer is still correct, because for |x| < 1 the series converges. So perhaps, instead of stating "that's wrong" and classifying my post as spam right on the spot, you could just say "that's incomplete", which is more appropriate. Don't you ever forget anything?

It's also interesting to note how some people here are so fast at classifying posts as spam, but leave most other posts of the same persons without any classification. If one post deserves classification, then all should deserve. At least be coherent.

I consistently (across all the various subfora I read on this forum) classify posts with a wrong answer and no explanation or commentary as spam.  I see no particular reason to make a personal issue out of it, and if you find that some of my posts are lacking in correctness or quality (or, less likely, if you find some of them to be particularly elegant or interesting), I invite you to rate them accordingly.

Incidentally, if I could rate my earlier post on this thread, I guess that it would receive a 3 or 4.  Had I also commented in that post that we both knew immediately that the interval in question had endpoints at $ \pm 1$ by the ratio test (or limit comparison test, or equivalent), I probably would rate it either a 4 or a 5.

I just gave most of the solution in my previous posts.  By the ratio test, the series converges whenever
\[ 1 >\lim_{n\to\infty}\left |\frac{\left(\frac{x^{n\plus{}1}}{2n\plus{}1}\right)}{\left(\frac{x^{n}}{2n\minus{}1}\right)}\right|\equal{}\lim_{n\to\infty}\left|\frac{x(2n\minus{}1)}{2n\plus{}1}\right|\equal{}|x|.\]
Thus, it converges on $ (\minus{}1, 1)$.  However, the ratio test is indeterminant when the limit is 1, so we must also check the cases $ x \equal{}\minus{}1$ (convergence by alternating series test) and $ x \equal{} 1$ (divergence by limit comparison with the harmonic series, for example).

For complex $ x,$ the series converges for $ |x|\le 1,x\ne 1.$

I've given the the general argument for this before, in case anyone wants to look for it.

http://www.mathlinks.ro/viewtopic.php?search_id=310364837&t=121138

Actually, there's a more general convergence criterion, which is proven using summation by parts, but I forgot the name.


If $n$ is an integer greater than $7$, prove that ${n \choose 7} - \left[ \frac{n}{7} \right]$ is divisible by $7$.

An idea...
[hide]Consider the intervals between two consecutive multiples of $7$.

In $8 \leq n < 14$ we have ${n \choose 7} \equiv 1$

In $14 \leq n < 21$, ${n \choose 7} \equiv 2$

In $21 \leq n < 28$, ${n \choose 7} \equiv 3$.

And in $7k \leq n < 7(k + 1)$ we have ${n \choose 7} \equiv k$.

Now consider the same interval for $\left[\frac{n}{7}\right]$.

In $7k \leq n < 7(k + 1)$ we obviously have $\left[\frac{n}{7}\right] \equiv k$.

Hence ${n \choose 7} \equiv \left[\frac{n}{7}\right]$.[/hide]

what is the proof for $7k \leq n < 7(k+1)$ we have ${n \choose 7}\equiv k$.
thanks!

[quote="srulikbd"]what is the proof for $ 7k \leq n < 7(k \plus{} 1)$ we have $ {n \choose 7}\equiv k$.
thanks![/quote]I'm reviving this so that somebody can supply a proof of this claim.

This result is generalized by [url=http://en.wikipedia.org/wiki/Lucas%27_theorem]Lucas' theorem[/url].  Prove it with the lemma $ (x \plus{} 1)^p \equiv x^p \plus{} 1 \bmod p$.

[b]My Solution[/b]
Let $ n\equal{}7q\plus{}r$ where $ q,r$ are integers, $ 0 \le r \le 6$

This implies that $ \left[\frac{n}{7}\right] \equal{} \left[q\plus{} \frac{r}{7}\right] \equal{} q$ since $ 0 \le r/7 < 1$

$ {n\choose 7}
\\\equal{} \frac{(n)(n\minus{}1)(n\minus{}2)(n\minus{}3)(n\minus{}4)(n\minus{}5)(n\minus{}6)}{7\cdot 6\cdot 5\cdot 4\cdot3\cdot2}
\\
\\\equal{}\frac{(7q\plus{}r)(7q\plus{}r\minus{}1)(7q\plus{}r\minus{}2)(7q\plus{}r\minus{}3)(7q\plus{}r\minus{}4)(7q\plus{}r\minus{}5)(7q\plus{}r\minus{}6)}{7\cdot 6\cdot 5\cdot 4\cdot3\cdot2}$

$ {n\choose 7} \equiv \left[\frac{n}{7}\right] (mod 7)
\\ \iff
q \equiv \frac{(7q\plus{}r)(7q\plus{}r\minus{}1)(7q\plus{}r\minus{}2)(7q\plus{}r\minus{}3)(7q\plus{}r\minus{}4)(7q\plus{}r\minus{}5)(7q\plus{}r\minus{}6)}{7\cdot 6\cdot 5\cdot 4\cdot3\cdot2} (mod 7)
\\ \iff (6\cdot 5\cdot 4\cdot3\cdot2)q \equiv \frac{(7q\plus{}r)(7q\plus{}r\minus{}1)(7q\plus{}r\minus{}2)(7q\plus{}r\minus{}3)(7q\plus{}r\minus{}4)(7q\plus{}r\minus{}5)(7q\plus{}r\minus{}6)}{7} (mod 7)$

Notice that in the numerator in the RHS , the term divisible by $ 7$ would be $ q$. Others would be
$ 6\cdot 5\cdot 4\cdot3\cdot2 (mod 7)$

Therefore ,
$ \frac{(7q\plus{}r)(7q\plus{}r\minus{}1)(7q\plus{}r\minus{}2)(7q\plus{}r\minus{}3)(7q\plus{}r\minus{}4)(7q\plus{}r\minus{}5)(7q\plus{}r\minus{}6)}{7} \equiv  (6\cdot 5\cdot 4\cdot3\cdot2)q  (mod 7)$
This was eqactly what we had to prove ...
We are done .....

$ Q.E.D.$

A lot easier: induction!!
Make sure to consider two cases, namely n congruent to 6 mod 7 and n incongruent to 6 mod 7. If f(n) is the given expression, show that 7 divides f(n+1) - f(n).

Sorry for bumping
Could someone present a concise solution

I meant whether someone could post the complete solution

We are going to prove the result by Induction.

1) As $n>7$, the base case forms $n=8$. For $n=8$, the expression becomes 7, which is divisible by 7.

2) Assume that this is true for $n=m$, so
\[7\mid\frac{m(m-1)\cdots (m-6)}{7!}-\bigg[\frac{m}{7}\bigg]=7q\]

where $q\in \mathbb{Z}$

3) For $n=m+1$, the expression becomes
\[\frac{((m-6)+7)(m)(m-1)\cdots (m-5)}{7!}-\bigg[\frac{m+1}{7}\bigg]\]

3.1) If,
\[\bigg[\frac{m+1}{7}\bigg]=\bigg[\frac{m}{7}\bigg]\]

Then the expression becomes,
\[7\bigg(q+\frac{m(m-1)\cdots (m-5)}{7!}\bigg)\]

So it is divisible by 7

3.2) If,
\[\bigg[\frac{m+1}{7}\bigg]=\bigg[\frac{m}{7}\bigg]+1\]
then $m+1=7q_1=m_1$.
The expression becomes,
\[\frac{(7q_1)(7q_1-1)\cdots (7q_1-6)-7!q_1}{7!}\]
which is again divisible by 7


calculus
integration
calculus computations

Notice $ I \equal{} \int^1_0 e^{x^{2}}x^{3}\,dx \equal{} \int^1_0 e^{x^{2}}.x.x^{2}\,dx$

Substitute $ u \equal{} x^2$. $ I \equal{} \frac{1} {2} \int^1_0 e^uudu$, then use integration by parts.


$ M \equal{} \left(
    \begin{array}{cc}
      A & B\\
      C & D
    \end{array}
  \right)$ is an $ 2n\times 2n$ matrix with each block being an $ n\times n$ matrix. Suppose $ A$ is invertible and that $ A$ and $ C$ commute. show that det(M) = det(AD-CB).

I need help with getting started. I've tried some ways, including induction, but not getting anywhere.

Let $ R\equal{}\begin{bmatrix}I&0\\\minus{}CA^{\minus{}1}&I\end{bmatrix}$; $ R$ has determinant $ 1$, and corresponds to the row operations needed to eliminate $ C$. $ RM\equal{}\begin{bmatrix}A&B\\0&\minus{}CA^{\minus{}1}B\plus{}D\end{bmatrix}\equal{} \begin{bmatrix}A&B\\0&\minus{}A^{\minus{}1}CB\plus{}D\end{bmatrix}$. The determinant is $ \det A\cdot\det(D\minus{}A^{\minus{}1}CB)\equal{}\det(AD\minus{}CB)$.

If we assume that $ A$ and $ B$ commute instead, we get $ \det(DA\minus{}CB)$.

maybe a stupid question but how do you know that the determinant equals $ \det A \times \det(D\minus{}A^{\minus{}1}CB)$?

[quote="daboa"]maybe a stupid question but how do you know that the determinant equals $ \det A \times \det(D \minus{} A^{ \minus{} 1}CB)$?[/quote]

induction on how many blocks there are


conics
geometry
geometric transformation
reflection
geometry proposed

Let $w=C(O,r),d$ be a circle and a line such that $O\in d$ and let $A,B,C,D\in (w-d)$ be the four points for which
$d\cap AC\ne\emptyset,\ d\cap AD\ne\emptyset,\ d\cap AB=\emptyset $.I note the intersections $M_1,N_1,M_2,N_2,M_3,N_3$  of the line $d$ with the lines $AB,CD,AD,BC,AC,BD$ respectively. Prove that $OM_1=ON_1\Longleftrightarrow OM_2=ON_2\Longleftrightarrow OM_3=ON_3.$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=18781

Given four points and a line which does not pass through any of them, the pencil of conics through the four points determines an involution on the line (i.e. if the conic $\mathcal C$ cuts the line in $P_{\mathcal C},P_{\mathcal C}'$, then the map $P_{\mathcal C}\mapsto P_{\mathcal C}'$ is a projective involution on the line).

Apply this to the four points $A,B,C,D$ and the line $d$. One of the conics in the pencil is the circle $\omega$. Assume it cuts the line in $M_4,N_4$. The involution is determined by the images of $M_1,M_4$, and if, say, $M_1,N_1$ are symmetric wrt $O$, the involution must be the reflection in $O$, since $M_4$ and $N_4$ are always symmetric wrt $O$.


function
real analysis
real analysis unsolved

Let $ (X,\Sigma)$ be a measurable space.
Prove that if $ f,g: X\to\mathbb{R}$ are measurable functions, then show that $ A\equal{}\{x|f(x)\equal{}g(x)\}$ and $ B\equal{}\{x|f(x)>g(x)\}$ are measurable sets.

