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Here is a very funny problem: find the exact value of the product of all numbers $\frac{n^4-1}{n^4+1}$ with $n>1$.

harazi, is it [hide]$e^{-\frac{180}{\pi^4}}$ -it seems really strange[/hide]

i wouldnt like to bring more confusion but something...i think could be done if
we tried to find only if the formula given is correct
so we start from back to front of problem to try to understand...
so we see that

[img]http://en.wikipedia.org/math/d/1/3/d13c47fdf7a104288bfc16b39dd00ba0.png[/img]

so if we make the replacement for x in this case that nasty exponent with $\pi^4$

now also we find in wikipedia this very strange formula(for me)

[img]http://en.wikipedia.org/math/9/1/0/9101c6a84ba1da3ebf07da7df0e1ab70.png[/img]
that big P there stands for prime number set i think

if we square that formula and multiply both sides with -5 we find the desired exponent...
now we can try to substitute this form of $-\frac{180}{\pi^4}$
in the power series of $e^x$ yielding some kind of monstruosity in wich there may be
some kind of order...

oh...i feel there is something correct about what i said because considering that product in
the hypothesis of the problem implies some simplifactions of that big fraction and then it would only resume to some prime numbers(actually all prime numbers wich is also what
that weird formula for $\documentclass \frac{6}{\pi^2}$ refers too)

there seems to be a strong connection between the formulas
but i cannot go further than that

Maple tells me it is  [hide]$\frac{\pi}{2}\frac{\sinh \pi}{\cosh^2 \frac{\pi}{\sqrt{2}} - \cos^2 \frac{\pi}{\sqrt{2}}} \approx 0.848054$[/hide]
without explaining how it got there  :?

but $e^{-\frac{180}{\pi^4}}\approx6,31142$
?!?!
one of you people posted a result
maple tells another
what is the truth then ?

Of course it was my error- the product al least can't be greater then 1.

ok seems it wasnt that complicated as i thought...
i hope n is an integer number....

$P=\prod_{n=1}^{\infty} \frac{n^4-1}{n^4+1}= \frac{\prod_{n=1}^{\infty} (n^4-1)}{\prod_{n=1}^{\infty}(n^4+1)}=\frac{\prod_{n=1}^{\infty} (n^2-1) \prod_{n=1}^{\infty} (n^2+1)}{\prod_{n=1}^{\infty}(n^4+1)}=$

$\frac{1}{\prod_{n=1}^{\infty}(n^4+1)} \dot$ $\prod_{n=1}^{\infty} (n^2-1) \cdot \prod_{n perfect square}^{\infty} (n^2+1) \cdot \prod_{n not perfect square}^{\infty} (n^2+1)$

now this product $\prod_{n perfect square}^{\infty}$ reduces with the
nominator $\prod_{n=1}^{\infty}(n^4+1)$ .
and what remains is
$P=\prod_{n=1}^{\infty} (n^2-1) \cdot \prod_{n not perfect square}^{\infty} (n^2+1)=+\infty$

is this true ?

its unexpected because all fracions in the product are of form $\frac{n^4-1}{n^4+1} <1$
so the product of subunitary fractions should be some $P<1$ and it tis verry weird
that tough this should happen for  the product to infinity int gives $P=\infty$

I think you made a pretty big mistake since $\prod \frac{\textrm{blah}_1}{\textrm{blah}_2} \neq \frac{\prod \textrm{blah}_1}{\prod \textrm{blah}_2}$. You can't really have $\displaystyle \frac{\infty}{\infty}$. From this problem and your own remark you should deduce why that isn't defined.

perfect_radio...now what do you think of this http://www.mathlinks.ro/Forum/viewtopic.php?t=80425 ???
is that correct there ?

...and i was checking these messages for help on my problem :roll: .   I just plugged in the prob in the spx2's link in my 89 (yes, for the first time, i didn't know the answer beforehand) and it indeed said the value was $\frac{1}{2}$.  Lemme try out this problem...

Why don't we try to find $\prod_{n>1}\frac{n^3-1}{n^3+1}$ first. That's a lot easier ;) On the other hand, $\prod_{n>1}\frac{n^2-1}{n^2+1}$ is a bit trickier.