I did it by using the fact that continuous binary operations are measurable, but I'm sure there must be an easier way to do this.

\[ B \equal{} \left\{ {x: f\left( x \right) > g\left( x \right)} \right\} \equal{} \bigcup\limits_{q \in \mathbb{Q}} {\left( {\left\{ {f\left( x \right) > q} \right\} \cap \left\{ {q > g\left( x \right)} \right\}} \right)} .
\]
\[ A \equal{} \left\{ {x: f\left( x \right) \equal{} g\left( x \right)} \right\} \equal{} \left\{ {f\left( x \right) \geqslant g\left( x \right)} \right\} \cap \left\{ {f\left( x \right) \leqslant g\left( x \right)} \right\}.
\]

[quote="bookworm_vn"]

\[ A \equal{} \left\{ {x: f\left( x \right) \equal{} g\left( x \right)} \right\} \equal{} \left\{ {f\left( x \right) \geqslant g\left( x \right)} \right\} \cap \left\{ {f\left( x \right) \leqslant g\left( x \right)} \right\}.
\]
[/quote]
small correction
\[ A \equal{} X\backslash (\left\{ {f\left( x \right) > g\left( x \right)} \right\} \cup \left\{ {g\left( x \right) > f\left( x \right)} \right\})
\]
:D


Find the least odd positive integer $ n$ such that for each prime $ p$ , $ \frac {n^{2} \minus{} 1}{4} \plus{} np^{4} \plus{} p^{8}$ is divisible by atleast four primes.

Should the primes be distinct ?

If no ,then $ n\equal{}1$

Yes the primes are distinct ....Sorry for not mentioning it

[hide="hint"] Set $ n\equal{}2k\plus{}1$  .
Then factorise the given expression as $ (p\plus{}k)(p\plus{}k\plus{}1)$
[/hide]

for $ p\equal{}2$ , we have $ (k\plus{}2)(k\plus{}3)$ dividable by $ 4$ distinct primes.
We're looking for the smallest value of $ k$ , then we look for the smallest primes

$ 2\cdot 3 \cdot 5 \cdot 7 \equal{}14\cdot 15 \Rightarrow k\equal{}12 \Rightarrow n\equal{}25$

@endbox ,I think you have made a mistake .
for $ p \equal{} 2$ ,we have the given expression $ \equal{} (16 \plus{} k)(17 \plus{} k)$ not $ (2 \plus{} k)(3 \plus{} k)$
actually the given question asks you to get the smallest value of $ n$,for $ k\equal{}4$ and $ p\equal{}2$ you get value of $ n \equal{}9$ ,you have to prove that  $ k\equal{}4$ is the least positive integer for an odd prime $ p$ also


$f : x\to x^3$ is a monomorphism from $G$ to $G$.
Prove that $G$ is commutative.

For what it's worth, liyi, putting spaces around your colon seemed to help.

$x^3y^3=(xy)^3$ for all $x,y$ $\Rightarrow x^2y^2=(yx)^2$, so $x^4y^4=(y^2x^2)^2=(xy)^4$, therefore $x^3y^3=(yx)^3$, hence $(xy)^3=(yx)^3$ and since $f$ is injective, $xy=yx$.


geometry
3D geometry
ratio
number theory
prime numbers
geometric sequence

Given three prime numbers, prove that the cube root of each cannot be part of a geometric progression (not necessarily consecutive).

edit:  my extreme bad, solution was completely bad.

Um, you mean $a - b$, not $a/b$, which does not lead to an immediate contradiction.

You also mean that it would imply u=1, but that's not a problem.

I think this is correct, however i am unsure because I didn't use the "cube root" part.  However, it looks right.

--
Let p,q,r be the 3 primes.  These numbers are greater than 1, so wlog the ratio of the geom series is greater than 1.

We have $\sqrt[3]{q/p} = x^a$ and $\sqrt[3]{r/q} = x^b$ for a real $x$, and integers $a$ and $b$.

So $(q/p)^{3b} = (r/q)^{3a}$

so that $p^br^a = q^{a+b}$ for integers a,b.  This happens only if a=b=0.  Contradiction (p,q,r are different)

Looks good to me.

In a way, the "cube root" part of the problem is irrelevant, because cubing each term of a geometric sequence still results in a geometric sequence.


we can't talk about questions. but we can talk about our scores.

no discussion on questions till november 21 after i think

[quote="indianamath"]Are we allowed to talk about it? I think every1 has taken it today.[/quote]

Some people take it on the 21st so we cant discuss it. :(

can we say what we think we got? lol

like score wise, so if someone said, i think i got a 5, u can say I think i got blah blah blah

Some school districts can't take it on the 14th. I haven't taken it yet.  :(

i think i got a [size=75]-snipped-[/size] i think :(

i will say i wasn't thinking
-jorian

same here. i wasn't thinking, but thats what happens when u have 2 minutes left and 5 problems to do.

No, it's still too early to talk about the contest, so this thread is locked.
I've also edited some of your posts. It is best not to discuss the contest at all, which includes posting what you think your score was.
(For more information, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119172]here[/url].)


This problem came to my mind when I read a similar one posted by [b]cancer[/b]:
A vertical cylinder of radius $R$ stands horizontally on an ideally smooth table and can rotate around its axis. A point mass is connected to the bottom of the cylinder by a string of lenght $l_{0}$. The cylinder begins to rotate around its axis with constant angular velocity $\omega$. Find the velocity of the mass as it hits the cylinder.

By cylinder you mean a roller device? Is the cylinder rigid, in the sense that it wont wobble or fall,
like having little height and being stable?

Yes, the cylinder is just a roller device - it's rigid and stable and is rotating with a constant angular velocity.


search
articles
number theory
modular arithmetic

What's modulus???? How can it be applied to problems, and where can I find more information on it???

I get really annoyed when there's a problem with modulus and I have no idea what they're talking about... :mad:

I really should know this by now.... :blush:

Search the math forum for modular arithmetic, there's a ton of stuff on it.Here's a nice wikipedia article:[url=http://en.wikipedia.org/wiki/Modular_arithmetic]en.wikipedia.org/wiki/Modular_arithmetic[/url]


set theory
superior algebra
superior algebra solved

Let $A$ and $B$ be sets.  Prove that if $\left|A\right|\leq\left|B\right|$ and $\left|B\right|\leq\left|A\right|$, then $\left|A\right|=\left|B\right|$.

Also known as the Cantor-Bernstein or Schröder-Bernstein Theorem. Try looking up that name, possibly in the index of a book on set theory or topology.

Hint: mother, father, grandfather, grandmother, ...

This is Bernstein theorem, you can find it in Halmos book of naive set theory, there is a simplification:

Note that yu can reduced it to: If there is $f: A\longrightarrow B$ inyective where $B\subseteq A$ then there must be a bijecction between $A$ and $B$.
This special case is kinda easier, if $f(A)\subsetneq B$ consider $C=B\setminus f(A)$ and $K=C\cup f(C)\cup f(f(C)) \cup\ldots$ now is easy to proof that $F: A\longrightarrow B$
\[ F(x)=\{\begin{array}{cc} f(x)& x\in K\\ x& x\notin K \end{array} \]
Is a biyection.

I'll look into that (I didn't really know the name of the theorem).  My instructor told me it's very difficult to prove, as opposed to some other easier problems where finding a 1-1 coorespondence is easy.

Thanks for the replies.


Prove that if a, b, c are the angles of a triangle then cotg2 a + cotg2 b + cotg2 c \geq 1.

so to prove that 1/sin² a + 1/sin² b + 1/sin² c  \geq 4
we know that each one is positive and that c = \pi - a - b
so to prove: 1/sin² a + 1/sin² b + 1/sin² (a+b)  \geq 4

The first solution that I had was simply deriving this for the minimum, calculate the value for that minimum, and saw it was for 60,60,60 so 4/3+4/3+4/3 = 4

It turns out to be a lot simpler than it may look on first sight, so I don't feel necessity to work on another solution. But I'm very sure there is one (almost had it but stopped cuz this was faster)

My own solution:
cotg2 a + cotg2 b + cotg2 c \geq cotg a cotg b + cotg b cotg c + cotg c cotg a = 1.

The identity cotg a cotg b + cotg b cotg c + cotg c cotg a = 1 is well known and easy to prove.
The fact that x2 + y2 + z2 \geq xy + yz + zx is well known too.
[It reduces to (x - y)2 + (y - z)2 + (z - x)2 \geq 0, obvious.]

i like that solution. too bad i didnt know that identity :(

neither did I

at least, I thought of it, but I didn't use it because I didn't know it was well-known :D


Does anyone know where i can find amc 12 prac tests? and does anyone know when the next round of the usamts starts?

[url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44]This[/url] is the most comprehensive AMC12 database that I know of, with more problems being added fairly consistently, but I know that there are more complete listings out there.

As for the USAMTS, check the [url=http://www.artofproblemsolving.com/Forum/index.php?f=164]USAMTS Forum[/url].


geometry
geometric transformation
geometry proposed

Let H the orthocenter of a triangle ABC with AB , AC not equal  and L the mid point of side BC. The perpendiculars to BC

and HL at L meet AB at M,N respectively.

   Prove that AM = BN

Babis

[color=red][b]edited[/b][/color]

Prove that AM = BN

[quote="vittasko"]( From [b]N. Kyriazis[/b] home ).

Do you mean $ AM \equal{} BN$ dear [b]Babis[/b]?

Greetings from Thessaloniki, Nikos Kyriazis and Kostas Vittas.[/quote]

Yes ! Thank you Kostas , I wish you to have a nice week  in this beautifull  city !

Babis

Let $ P,Q$ be the middles of $ AB,AC$ respectively.
So it is enough to prove that the pencil (1) $ L,LM,LP,LN,LQ$ is harmonic.
Let $ S,T$ be the feet of the perpendiculars from $ L$ to $ AC,AB$ respectively.
Since the rays of the pencil (2) $ L,LC,LS,LH,LT$ are perpendiculars to the rays of pencil (1), we must prove that pencil (2) is harmonic.
Let the altitudes of the triangle $ ABC$ be $ AA',BB',CC'$.
Let $ R$ be the intersection point of $ BC$ and $ B'C'$.
It is well known that $ (R,B,A',C) \equal{} \minus{} 1$.
So the pencil (3) $ (A,AR,AB,AA',AC)$ is harmonic.
However the rays of pencil (2) are perpendiculars to those of pencil (3)
and our problem is solved.

Many greetings to Mr. Stergiu and Mr. Vittas.

Denote $ E \equal{} CH\cap AB$
Suppose the  pependicular line to $ CE$ at $ C$ meets $ ML$ at $ S$.
Since both $ \angle HEL \equal{} \angle LCE \equal{} \angle HAN$ and $ \angle HLE \equal{} \angle HNA$, we see two triangles $ HEL$ and $ HAN$ are similar.It follows that  triagles $ AEH$ and $ NLH$ are similar.
But,triangles $ AEH$ and $ SLB$ are similar ,so triangles $ NLH$ and $ SLB$ are similar.
Then the spiral similarity $ S(L,90^0,\frac {HL}{LN})$ maps $ N\to H,S\to B,SN\to HB$
So $ SN\perp BH$,hence $ SN//AC$.
Therefore $ BM \equal{} CS \equal{} AN$ ,so $ AM \equal{} BN$.

@[b]Sakis[/b] :We can show that both $ LT$ passes through the midpoint of  $ BH$ and $ LS//BH$,so the pencil (2) is harmonic.
Thank you for nice solution. :)


trigonometry
geometry
cyclic quadrilateral
angle bisector
geometry solved

P,Q,R are feet of perpendiculars on to the sides of triangle ABC  from any point X on the bisector of angle BAC

Prove that PX and QR meet on a median of the triangle.

Suppose the intersection of the median AD and QR is Y ,and the intersection
of PX and QR is Z.Then by the theorem of sine
QY/YR=(AQ/AR)*(sinQAD/sinRAD)=sinBAD/sinCAD=(BDsinB/AD)/(CDsinC/AD)=sinB/sinC
QZ/ZR=(QX/XR)*(sinQXZ/sinRXZ)=sinQXZ/sinRXZ
But QXPB is a cyclic quadrilateral,QXZ = B.Similar, RXZ = C.So QY/YR=QZ/ZR
,that means Y = Z.So the conclusion is correct.

Another approach:

Let $T$ be the point where $PX$ cuts the $A$-median of $ABC$, and let $B',C'$ be the points where the parallel through $T$ to $BC$ cuts $AB,AC$ respectively. If we prove that $QR$ passes through $T$, we're done.

$XRB'T$ is cyclic, so $\angle XTR=\angle XB'R\ (1)$. In the same way we get $\angle XTQ=\pi-\angle XC'Q\ (2)$. $(1),(2)$, together with the observation that the triangles $XRB',XQC'$ are congruent, give the conclusion $\angle XTR+\angle XTQ=\pi$, i.e. $Q,R,T$ are collinear, as desired.

Let N be the point of intersection of the lines PX and QR. Then, since $XP\perp BC$, the point N is the point on the line QR such that $XN\perp BC$.

Since the point X lies on the angle bisector of the angle BAC, it is equidistant from the sides AB and CA of this angle, i. e. we have XQ = XR. Since $XQ\perp CA$ and $XR\perp AB$, the circle with center X and radius XQ = XR touches the sides CA and AB of triangle ABC at the points Q and R, respectively.

So we have a triangle ABC, a circle with center X touching its sides AB and CA at the points R and Q, and a point N on the line QR such that $XN\perp BC$. Hence, by Theorem 2 from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124774#p124774]http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 post #4[/url], if the lines AN and BC intersect at W, then the point W is the midpoint of the segment BC. Hence, the line AW is the A-median of triangle ABC; but since the line AW coincides with the line AN, it follows that the line AN is the A-median of triangle ABC. Hence, the point of intersection N of the lines PX and QR lies on the A-median of triangle ABC. This solves the problem.

  Darij


symmetry geometry Gauss vector calculus integration trigonometry

just use symmetry arguements...complete the quarter cylinder with other 4 parts one to the side of it and other aside it such that the charge is at the centre of the cylinder
by symmetry the flux through each of them is $ \frac{Q}{4\epsilon_{0}}$

I think you must take the two surfaces  with area $ \pi r^2$ into account.

i don think so cos in that area. E and dS will be perpendicular to each other....
so flux trough that surface would be 0

[quote="Nihat"]Find electric flux passing through the surface shown in the figure.[/quote]

As written the question is trivial, because Gauss' law gives us the electric flux:

\[ \Phi_E \equal{}\iint_S \vec{D}\cdot\hat{n}\, da \equal{} Q \].

In as much as the total charge is $ Q$,  that's the answer.

An alternative definition of electric flux still from Gauss' law is:

\[ \Phi_{E}^{\prime} \equal{} \iint_S \vec{E}\cdot\hat{n}\, da \equal{}\frac{Q}{\epsilon} \].

I believe the point of the question is more at, "What is the electric flux density?"

The answer to this question is: $ \frac{d\Phi_E}{d\vec{A}} \equal{} \vec{D} \equal{} \frac{Q}{A}\, \hat{n}$.

Now, the question is what is $ A$ ?

You can split this into two parts:

1) $ \vec{A_1} \equal{} \frac{\pi}{2} r L\; \hat{r}$, the surface of the quarter cylinder.

2) $ \vec{A_2}$, the combined surface of the quarter circle end-plates, but look what happens to these.

\[ \vec{A_2} \equal{}\frac{\pi}{4}r^2\,\hat{\ell}\minus{}\frac{\pi}{4}r^2\,\hat{\ell} \equal{} 0 \]

The contributions of the ends cancel each other.

Therefore,

\[ \boxed{\vec{D} \equal{}\frac{Q}{\frac{\pi}{2} r L}\, \hat{r}} \]

sorry u need 7 more...but that doesn't take the  heart out of my proof.... :D
consider any such part the flux $ \Phi$ through it 'assuming outwrd normal to be positive is the same...when all such fluxes are aded  at the common surfaces since the normals are oppositely directed they cancel out each other so on addition what remains is the flux through the entire cylinder so the answer being $ \frac{Q}{8 \epsilon_{0}}$

[quote="shreyas_patankar"]The charge is a point charge isn't it?
Then you need not four, but eight such surfaces to complete the cylinder (on both sides of the charge), and the flux through the flat faces is not zero. The problem doesn't seem trivial.[/quote]

I hadn't considered the flux through the back or bottom. [i]Interesting...[/i]

In any case, the net flux itself is just $ Q$, the vector directions of the components of flux density notwithstanding.

[quote="Immanuel Bonfils"]By integration, again, we get  $ \;{{QL}\over{8\epsilon_0 \sqrt {L^2 + r^2}}\;}$.

Note that it goes to $ \;{Q \over{8\epsilon_0}\;}$  when L goes to infinity.[/quote]

No. In this case there actually [b]is[/b] a point charge $ Q$.

The surface over which the flux density shines can calculated piecewise as:

The two – quarter circle end plates: $ 2\cdot\frac {\pi}{4}r^2 = \frac {\pi}{2}r^2$.

The bottom + the back plate: $ 2 r L$.

The quarter cylinder shell: $ \frac {\pi}{2} r L$.

Summing it all up, $ A = \frac {\pi}{2}r^2 + 2 r L + \frac {\pi}{2} r L$.

Therefore the magnitude of the flux density: $ \left|\vec{D}\right| = D = \frac {Q}{A} = \frac {Q}{\frac {\pi}{2}r^2 + 2 r L + \frac {\pi}{2} r L}$.

$ \Rightarrow \vec{D} = D\left\{\hat{z} - \hat{z} + \hat{r} - \hat{x} - \hat{y}\right\}$

$ = D\left\{\left(1 - \left[\cos\theta + \sin\theta\right]\right)\hat{r} + \left[\sin\theta - \cos\theta\right]\hat{\theta}\right\}$
\[ \boxed{\therefore\vec{D} = \frac {Q}{\frac {\pi}{2}r^2 + 2 r L + \frac {\pi}{2} r L}\left\{\left(1 - \left[\cos\theta + \sin\theta\right]\right)\hat{r} + \left[\sin\theta - \cos\theta\right]\hat{\theta}\right\}}
\]

Yes, but there [i]is[/i] no cylindrical symmetry for a point charge. All you have is spherical symmetry.

there may not be a cylindrical symmetry but for the point charge at the centre of a cylinder the flux through all the 8 portions is the same :)
That's what i require

[quote="pardesi"]i think my proof using symmetry was pretty clear to find the answer  :huh:[/quote]

I would agree with your attempt to use symmetry, but for three things:

1) There aren't eight sectors, there are only four. The curved portion is a quarter cylinder.

2) The back side and the bottom are rectangular, not cylindrical and

3) $ Q$ is a point charge, not a sheet charge; integration is not necessary.

yes i agree it's a quarter cylinder but i will join two different cylinders so that the charge $ Q$ is the centre of the big cylinder(that is what i had written before)
now the flux thru the entire new arrangement is $ \frac{Q}{\epsilon_{0}}$
so by symmetry and the fact that for cylinders having common faces the fluxes get cancelled at the common faces we have flux thru ecah one of such part is $ \frac{Q}{8 \epsilon_{0}}$

The [u]total[/u] flux through eight curved and two flat faces is $ \frac{Q}{\epsilon_0}$
By symmetry, the flux through each of the eight curved faces and two flat faces is [u]equal[/u], both in magnitude and sign. The flux through the flat faces does not get "cancelled", and so only way to calculate flux through the curved surface is to actually calculate the flux through eitherthe curved surface or one of the flat surface. (The latter is perhaps more convenient)

yes i am speaking of the total flux thru the cylinder of length $ 2l$ is $ \frac {Q}{\epsilon_{0}}$
now this is composed of 8 equal cylinders thru which the fluxes is the same assuming otward normal positive.so upon adding the flux thru the 'bigger' flat surfaces and the 'smaller' flat surface containing the charge cancel out and what u get is the sum of the fluxes of the 8 'partial'cylinders is equal to that of the bigger cylinder o length $ 2l$
hence the answer being $ \frac {Q}{8\epsilon_{0}}$

[quote="pardesi"]now this is composed of 8 equal [b]cylinders[/b] thru which the fluxes is the same assuming otward normal positive.[/quote]
I am assuming you meant "cylindrical surfaces"

The flux through both the flat surfaces is (using your convention) positive, and hence does not cancel.
(I just realised, which cylinder are you using in your diagram - the one with the charge at the center of a face or the one with charge at the centre of the cylinder? The flux through the flat faces does not cancel in either case)

yes i meant the 'partial cylinders' and not only cylindrical surfaces....
and why don't they cancel out .
consider two such 'partial cylinders' side by side .They have a common flat side.
the outward normals to both are oppositely directed(assuming the outward normal positive convention while deriving the Gauss's law..or more so the Divergence theorem)
so for a point common to both of then one will have positibve flux while the other will have negative of equal magnitude :)

[quote="pardesi"]diagram of what....
consider a cy;inder of length $ 2l$ cut it into 8 equal parts....and one of the parts will be the one given by nihat[/quote]

Why not just work the problem as given? It is unclear how what you propose simplifies the calculation, and without a diagram it is difficult to visualize what you're saying.