Wherever you're getting this from, I think I made some progress on the "easier" product, here:

Let $y=\prod_{n=2}^{\infty} \frac{n^3-1}{n^3+1}$
Note that the sum and difference of cubes factors. Taking the natural log of both sides,
$\ln {y} = \ln {(2^3-1)}-\ln{(2^3+1)}+\sum_{n=3}^{\infty} \ln{(n-1)}+\sum_{n=3}^{\infty}\ln(n^2+1+n) -\sum_{n=3}^{\infty}\ln{(n+1)}-\sum_{n=3}^{\infty}\ln{(n^2+1-n)}$
(I don't want to factor out the $2^3-1$ from the $ln$ because then there would be an $ln(2-1)$ term, which is zero, and i don't want to deal with that.) Now,
$\ln{y}= \ln{\frac{7}{9}}+\ln{(3-1)}+\left (\sum_{n=4}^{\infty} \ln{(n-1)}-\sum_{n=3}^{\infty}\ln{(n+1)}\right)+\sum_{n=3}^{\infty}\ln(n^2+1+n)-\sum_{n=3}^{\infty}\ln{(n^2+1-n)}$

Of course, the stuff in the brackets cancels each other out so
$\ln{y}=\ln{\frac{14}{9}}+\sum_{n=3}^{\infty}\ln(n^2+1+n)-\sum_{n=3}^{\infty}\ln{(n^2+1-n)}$

It kinda looks simpler, especially if you take raise each side as a power of $e$ to undo the $ln$.  Do with this half-baked result as you wish...

It seems i have an idea to solve the original problem:
From the Euler formula $sin(\pi \sqrt{x})=\pi \sqrt{x} \prod_{n=1}^{\infty}(1-\frac{x}{n^2})$ (*).Now if we put in (*) $-x$ instead of $x$ we get $sin(i\pi \sqrt{x})=i \pi \sqrt{x} \prod_{n=1}^{\infty}(1+\frac{x}{n^2})$ (**) and let's mulltiply (*) and (**) so that $sin(\pi \sqrt{x})sin(i \pi \sqrt{x})=i \pi^2 x \prod_{n=1}^{\infty}(1-\frac{x^2}{n^4})$ (***). And it is left to put $ix$ in (***) and divide (***) by this equality  to get $\frac{sin(\pi \sqrt{x})sin(i \pi \sqrt{x})} {sin(\pi \sqrt{ix})sin(i \pi \sqrt{ix})}=-i \prod_{n=1}^{\infty}\frac{n^4-x^2}{n^4+x^2}$ and to this end it is left to take $x \to 1$ to get the required value if haven't done stupid mistakes(but it seems the idea works).

[quote="perfect_radio"]That problem is a classic: See what happens with $(n+1)^2-(n+1)+1$.

On the other hand, is somebody gonna solve the original problem? It's too hard for me  :([/quote][url=http://www.amazon.com/gp/product/1568811365/ref=nosim/103-2611228-7327014?n=283155]Here[/url] is a discussion. Click on "Search Inside" and continue clicking until you get to formula (1.12) in "Two Products". The solution involves product representations of the $\Gamma$-function.

eugene, your "solution" (i'm thinking about calling it "work" because I don't see a final answer) looks fine from here but why would you take $\lim_{x \to 1}$ what would your final result simplify to?

[quote="eugene"]Hpe,Is my 16'th post wrong?[/quote]Formula (***) seems OK. I 'm not sure if you can cancel the $n^4$ terms in the next step. In any case, in the final result you'd have to divide by $\frac{1-x^2}{1+x^2}$ and then take the limit as $x \to 1$.

Well... http://mathworld.wolfram.com/InfiniteProduct.html gives closed forms but no proof, just a citation.  Borwein [i]et al.[/i] 2004, pp. 4-6.

(I fail to see the humor.   :huh: )

I'm taking a class with Dr. Borwein this semester :)