@[b]Dr. No[/b], it was my suggestion that we should use a cylinder of length $ 2L$ so that symmetry can be preserved, but it doesn't seem to help the problem.
By actual integration, Flux through the flat face is $ \frac{Q}{2\epsilon_0}\frac{r}{\sqrt{r^2\plus{}l^2}}$
So the answer to the original question is $ \frac{Q}{8\epsilon_0}\left( 1\minus{}\frac{r}{\sqrt{r^2\plus{}l^2}}\right)$

shreyas i used symmetry in the right way but couldn't express myself wrote 4 cylinders wrongly...
i don't think drawing picture will be of any help...

here is a detailled proof
1)Consider a cylinder of length $ 2l$ and radii $ r$ with achrge at the centtre
2)Through it'se centre first cut along the cross section symmetrically
3)then now cut along it's axis
4)We get 8 pieces which look similar to the photo given by Nihat
5) Now by symmetry the flux thru each such part is the same
6)As by Gauss's laws convemtion we assume outward normal positive
7)Consider any point on the 'plane surface' common to two such parts the positive normals are oppsitely directed w.r.t both of them but since the field fro both of them at that point is the same so if we add their fluxes they will cancel out
8)now add the fluxes due to all the 8 parts ...since they have the flat parts common so the flux thru these flat parts(minus the outermost flat part) will cancel out leaving only the flux thru the bigger cylinder which we know is $ \frac{Q}{\epsilon_{0}}$
So flux thru each such part is $ \frac{Q}{8\epsilon_{0}}$

There is ambiguity in the question, as to should we use the curved surface only; or the surface of the entire 'solid'. I have throughout been assuming that we have to calculate the flux only through the curved part. If we have to consider the entire surface, the problem is trivial and pardesi's anwer $ \frac{Q}{8\epsilon_0}$ is correct.

To calculate flux for the curved part, we note that the flux through each of the planes containing the charge is zero. So the flux through the curved part is the  flux through the 'farther' flat face subtracted from the total flux $ \frac{Q}{8\epsilon_0}$

[quote="shreyas_patankar"] By actual integration, Flux through the flat face is $ \frac{Q}{2\epsilon_0}\frac{r}{\sqrt{r^2\plus{}l^2}}$ [/quote]

By flat face, did you mean the circle basis?

Yes, it is the circular face of radius $ r$ at a distance $ l$ from the charge. The surface in the question includes only one quarter of this face.

This should be
\[ {Q \over {2\epsilon_0}}(1\minus{}{l\over\sqrt{l^2\plus{}r^2}})\]


limit
number theory
prime numbers
number theory theorems

[size=150][b]About the equivalence between non trivial zeros of Riemann Hipotesis and prime numbers:[/b][/size]

[b]1º non trivial zero of Riemann Hipotesis    (0.5; 14.096) ->  number  2[/b]
\[ \sum\limits_{x \equal{} 1}^{60} {\frac {1}{{x^{\frac {1}{2}} }} \equal{} 14.096}
\]
14/2=7

[b]2º non trivial zero of Riemann Hipotesis   (0.5; 20.494159) ->  number 3[/b]
\[ \sum\limits_{x \equal{} 1}^{120} {\frac {1}{{x^{\frac {1}{2}} }} \equal{} 20.494159}
\]
20/3 = 6,...

[b]3º  non trivial zero of Riemann Hipotesis  (0.5; 25.4097) -> number 5[/b]
\[ \sum\limits_{x \equal{} 1}^{180} {\frac {1}{{x^{\frac {1}{2}} }} \equal{} 25.4097}
\]
25/5 =5

[b]4º non trivial zero of Riemann Hipotesis  (0.5; 29.555)  ->  number 7[/b]
\[ \sum\limits_{x \equal{} 1}^{240} {\frac {1}{{x^{\frac {1}{2}} }} \equal{} 29.555}
\]
29/7 =4,...

[b]5º non trivial zero of Riemann Hipotesis  (0.5; 33.2095) -> number 11[/b]
\[ \sum\limits_{x \equal{} 1}^{300} {\frac {1}{{x^{\frac {1}{2}} }} \equal{} 33.2095}
\]
33/11=3

[b]6º  non trivial zero of Riemann Hipotesis  (0.5; 36.5133) -> number 13[/b]
\[ \sum\limits_{x \equal{} 1}^{360} {\frac {1}{{x^{\frac {1}{2}} }} \equal{} 36.5133}
\]
36/13= 2,...

[b]7º non trivial zero of Riemann Hipotesis  (0.5; 39.5518) ->  number 17[/b]
\[ \sum\limits_{x \equal{} 1}^{420} {\frac {1}{{x^{\frac {1}{2}} }} \equal{} 39.5518}
\]
39/17=2,...

[b]8º non trivial zero of Riemann Hipotesis (0.5; 42.38) -> number 19[/b]
\[ \sum\limits_{x \equal{} 1}^{480} {\frac {1}{{x^{\frac {1}{2}} }} \equal{} 42.38}
\]
42/19=2,...

etc.

[size=150][b]my conjecture:[/b][/size]
\[ \mathop {\lim }\limits_n (\frac {{{\mathop{\rm int}} (\sum\limits_{x \equal{} 1}^{60 \times n} {\frac {1}{{x^{\frac {1}{2}} }}} )}}{{prime_n }}) \equal{} 2
\]
:(

The numerator and denominator have well-known (different) growth rates; the limit is in fact zero.  This has very little to do with the Riemann Hypothesis.


A Force and motion problem I'm working on:

The solution informs me:

$ \dfrac{dm}{dt} \equal{} \dfrac{T}{G}(cosx \minus{} \dfrac{sinx}{y}) \equal{} 0$

leads to:

$ tanx \equal{} y$

(I've used x and y in place of the physics variables.)
Can someone please tell me the steps to get that solution? Thanks in advance!

Think a bit, guy. If  $ cos(x) \neq 0$ and  $ T \neq 0$ it is too trivial !!!

Oops, you're right, should have seen that :(


Hi everyone Romanian problem solvers:
 I was very surprised when I saw the tests of Junior team selection test

 How you select  you junior team ? by 4 testes :o

Hi !!! The tests are very hard because there are a lot of good students who may represent our country. I think that the number of the tests is OK. The first test or the first two tests are hold at the National Mathematical Olympiad. The juniors have a week of preparation in Bucharest, and after this period they have the rest of the tests.


1. Suppose that every Caucy sequence in R converges to a point in R. Show that every monotone increasing sequence bounded above converges in R.

2. Let A be a nonempty set in R. Suppose sup A exists. Show that there exists a sequence <xn> in A such that the sequence <xn> converges to sup A.

I take the analysis lately. It is hard for me. :P

2. use the definition of sup... for instance, let x_n in $ ]\sup A \minus{} \frac 1n, \sup A]$.

[quote="§outh§tar"]pic, what if A is a singleton? ;)[/quote]

ok I edited :)


So I have these three problems for homework, and I am just stumped on a couple. Wondering if anyone could help me.
For number 42, all I have to do is B.
And I think I did 36 right... but would like someone to double check!

If you have any free-time I would really appreciate it!
[img]http://img187.imageshack.us/img187/9933/0309071535yy0.jpg[/img]
[img]http://img187.imageshack.us/img187/3323/0309071549sn7.jpg[img]

[/img]http://img119.imageshack.us/img119/7576/0309071533dl7.jpg[img]
[/img]http://img250.imageshack.us/img250/5686/0309071536dl2.jpg[img]

[img]http://img151.imageshack.us/img151/499/0309071710lh1.jpg[/img][/hide]

oh comon now! i see problems i cant even fathom in my head on these forums!!! This one should be easy!!!! I really need it by tomorrow morning!

It means a difference between a 96 or 95 in calculus!!
Thanks again

Posting scanned files is considered bad form around here - take the trouble to type things up for yourself. (Learning how to do it in $\LaTeX$ is a bonus there). I'm on a dial-up connection right now and am having trouble even seeing what you have.

[quote="orangedude654"]oh comon now! i see problems i cant even fathom in my head on these forums!!! This one should be easy!!!! I really need it by tomorrow morning![/quote]
Of course it's easy. That's one reason for it to be ignored- we're just not very interested. Also, large pictures are not a very good way of posing your question. (Some of them aren't there at all)
This forum does not provide immediate answers in general. While a problem might be answered any hour of the day, it is also likely to wait hours or days for a response. I've had many exchanges with European forum members- those threads could go on for quite a while at a rate of two or three posts per day.

[quote="orangedude654"]It means a difference between a 96 or 95 in calculus!!
Thanks again[/quote]
Why should we care? You're doing fine. If a student in one of my classes on track for that sort of grade came to me, I wouldn't spend much time on routine homework problems- I would bring up more interesting questions and try to get at a deeper understanding.

Sorry, i am very new to the thread... and didnt even know one had the options to write it all out with the symbols...

Sorry if i offended anyone


When I put a quote in my blog, it leaves a ton of space between the quote and the text above it. And I mean a ton of space. about 2 screens maybe.

Yet, on the entry preview, it doesn't show the space at all.

Resize your browser window (widen it).  Does that make the problem go away?  (Yes, that's still a bug, but this will help us hopefully figure out how to fix it when we dive into the blog code later this year.)


analytic geometry
ratio
geometry
MATHCOUNTS
quadratics
rectangle
Pythagorean Theorem

"Square ABCD has side lengths of 13 units. Point E lies in the interior of the square such that AE = 5 units and BE = 12 units. What is the distance from E to side AD? Express your answer as a mixed number."

How do they expect middle schoolers to solve this?

[hide="Solution"]
First of all, we know AEB is a right angle. Let point F be a line on AB such that EF is perpendicular to AB. Let point G be a point on AD such that EG is perpendicular to AD. We need to find EG.

Triangle AEB is similar to triangle AEF. We can set up the ratio of sides, where x is GA or FE:

$ \frac {13}{5}\equal{}\frac{12}{x} \implies x\equal{}\frac{60}{13}\equal{}GA$

Now, to find EG. Using the Pythagorean theorem:

$ GE\equal{}\sqrt{ 25\minus{}\frac{3600}{169}}\equal{}\sqrt{\frac{625}{169}}\equal{}\frac{25}{13}$

So, as a mixed number, the answer is $ \boxed{1\frac{12}{13}}$.
[/hide]

[quote="AIME15"]You can use either coordinates or similar triangles.[/quote]

dangit, I bashed soooooo much with like 3 variables...

[quote="12markkram34"]A 2009 school team problem involved arcs and tangents. Since when were high school geometry topics used in Mathcounts?[/quote]

Er, they've always been part of it...what about all the similar triangles problems in MATHCOUNTS?

I've never seen Mathcounts problems involving the measures of angles that intersect circles. That's high school geometry -_-. Why should that be in a school round? That should be state at least.

Similar triangles are covered in the same year as arcs and tangents...

Anyway, school/chapter/state/national problems are not decided based on the topics covered, but the difficulty of the problems (or lack thereof).

UM I also think u should use similar triangles  :D  :D

iirc, Mathcounts is anything up to Geometry level.

I believe some Algebra II is covered as well. (fractional exponents, special arithmetic/geometric sequences, and special quadratic equations)

Can Mods lock this thread? The answer and solution was given

Wait first can someone show me how to use coordinates to solve this?

actually, MAthcounts topics go to geometry and somewhat beyond it.
I guess you can say it was a hard problem for school, but there's always one of those there.
By the way, on the state sprint this year, problem 30, it involved arcs and circles.

also, mathcounts problems can use trig. to deal with problems in an easier way.

dude problem 30 of states should have been like problem 15. The circle part didnt really come into play at all (well maybe once or twice), and the question was definitely middle school level. (yet I still messed up :P )

So anyways, can someone show me how to use coordinates on this question?

Sorry, I had this confused with another old School Round problem.  You still can use coordinates, but it's very messy (you use the distance formula to get some ugly radical point for E, then you find the distance to AD).

No no no, if you coordinate bash, you set the top of the rectangle to the x-axis and draw two circles.

Circles are pretty much exactly the same as the distance formula...

I think that applies, but geometry is a really wide subject, right? So some schools might have gotten to that part of curriculum earlier if they were in say a different state or region, so Mathcounts probably uses the most common.


geometry
perimeter
analytic geometry
calculus
integration

What is the center of mass of triangle centered at (0,0), (3,0), (1,2)?

[hide="Solution"]The center of mass of a triangle is the centroid. To find it, the intersection of two medians must be found.

The median from $ (0,0)$ to $ (2,1)$ has an equation of $ x\minus{}2y\equal{}0$.

The one from $ (1,2)$ to $ \left(\tfrac{3}{2},0\right)$ is represented by $ 4x\plus{}y\equal{}6$.

The intersection of them is at $ \boxed{\left(\tfrac{4}{3},\tfrac{2}{3}\right)}$.[/hide]

[hide="Answer"]
The center of mass is "the specific point at which, for many purposes, the system's mass behaves as if it was concentrated," where the three medians intersect. This is the centroid. (See attached for a picture)

Anyway, we'll just find the intersection of the three lines.

These lines are $ 4x\plus{}y\equal{}6$, $ 5x\plus{}12y\equal{}15$, and $ x\minus{}2y\equal{}0$. Then, just find the value $ (x,y)$ that fits all of this. Just use substitution on any two to get $ \left(\frac{4}{3},\frac{2}{3}\right)$.[/hide]

[quote="shake9991"]triangle centered at (0,0), (3,0), (1,2)?[/quote]

What does this mean?  If you mean a triangle whose vertices are those three, then the center of mass of this homogeneous triangle is just

[hide]${ \frac{1}{3}[(0, 0)+(3, 0)+(1, 2)}=(\frac{4}{3},\frac{2}{3})$.  This has to be where the medians intersect, because each median trisects the other two.  The intersection of the medians has to be the centroid, because if you draw a median, each of the two resulting regions provides the same torque: hence, the center of mass has to lie on each median.[/hide]

Guys, please do remember how the center of mass is defined.  It's much much easier than you're making it out to be.

[quote="sludgethrower"]Erm, you haven't specified whether the triangle is a plane, or a "wire" perimeter.[/quote]

This is true.  If a wire perimeter was intended, then it is

[hide] the point of concurrence of the lines which each go through the midpoint of one side and are parallel to the angles bisector from the opposite vertex[/hide]

thanks.
i meant to say:
triangle with vertices at (0,0) (3,0) (1,2).
and there was no specification as to the triangle being a plane surface of with a wire boundary.

i did indeed get (4/3, 2/3)... but the answer key to the set from which I was working said (4/3, 38/27)... erm wtf?

also, is there a way to compute the coordinates of the center of gravity of a planar region in xy-plane without having to use iterated integrals?


I had mine yesterday (8th) in MA. I'ts finally over :)

Ours was SUPPOSED to be yesterday, but it was postponed until the 29th or something.

There was already a post on this. :wink:

Mine is next Saturday, I think. The 15th.

Mine is the 15th also. When is the latest state competition? The 29th?

[quote="zolojetto"]mine is like not next saturday but the one after that...i beleive its the 23rd!!!
Im pretty nervous![/quote]

29th

OMFG the 29th....
can't.... wait....
gah so much pressure

I think a lot are on the 15th.  I know Minnesota and California are then.

Next year, March 14th (Pi Day) will fall on a Saturday.  I bet a lot of states will get to hold tournaments that day.  :)

March 14 ($ \pi$ day) :lol:  We get to miss school :D


geometry
3D geometry
analytic geometry
combinatorics proposed
combinatorics

Colour 7 out of the 8 corners of a cube red. A point $C$ in space can be turned red if there exist red points $A$ and $B$ such that $C$ is the symmetric image of $A$ through $B$.
Can the $8^{th}$ corner of the cub be turned red?

Regard the vertices of the cube as the points $(x,y,z)$ where $x,y,z\in\{0,1\}$, and asume the vertex which is not turned red initially is $(0,0,0)$. Notice that if we regard two points with the same reductions of the coordinates modulo $2$ as equivalent, then the operation described for turning $C$ red requires that it be equivalent to some red point $A$. Since we don't have any points equivalent to $(0,0,0)$ initially, we can't get any as we perform the described moves, so the answer is negative.


We were wacthing a video in class today, and it said the following about probability:

[u][b]Q:[/b][/u] On a game show, there are three boxes, Box 1, Box 2, and Box 3. Under two of them are goats and under one is a car. To win the car, you must choose the box with the car. Say, for example, you choose Box 3. Then Box 1 is revealed and it is a goat. Then you are given the chance to switch boxes. Is there a better chance of winning the car if you switch to Box 2?
[u][b]A:[/b][/u] Yes

I don't understand why. It says it has something to do with Bayes Theorem. I don't know what that is either. Could someone explain this to me?

This is an example of the Monty Hall Paradox.
Since there is no page on the wiki, I shall explain it :D

Once you choose what you have chosen, there is a $ \frac{2}{3}$ chance you didn't get the car. Even if one of the other choices that you didn't choose is eliminated, the remaining choice is left with a $ \frac{2}{3}$ probability of having the car. On the other hand, you have a $ \frac{1}{3}$ chance of winning the car on the first try. \[ \frac{2}{3}>\frac{1}{3}\] so it is advised to switch.

Another way to think about it. If there were $ 1000000000$ choices where $ 1$ of them has the car, and you chose one. The "host" then eliminates $ 999999998$ choices that are goats, wouldn't you switch to the other one?

If this confuses you, don't worry!

Yep, the reason why this is called a paradox is because it's pretty confusing and very baffling to common sense.

http://en.wikipedia.org/wiki/Monty_Hall_problem

And the strange thing is that if the host knows where the car is, then you get a $ 2/3$ chance. If he doesn't know where it is and guesses and opens a door with a goat, then you still have $ 1/3$ chance.


number theory
combinatorics solved
combinatorics

This is a mixture of number theory and combinatorics - not too difficult: :)

A positive integer is called monotonic if its digits are in non-decreasing order. Prove that for any positive integer n, there exists a monotonic integer, having n digits, which is a perfect square.

took a long while, but I found the other one as well

for all n>1:
(_ indicates where repeting parts start/stop)

1_6..6_7² = 2_7..7_8..8_89
6..6_7² = 4_4..4_8..8_9

so we can make all numbers now, odd and even amount of digits
just n=1 needs to be done, but that's no problem: just take 9,4,1 or even 0.

PS: i don't know if this is coïncidence that the numbers look a little parallel? is this some special trick?


Here is a very very beautiful problem that became classical for the oral examination at ENS:
   Let $a_n$ be a sequence of nonnegative numbers such that $\frac{a_1+...+a_n}{n}$ goes to 0 when n is large. Prove that there exists a set $A={n_1<n_2<...}$ of density 1 such that $a_{n_i}$ tends to 0 when $i$ is large. Prove also the converse.

[quote="harazi"]Prove also the converse.[/quote]
The converse is false: just put $a_n=n$ if $n$ is a power of $2$ (or square, or...) and $0$ otherwise. The direct implication is true and, indeed, a good exercise in analysis.  ;)

I think I've modified a little bit the statement. What if we say that there exists a set $A$ of zero density such that $ (a_n)_{n\in{N-A}})$ tends to 0? Is it ok now? Anyway, I will try to find the exact statement.

Sorry, harazi, how do you define "a set of density one"? Many thanks!

Regards,
jivko

We say that the set A of natural numbers has density a if the limit of $\frac{x_n}{n}$ is a, where $x_n$ is the number of elements of A smaller than n.

Let $s_n$ be the sum $a_1 + \cdots + a_n$.  Let $m_n$ be the mean $s_n / n$.  We are given that $m_n$ approaches 0.  Unfortunately, $m_n$ might not be decreasing.  To "force" it to be, define $M_n = \max(m_n, m_{n+1}, \ldots)$.  Then $M_n$ is decreasing and approaches 0.

Let $A$ be the set of indices $n$ such that $a_n \le \sqrt{M_n}$.  We need to prove that $A$ has density 1 and that $a_n$ for $n \in A$ approaches 0.  First, the density argument.  By the definition of $A$, we can show that the number of indices from 1 to $n$ not in $A$ is at most $s_n / \sqrt{M_n}$.  From there, we see that the density of the complement of $A$ is 0, which means that the density of $A$ itself is 1.

The claim that $a_n$ for $n \in A$ approaches 0 follows from the definition of $A$, because $M_n$ approaches 0.

For each $k$, the fraction $\frac {x_n(k)}n$ tends to $1$ as $n\to\infty$, where by $x_n(k)$ we denote the number of elements of the sequence with index $\le n$ which are $\le\frac 1k$. Let $(n_k)_k$ be an increasing sequence s.t. $\frac{x_n(k+1)}n>1-\frac 1{k+1},\ \forall k,\ \forall n\ge n_k$.

Now let $S_k$ be the set of terms of the sequence consisting of those terms which have index $\le n_k$ and which are $\le\frac 1k$, and let $S=\bigcup_{k\ge 1}S_k$. I believe this set (whose terms form a subsequence of our initial sequence) satisfies the requirements.

Does this work?

Edit: Looks like Ravi B beat me to it :)


percent
combinatorics unsolved
combinatorics

Hi everybody

Do you have any idea to solve this problem:

[b]Haw many ways can n married couples be seated about a round table in such a manner that there is always one man between two women and none of the men is ever next to his own wife?[/b]

Please tell me your ideas for this problem.

let us choose a woman. to arrange the extra n-1 women in a line!!! : (n-1)!.
to arrange the men between all the women is n!. all together n!(n-1)!.
this is without the prohibition that a man cannot seat next to his wife.
now, let's use the Inclusion-Exclusion Principle:
let the A(1) be the attribute: the 1st man seats next to his wife: (n-1)! to arrange the woman, (n-1)! to arrange the men. 2! the man and his wife can switch places beteen them. all together : (n-1)!(n-1)!2!.

let the A(2) be the attribute: the 2nd man seats next to his wife : the same as the first man
.........................................................................................
.........................................................................................
let the A(n) be the attribute: the 'n'th man seats next to his wife  : the same as the first man

let A(i,j) - the ith man seats next to his wife and the jth man seats next to his wife -
..........................
........................
A(i,j,k,......,n) - the ith man seats next to his wife and the jth man seats next to his wife and the kth man seats next to his wife ..........
and the nth man seats next to his wife

now use the principle:

|A| = n!(n-1)!. |A(i)|=n*(n-1)!(n-1)!2!.........

your answer will be: S = |A|-|A(i)|+|A(i,j)|-|A(i,j,k)||+.....(-1)^n|A(i,j,k,....,n)|

i'm one hundred percent not sure if the answer is correct and take no responsibility.

p.s. do anyone knows where i can find more questions of this type ( married couples be seated about a round table...)?

Each man may sit in right or left side of his wife but if the man of 2nd female is in right side of her then clearly the husband of 3rd female cant sit in left side of his wife . and this cause the different between A1 and A2.
so what you said cant be true.

:rotfl:  :rotfl:
I didnt know you asked this ,Arman ,But this was supposed to be your research subject ,So its not form just $\text{A PAPER}$.
Any way man ,you have got your gold medal ,So its not important anyway.
Well done for getting gold Mr.Fazeli .but you shouldnt had asked this problem here ,Should you  ?

Hope you would make the way to the team ,when will the courses begin?


Prove that $ \binom{n}{3} \equal{} \binom{i}{3} \plus{} \binom{k}{3} \plus{} i\binom{k}{2} \plus{} k\binom{i}{2}$ where $ n \equal{} i \plus{} k; i,k\geq2$.

[hide="Hint"]Rewrite as:

$ \binom{n}{3} \equal{} \binom{i}{3} \binom{k}{0} \plus{} \binom{i}{2} \binom{k}{1} \plus{} \binom{i}{1} \binom{k}{2} \plus{} \binom{i}{3}\binom{k}{3}$

What does this assert about the possible ways of picking 3 people from a group of n, if we partition the group into 2 parts, one of size i and one of size k?
[/hide]

Someone should provide the generalization suggested by the above hint.


I have been thinking till headache and still couldn't get to solve it:

Prove that the equation:

$ \sqrt {2 \minus{} x^2} \plus{} \sqrt [3]{3 \minus{} x^3} \equal{} 0$

has no real roots.

Thanks in advance!

This seems a little bit too obvious, but...

note that $ 0 \leq x \leq \sqrt {2}$, and then note that for those $ x$, $ 0\leq x^3 \leq 2\sqrt {2} < 3$. You then have a sum of two positive real numbers equal to zero. Contradiction.

Wow...never thought of using inequality. I have been expanding for ages. Thanks!

It would seem that someone forgot the case if

$ \minus{}\sqrt{2}\le x\le0$

Then

$ \minus{}2\sqrt{2}\le x^3\le0$

(Still a sum of two positives, so no solution)

Yep, forgot to write that one (the "more obvious" one...  :lol: )


I watched this totally drugged-out cartoon at http://www.ebaumsworld.com called "Shfifity-five" and i noticed something odd about their system of math.

I think that "Shfifty-five" is NOT equal to the real 55, let's call it sh55. There are little things the characters say like "5 + 5 is sh55!" Since 5 + 5 also equals 10, then sh55 = 10.

I remember from a math problem i did a while back that 1+2+3+4+5+6+7+8+9+10 = 55. So my conjecture is that if you put a "sh" in front of a number, to find the number of terms you must add in the sequence 1+2+3... to equal that number. So sh55 = 10, sh36 = 8, and sh210 = 20. (I know this because at the beginning the stick-guy counted ten numbers.)

In equations:

sh(x) = y     if...    x = y(y+1)/2

So my problem is, is there an exact answer for numbers between exact sums of the squence? For example we know that sh55 = 10, but what is sh56? sh57? sh42? What is sh1000?

Just something i thought would be fun to figure out. Turns out those idiots who made the cartoon might have mistakenly brought out a really cool math formula.

I got something like that... Its a function sh(x). so i figured it out like

sh(x) = sqrt(2x + 1/4) - 1/2

Heres another thing ive been trying to figure out.

Square-rooting something is the same as raising it to the power of 1/2.
Is there any way to simplify (2x + 1/4)^1/2?

[quote="Hexaditidom"]
Square-rooting something is the same as raising it to the power of 1/2.
Is there any way to simplify (2x + 1/4)^1/2?[/quote]

Hmm.. I'm not quite sure what you're looking for, but the only thing I can think of doing is to factor out 1/4: [(1/4)(8x+1)]^(1/2).  Taking the positive root, we have (1/2)[(8x+1)^(1/2)], or  [(8x+1)^(1/2)] / 2.  I don't really think this is any simpler, though ;)

Err.. I should learn TeX.


Euler
analytic geometry
graphing lines
slope

The uniform scalene triangular lamina ABC is at rest in equilibrium freely suspended from a point K by three equal light inextensible strings KA.KB,KC.

Prove that the Euler line of the triangle ABC is a line of greatest slope of the plane ABC

Wow, this problem is sooo nice :). I found the solution but I won't tell unless you badly need it. Try it by yourself, it's really simple.


Brute force method:

[hide]He must mess up on 6 or less lines.  The probability of messing up on exactly 6 lines is $\left(\frac13\right)^{6}\cdot\left(\frac23\right)^{14}\cdot\binom{20}{6}$.
Five lines:
$\left(\frac13\right)^{5}\cdot\left(\frac23\right)^{15}\cdot\binom{20}{5}$
Four lines:
$\left(\frac13\right)^{4}\cdot\left(\frac23\right)^{16}\cdot\binom{20}{4}$
Three lines:
$\left(\frac13\right)^{3}\cdot\left(\frac23\right)^{17}\cdot\binom{20}{3}$
Two lines:
$\left(\frac13\right)^{2}\cdot\left(\frac23\right)^{18}\cdot\binom{20}{2}$
One line:
$\frac13\cdot\left(\frac23\right)^{19}\cdot20$
Zero lines:
$\left(\frac23\right)^{20}$

Add them up to get some big fraction I will figure out later  :D [/hide]

[quote="bpms"]There is a $\frac13$ chance I mess one line of latex. If I do twenty lines, what is the probability that I mess up less then one third of the lines?Express your answer as a percent.[/quote][hide] You would have to mess up only 0, 1, 2, 3, 4, 5, or 6 lines. The probability of this is $\binom{20}{0}*(\frac{1}{3})^{0}*(\frac{2}{3})^{20}+\binom{20}{1}*(\frac{1}{3})^{1}*(\frac{2}{3})^{19}+\binom{20}{2}*(\frac{1}{3})^{2}*(\frac{2}{3})^{18}+\binom{20}{3}*(\frac{1}{3})^{3}*(\frac{2}{3})^{17}+\binom{20}{4}*(\frac{1}{3})^{4}*(\frac{2}{3})^{16}+\binom{20}{5}*(\frac{1}{3})^{5}*(\frac{2}{3})^{15}+\binom{20}{6}*(\frac{1}{3})^{6}*(\frac{2}{3})^{14}$, which I don't feel like figuring out. :D[/hide]

Um, unless someone finds a clever solution, this is officially a target round problem :D.

[quote="lotrgreengrapes7926"][quote="bpms"]There is a $\frac13$ chance I mess one line of latex. If I do twenty lines, what is the probability that I mess up less then one third of the lines?Express your answer as a percent.[/quote][hide] You would have to mess up only 0, 1, 2, 3, 4, 5, or 6 lines. The probability of this is $\binom{20}{0}*(\frac{1}{3})^{0}*(\frac{2}{3})^{2}0+\binom{20}{1}*(\frac{1}{3})^{1}*(\frac{2}{3})^{1}9+\binom{20}{2}*(\frac{1}{3})^{2}*(\frac{2}{3})^{1}8+\binom{20}{3}*(\frac{1}{3})^{3}*(\frac{2}{3})^{1}7+\binom{20}{4}*(\frac{1}{3})^{4}*(\frac{2}{3})^{1}6+\binom{20}{5}*(\frac{1}{3})^{5}*(\frac{2}{3})^{1}5+\binom{20}{6}*(\frac{1}{3})^{6}*(\frac{2}{3})^{1}4$, which I don't feel like figuring out. [/hide][/quote]

When using exponents with more than one digit, put braces around them like \frac13^{14}.  Also, use \cdot or \times for * because it looks much nicer  :)

This is too complicated.  It requires too much finger pressing on the Calculator.  Someone else can do it.

If the problem was, what is the probability that I mess up on an even number of lines or an odd number of lines for Latex, I would get it right without using a calculator.

Let me make it team then. That way you have 8 minutes not 3. :D


The formula can be used to determine the degree measure $ a$ of each interior angle of a regular $ n$-sided polygon. What is the degree
measure of each interior angle of a regular 12-sided figure?
\[ a \equal{} 180 \minus{} \frac{360}{n}\]

$ 180\minus{}\frac{360}{12}\equal{}180\minus{}30\equal{}\boxed{150}$

or, each ext angle is $ \frac{180}{12}\equal{}30$, so each int is $ 180\minus{}30\equal{}\boxed{150}$


Let ABC be a triangle. Prove that
  4(tg(A/2)+tg(B/2)+tg(C/2))*(cotgA+cotgB+cotgC)<=(1/(sinA)+1/sinB+1/sinC)^2

tan(A/2)=(-1+ \sqrt (1+tan^2A))/tanA = (-1+(1/cosA))/tanA=(1-cosA)/sinA

So LHS=4((1-cosA)/sinA + (1-cosB)/sinB + (1-cosC)/sinC)(cosA/sinA + cosB/sinB + cosC/sinC)

<=((1-cosA)/sinA + (1-cosB)/sinB + (1-cosC)/sinC + cosA/sinA + cosB/sinB + cosC/sinC)^2

=(1/sinA + 1/sinB + 1/sinC)^2, as desired.

(Because of 4ST<=(S+T)^2 by AM>=GM)

Thanks.


Let be $ 0<a,b<\frac{3}{2}$ . Prove that :

$ \sqrt{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{a\plus{}b\plus{}3}}\plus{}\frac{1}{\sqrt{a}\plus{}\sqrt{b}}\le \frac{1}{\sqrt{a}}\plus{}\frac{1}{\sqrt{b}}$

Is that $ 0<a,b \le \frac{3}{2}$

[hide="A Bash"]
$ \sqrt{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{a\plus{}b\plus{}3}}\le\frac{1}{\sqrt{a}}\plus{}\frac{1}{\sqrt{b}} \minus{} \frac{1}{\sqrt{a}\plus{}\sqrt{b}}$

From our condition we have;
$ \sqrt{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{a\plus{}b\plus{}3}} \le \sqrt{\frac{a\plus{}b}{ab}\plus{}\frac{1}{2(a\plus{}b)}}$

And we have to show that
$ \sqrt{\frac{a\plus{}b}{ab}\plus{}\frac{1}{2(a\plus{}b)}} \le \frac{\sqrt{a}\plus{}\sqrt{b}}{\sqrt{ab}} \minus{} \frac{1}{\sqrt{a}\plus{}\sqrt{b}}$

Squaring both sides gives
$ \frac{a\plus{}b}{ab}\plus{}\frac{1}{2(a\plus{}b)} \le \frac{a\plus{}b}{ab} \plus{} \frac{1}{(\sqrt{a}\plus{}\sqrt{b})^2}$

Which becomes;
$ a\plus{}b \ge 2\sqrt{ab} \Longrightarrow (\sqrt{a}\minus{}\sqrt{b})^2 \ge 0$ which is true by the trivial inequality

Equality occurs when $ a\equal{}b\equal{}\frac{3}{2}$[/hide]


calculus geometry trigonometry MATHCOUNTS AMC AIME USA(J)MO

[quote="EverLane"]I just want to know what grade everybody's taking it and what classes they had before it, yes it's the scheduling time of year again[/quote]

hehe i voted 9th because im takingit unofficially right now (im in 8th grade taking alg 2 and i didn't take geometry yet)...i know some basic trig and some basic calculus but i should be done basic calculus by the end of the school year hopefully  :maybe:

I'm currently planned to take it in 11th, but I'm trying to convince the school to let me take it in 10th. The school board is surprisingly rigid though: I know of one person in the district who will be taking it in 10th. I'm currently taking hnrs alg 2 in 9th.

7th for AB and 8th for BC, which left me with a lot of math classes not to take in High school- By the time I graduate I will have had math classes 2 trimesters out of the last 9, and neither of those classes really covered a lot of new material.

Mmm... I'm not in US...
But in my high school in Macau, the science students must learn calculus when they are in 12th grade, and the arts students just learn differentiation. The textbook we (science students) used is originally for F5 (11th grade) students who take the Additional Mathematics. I think the level is about the AP Calculus AB.

Well, look at the posted time of my previous post. :roll:
There are some error about the time system...

[quote="Kent Merryfield"]7h3.D3m0n.117: You've made several posts recently in calculus threads that make me wonder if perhaps you're pushing too hard to "get through" calculus too early. Are you really sure you're ready for it?[/quote]

Well there are several reason why I want to take calculus

1) A head start always help
2) i'm not actually looking to graduate but to at least understand the stuff i'm going to be learning in 11th and 12th grade
3) algebra 2 is extremely boring so im looking for more challenging material
4) my family is pretty much math oriented and getting a B in calculus or trig would certainly make me quite umm well i won't go there

I'm considering (as a soph next year) to take precalc (math analysis) next year then ap calc ab junior, then I thought I might go for the precalc in the summer then go straight for AP I'm just really shaky on my success rate, all well go figure that's life

I'm taking it as a freshman.  I could have taken it this year, but I decided that it was better to take a real class for a change and do my own math (AoPS courses).

I will do it in grade 8 because I am doing grade 10 math in grade 7.

I'm a senior now, but I took AB/BC Calc as a junior, with a small class of 3 juniors (including me) and about 12 seniors, so I'm quite impressed that so many people are able to take calculus even earlier.  I was lucky enough to have a magnificent teacher (11/13 students who took the test got 5s and she's never had a student get lower than a 3) and I really think that makes all the difference.

12th grade. It's the only year we can take it at our school.

[quote="the future"]Calculus is just a bunch of memorization of formulas/methods.[/quote]

By this, I sincerely hope you are referring to AP Calculus, and not to the subject  :wink:

They made me take Calculus in 10th grade... I wanted to take 2 science courses and save calc for junior year when they would offer BC, but they were like NO YOU HAVE TO TAKE ONE MATH ONE SCIENCE so I took Calc AB lol... it was a party with the seniors though, it was lots of fun! The AP test was fun too, I was sleeping half the time. But for real, calculus is much more interesting when you're not doing it in a [b][color=red][X][/color][/b]ing lame-[b][color=red][X][/color][/b] high school class.

If you know what's good for you, make sure not to skip multivariable calculus when you go to college unless you're a god.

If you want something with something totally new, try a theoretical approach to Linear Algebra :) it will bring about new vistas on math as a whole.

I would recommend [i]Linear Algebra Done Right[/i] by Sheldon Axler.

I'm taking BC right now as a freshman.  It's waaaay too easy.

I'm taking Algebra II now in 8th grade and Pre-calc in 9th. Then I'll take Stat in 10th, Calc BC in 11th and Multivariable calc in 12th.

Well our math class is moving so fast that we will finish BC Calculus before 1st semester of 12th grade ends, then we will be doing Multivariable. But this sucks because it doesn't count on our transcript and we'll be screwed for the AP exam

That's kind of a bad idea. There's a reason it's a year-long course. Admittedly, we finished ahead of schedule too (end of March) but covering it in less than a semester...  :ninja:

I don't know what all these AB-BC thingies are, but I would be taking AP Calculus in 9th, but it requires Pre-Calc as a prerequisite, and my stupid principal won't let me take Pre-Calc this year (8th), even though I took Algebra 2 the year before.
So for me, not until 10th grade.  :(  :(  :(

Calc AB is where you learn most of the methods of calculus.

I believe Calc BC is where you integrate the methods in applications, mostly in physics. (As well as learn some more advanced techniques but I think it's mainly application).

[quote="erabel"]even though I took Algebra 2 the year before.[/quote]
then what are you taking now?

Calculus AB covers about two quarters worth of a typical college sequence: limits, differentiation, and integration. Calculus BC is about a year's worth, and sequences/series are the main additional topic. Many high schools do the AB as a year class, and the BC as a second year; since the additional BC material is easily covered in half of the year, there's a lot of room to branch out and cover material not on the AP test.
If you go to a semester school in college, AB will give you one semester worth and a leg up on the second. At a quarter school, it's two quarters, but a little thin at the end. The AP test is a decent standardized measure, and transfers well. If you have a passing score, everyone will know what it means.

[quote="the future"]Since Im pretty sure people that qualify for USAMO can all get 5s on the AP Calc test but not all people that get 5s on the AP Calc test can pass the AMCs[/quote]It's not that simple. I qualified for the USAMO as a 9th grader, and I wouldn't have passed the AP then- because I simply didn't know calculus yet. I got my 5 the next year, after a full year's class. Classroom material may be "easy" compared to contests, but only if you actually have the class.
As for the second part? Absolutely true. My old high school has more 5s on the AP than AIME qualifiers every year.

Please ignore the stuff in small font. I'm just ranting...

I'm guessing a five is a perfect score...or at least pretty good.

[size=42]....Let's just say that my parents never wanted to accelerate. (I could've taken a Pre-Algebra/Algebra I) class in 4th grade.....all we learned about were ratios and solving for x with...get ready for this...really brace yourself...DECIMALS!! And....FRACTIONS!!!

...-_- I was bored to tears. I lost interest in math for a brief time. It was dismal.

And because of that, my district doesn't think I can accelerate...so[/size] I can't take AP Calc AB or BC 'till....*goes to check course book*...erm. Until 11th grade. And, to add to that, we can't take AP Calc AB if we got a grade higher than a B. Hmmm....maybe the course is different than the standard AP Calc AB. The course also includes 'Analytic Geometry'. Not entirely sure what that is, but whoo. :D

At my (future) high school, they don't really seem to differentiate between the two, other than one' harder and one's easier. (I.e., one's for those who aren't 'accelerated' and one for those who are.)

.....

I may just learn Calc by myself, as I currently am. I also am not allowed to take Discrete Math or AP Statistics until 12th grade. :wallbash_red:

[size=42]And the delicious irony of it all now? My parents now want me to do high school in 2 or 3 years instead of 4, if I'm "really that bored". -_-.....where was this when I was allowed to skip practically every grade in elementary school? Anyone?[/size] :wallbash:

uhh dont worry about it, school math is pretty useless anyways.

if you are really interested in math and want to "accelerate", you should participate in various math competitions, like MATHCOUNTS (i assume you are still in middle school), AMC/AIME/USAMO, ARML, and any other local or regional competitions i have not mentioned yet, and also research...and you can do other competitions, such as USNCO or something incase you feel bored in science...

my school doesnt allow us to take Calc BC until 11th grade also, but if you can "pass" the pre-calc test you can get into it during 10th grade (and 9th if you are really really good, like these two 8th grade USAMO/MATHCOUNTS national qualifiers)...

but what do you expect, its a public school.

[quote="lifesapain"]Heck no, we don't do it all of it in one semester. That'd be ridiculous. We start BC Calculus sometime in junior year, probably 3rd quarter.[/quote]
My school does the same thing.  We speed through precalc, and then start the calculus curriculum in March.

My school doesn't allow you to take it until 11th grade, I think.
Calculus isn't too hard. Lots of memorization though.


real analysis
calculus
integration
limit
inequalities
derivative
function

Suppose $f \in L(\mathbb{R}^1)$ and $F(x) = \int_0^x f(t)dt, x\in \mathbb{R}^1$. If $F(x)$ is increasing on $\mathbb{R}^1$, show that $f(x)\geq 0$, a.e., $x\in \mathbb{R}^1$.

I'm assuming the $\sigma$-algebra here is the usual Borel $\sigma$-algebra?

Take $S=\{x\in\mathbb R|f(x)<0\},\ S_n=\{x\in\mathbb R|f(x)<-\frac 1n\}$. Assume that for some $n,\ S_n$ contains a bounded subset $E$ with positive measure. Then we can find a sequence of unions of disjoint intervals $(\bigcup_{m\ge 1} I_k^m)_k$ s.t. $\bigcup _{m\ge 1}I_k^m\to E$ as $k\to\infty$. We then have $-\frac 1n\cdot m(E)>\int_E f dx=\lim_{k\to\infty}(\sum_{m\ge 1}\int_{I_k^m}f dx)\ge 0$, contradiction (the last inequality holds because the integral is $\ge 0$ on each interval, and the equality before the last inequality holds because $f$ is $L$).

I'm sorry for any stupidities I might say. As I have already mentioned in another thread, I know next to nothing about this subject (even though I have the feeling these problems are among the easiest one can find on the topic)  :blush:

There is the Lebesgue theorem on the differentiation of the integral which holds that $F'(x)$ exists and equals $f(x)$ almost everywhere. Since the derivative of an increasing function, when it exists, must be nonnegative, that gives us what we need.

Kent Merryfield, can you avoid using Dini derivatives and the differetiation of Lebesgue's integral?
This problem is proposed as an exercise of the section of "the relation of lebesgue's integral and riemann integral"

We have to prove that for every measurable set $E$
 $\int_{E}fdx \geq 0 $ (*)
 then the result will follow

 Firstly : (*) holds for every open interval $(a,b)$
 Since every open set is disjoint union of at most countable such open intervals, (*) holds for every open set
 If $G$ is a $G_{\sigma}$ set, which means $G$ is a decreasing limit of open sets, then (*) also holds for $G$ ( due to the continuity of Lebesgue integral )
 Finally, every measurable set $E$ is infact a $G_{\sigma}$ differing in a zero measure set, we have (*) holds for $E$.

 Now let $E$ be a the set where $f$ is negative. Since the integral of $f$ over $E$ is nonnegative, we must have $m(E)=0$

If I am not mistaken, that's what I did.. :)

[quote="Myth"]I think, Grobber used the same argumentation as you, dickchimney.[/quote]

 Sorry, I was too lazy to read :blush:  :D
 So is it solved or still not?


As an alumnus of the Upstate NY team (I grew up in Batavia, and went to ARML in 1988), I'm curious how the team is selected these days.

Two decades ago, it seemed really ad hoc: something like 10 or 11 of the 15 members of the team were from the Rochester area, and none(!) were from the Buffalo area (unless you count me, but I got on the team through my Rochester-area connections).  Is this still the case, or is the state more thoroughly represented these days?

(Now of course I work for AoPS here in San Diego.  I think I'm going to miss snow...)

Sorry, I'm a stupid Albany area person, and can't tell the difference between Buffalo and Rochester. ;-)  I think most of the people on this forum are from the eastern/central part of the state, but that's not entirely representative.  (There may have been less people from out west this past year because to cut costs we had a single bus that started in Albany and didn't make any stops west of Corning.)  Still, there's definitely more representation from the eastern part of the state than back then.

As for team selection, you'd have to ask Tom, because he has access to the coaches' secret formula ;-)  I believe it involves a combination of AMC/AIME scores and participation/performance in other math competitions.

[quote="Alison"]Nope, none of the full coaches responded to my e-mail to the yahoogroup (are the coaches even on that list?).[/quote]

Mrs. Ohlerich should be, I think, since she occasionally sends emails via it.

I would check the list right now, but that site is considered a "Chat Site" and is thus blocked from work.

[quote="intrepia"]
Mrs. Ohlerich should be, I think, since she occasionally sends emails via it.[/quote]

Yes, she is.  I didn't see any of the other coaches there.

[quote="tom11784"][quote="Alison"]As for team selection, you'd have to ask Tom, because he has access to the coaches' secret formula ;-)  I believe it involves a combination of AMC/AIME scores and participation/performance in other math competitions.[/quote]

I do?  Funny; I didn't know that.
However, Mr. Reuter plans on re-writing Mr. Schneider's old selection formula, and perhaps I can help influenece that.

But I believe it was something along the lines of:
AMC (better of last 2 years) + 10*AIME + k1*previous_NYSML_scores + k2*previous_ARML_scores + k3*NYSML_appearances + k4*ARML_appearances for some k1,k2,k3,k4

However, this year to qualify all you had to do was send in an app and the cost of attending, but they determined the team of 15 from the 22? attendees using that formula.

Do any of the "full" coaches know about this forum?
They would be the ones to ask about the full formula.  Also, does anyone have Mr. Neuter, I mean Reuter's email?  If so, could you please PM it to me or email me at zinkt@rpi.edu[/quote]

I don't have any clues on the formula ... but I'd say that excepting this year we've been fairly representative. One of the busses used to start down near Westchester county and another bus would start near Buffalo. People from the very Northern parts of NY would meet the southern bus @ Binghamton.

Tom: Isn't his e-mail the one that looks like it should be his on the coach e-mail list?

I know that my first year, Mr. Schneider said that the formula involved curving AMC 10/12 scores so that scores from the different tests would be comparable.  That's all I know.

[quote="tom11784"][quote="Joecom3"]Tom: Isn't his e-mail the one that looks like it should be his on the coach e-mail list?[/quote]

Yes - except there are like 12 emails addys on there and I can only pick out 4 or 5 as specific people.[/quote]

IM me when you get back in ...

wow - this is the first forum i've been on that i can't delete my own posts if they've been replied to

uh.. I think that the first placing A and B team get to go. Last year(my first) I was on the B team. Too young to go to ARML, all I could do is sulk at home. :P


the formula for Work.

"Three postal workers can sort a stack of mial in 25 minuites, 25 minuites and 50 minuites respectively. Find how long it takes them to sort the mail if all three work together."

I was wondering if I use the formula of work to solve this problem.

$ t_1\equal{}25'\equal{}A$ and $ t_1\equal{}50'\equal{}2A$
$ t_2\equal{}25'\equal{}A$ and $ t_2\equal{}50'\equal{}2A$
$ t_3\equal{}50'\equal{}A$ and $ t_3\equal{}50'\equal{}A$

$ (t_1\plus{}t_2\plus{}t_3)50'\equal{}5A$
$ (t_1\plus{}t_2\plus{}t_3)50'/5\equal{}A$
$ (t_1\plus{}t_2\plus{}t_3)10'\equal{}A$

So the answer is 10 minutes

I can not andertand what you can find with the work formula $ A\equal{}F*l$ ? :huh:

i have no idea what the formula for Work is
because i have never heard of it
but i can help u solve it

answer and solution
[hide]
im naming the workers A,B, and C, to avoid confusion

A can sort the mails in 25 minutes, so in 1 minute, he can sort 1/25 of the mails.
B can also sort the mails in 25 minutes, so same as A, 1/25.
C can sort the mails in 50 minutes, so in 1 minute, he can sort 1/50 of the mails.

add A and B's workforce for 1 minute. Together, A and B can sort through 2/25 of the mails in 1 minute. add C's workforce to that.
1/50 + 2/25 = 5/50 = 1/10
so A, B, and C can work through 1/10 of the mails in 1 minute.

therefore, A, B, and C sort the mails in 10 minutes when they work together.[/hide]

You don't need formula for this kind of problem.

All you have to do is what spiders_hate_chestnut did.
That's the way that I would suggest, but if you insist, use formula if that is more comfortable. :lol:

I didn't use formula to solve that :mad:

I didn't say that you did... I was just telling EDgar what I would do...

If that offended you, I am sorry...

The formula for work is $ Work \equal{} Force \cdot Distance$ or $ Joules \equal{} Newtons \cdot Meters$ This is a physics formula though

The way of doing this is the way that Elio(n) and Spiders_hate_chestnut did. The reason you can't use the work formula is that when the workers are not going a distance.

The work formula (if you are referring to the one from physics) is completely irrelevant.  "Work" as defined in physics is very specific and not necessarily related to the common definition of "work."  The only "formula" you need to solve this problem is that

$ \text{Rate of mail\minus{}sorting} \equal{} \frac{ \text{Number of stacks sorted} }{ \text{Time taken to sort} }$

and the only concept you need is that when people work together, their rates are added.

An elementary way to think about this problem:  In $ 50$ minutes, the first postal worker sorts $ 2$ stacks, the second sorts $ 2$ stacks, and the third sorts $ 1$ stack, so all three workers working together can sort $ 5$ stacks in $ 50$ minutes, or $ 1$ stack in $ 10$ minutes.


I am writing a program, and have run into a dead end and need some help to get out.

I simplified it this far, but have no idea where to go.

I need to solve for X but have no idea how to go about it because of the Cos.  My calculate can solve it when i plug in numbers, but i need to make it ino X= with the variables in it still.  Here is the equation, any help is greatly appreciated.
 X = A + B * Cos( C * X )

There is no "nice" solution to that equation, unfortunately.

Is there a "ugly" solution? or do i need to attack this problem from a completely different angle?

It's a program, right?  Brute-force and approximate the solution, there isn't really a nice answer.  If you've already learned calculus, try

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

Or:  tell us where the problem came up.  There might be another way to do what you want without solving this equation.

Yeah, i have decided to just brute force it, it will prolly be more efficient than doing that many equations anyway.  Thanks for the help guys

There is an ugly solution in that you can numerically approximate solutions.  That's really the best you're going to do, I'm afraid.

Another option may be taylor series, I haven't looked at it too much, but it may help. If not in all cases, then definitely for certain specail values of A, B, and C.

Also, the special case where C=0 is easy to solve.


Let $ a,b,c \in R, a\minus{}2b\plus{}3c\equal{}16$. Find min:
$ F\equal{}2(a^2\plus{}b^2\plus{}c^2)\minus{}4(a\plus{}b\plus{}c)\plus{}15$

Let $ a \minus{} 1 \equal{} A,\ b \minus{} 1 \equal{} B,\ c \minus{} 1 \equal{} C$ we have $ A \minus{} 2B \plus{} 3C \equal{} 14,\ A,\ B,\ C\in\mathbb{R}$ and $ F \equal{} 2(A^2 \plus{} B^2 \plus{} C^2) \plus{} 9.$
Cauchy-Schwarz gives $ \{1^2 \plus{} ( \minus{} 2)^2 \plus{} 3^2\}(A^2 \plus{} B^2 \plus{} C^2)\geq \{1\cdot A \plus{} ( \minus{} 2)\cdot B \plus{} 3\cdot C\}^2$
$ \Longleftrightarrow 14(A^2 \plus{} B^2 \plus{} C^2) \geq 14^2$, yielding $ A^2 \plus{} B^2 \plus{} C^2\geq 14.$
$ \therefore F\geq 2\cdot 14 \plus{} 9 \equal{} \boxed{37}$. The equality holds when $ (1,\minus{}2,\ 3)\parallel{}(A,\ B,\ C)$ and $ A\minus{}2B\plus{}3C\equal{}14 \Longleftrightarrow (A,\ B,\ C) \equal{}(1,\minus{}2,\ 3)$, $ (a,\ b,\ c)\equal{}(2,\minus{}1,\ 4).$