Using eugene's idea of post #16, it' s possible to solve this problem completely and in fact generalize it. Here's eugene's post, slightly modified:
[quote="eugene"]Use the Euler formula $\sin(\pi z)=\pi z \prod_{n=1}^{\infty}(1-\frac{z^2}{n^2}) = \lim_{N\to \infty} \pi z \prod_{n=1}^{N}(1-\frac{z^2}{n^2}) = \lim_{N \to \infty}P_N(z)$. (*), where $P_N(z) = \pi z \prod_{-N \le n \le N, n \ne 0}\frac{n-z}{n}$.[/quote]
This is true for all complex $z$.  Since $\frac{n^4-z^4}{n^4 + z^4} = \frac{(n-z)(-n-z)(n-iz)(-n-iz)}{(n- e^{i \pi/4} z)(-n-e^{i \pi/4}z)(n-e^{i 3\pi/4}z)(-n-e^{i 3\pi/4}z)}$, we obtain
\[ \prod_{n \ge 1}\frac{n^4-z^4}{n^4+z^4} = \lim_{N \to \infty} \frac{i P_N(z) P_N(iz)}{P_N(e^{i \pi/4}z)P_N(e^{i 3\pi/4}z)} = i \frac{\sin(\pi z) \sin(i \pi z)}{\sin(e^{i \pi/4}z) \sin (e^{i 3\pi/4}z)}= \frac{\sin(\pi z) \sinh (\pi z)}{\cosh (\frac{z}{\sqrt{2}})^2 - \cos (\frac{z}{\sqrt{2}})^2} . \]
Now $\lim_{z \to 1} \frac{1^4+z^4}{1^4-z^4} \sin (\pi z) = \frac{\pi}{2}$ and therefore
\[ \prod_{n \ge 2}\frac{n^4-1}{n^4+1} = \lim_{z \to 1} \frac{1-z^4}{1-z^4}\frac{\sin(\pi z) \sinh (\pi z)}{\cosh (\frac{z}{\sqrt{2}})^2 - \cos (\frac{z}{\sqrt{2}})^2} = \frac{\pi \sinh \pi}{2(\cosh (\frac{2}{\sqrt{2}})^2 - \cos (\frac{2}{\sqrt{2}})^2}. \]

For a generalization, let $r > 1$ and consider
\[ \frac{n^{2r}-z^{2r}}{n^{2r} + z^{2r}} = \prod_{k=0}^{r-1} \frac{(n- \omega_{2k}z)(-n-\omega_{2k}z)}{(n- \omega_{2k+1}z)(-n-\omega_{2k+1}z)}\, ,  \]
where $\omega_m = e^{i m \pi/{2r}} = \omega_1^m$ . By the same argument as above, one obtains
\[ \prod_{n \ge 1}\frac{n^{2r}-z^{2r}}{n^{2r}+z^{2r}}= \prod_{k=0}^{r-1} \frac{\omega_{2k} \sin(\pi \omega_{2k}z )}{\omega_{2k+1}\sin(\pi \omega_{2k+1}z)} = -\sin (\pi z) \prod_{k=1}^{r-1}\sin(\pi \omega_{2k}z ) \big / \prod_{k=0}^{r-1} \sin(\pi \omega_{2k+1}z) \]
Since $\lim_{z \to 1} \frac{1^{2r}+z^{2r}}{1^{2r}-z^{2r}} \sin (\pi z) = \frac{\pi}{2r}$, we therefore obtain a formula for $\prod_{n \ge 1}\frac{n^{2r}-1}{n^{2r}+1}$, although in general not in terms  of radicals and elementary functions.

[i] This argument follows that of Borwein. but it does not use the $\Gamma$-function. [/i]

can we at least get some hints on this problem because its not at all an easy one
i mean...i think its a pretty ugly solution..i understood it but i dont like it.
i would like it if this problem could be solved in an elementary way,using
number theoretic stuff

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

How about we not reinvent something and waste, oh, say 7 years of our life, and just look it up?  The later products( like $\prod_{n=2}^{\infty} \frac{n^5-1}{n^5+1}$) get pretty nasty...

There's value in doing stuff yourself (and particularly the underlying methods).  If you don't think so, don't bother posting in the thread.


[quote="max1546"]a,b,c,d are real numbers
f(x,y)=(ax+by,cx+dy)
If f(x) is a one to one function for all integers x,y so that ax+by and cx+dy are integers
what are the if and only if conditions for a,b,c,d?[/quote]

Rather ugly, ... and I hope I made not so much mistakes  :blush:

Let $ A\equal{}\{(x,y)\in\mathbb Z^2$ such that $ ax\plus{}by\in\mathbb Z$ and $ cx\plus{}dy\in\mathbb Z\}$
Let $ f_r : A\to\mathbb Z^2$ be the restriction of $ f(x)$ on $ A$.
We are looking for the conditions to have $ f_r(x,y)$ an injective function.

1) $ a\equal{}0,b\equal{}0,c\equal{}0,d\equal{}0$ : $ f_r(0,0)\equal{}f_r(1,1)$ and so $ f_r(x)$ not injective

2) $ a\equal{}0,b\equal{}0,c\equal{}0,d\ne 0$ : $ f_r(0,0)\equal{}f_r(1,0)$ and so $ f_r(x)$ not injective

3) $ a\equal{}0,b\equal{}0,c\ne 0,d\equal{}0$ : $ f_r(0,0)\equal{}f_r(0,1)$ and so $ f_r(x)$ not injective

4) $ a\equal{}0,b\equal{}0,c\ne 0,d\ne 0$ : $ f_r(x)$ injective $ \iff$ $ \frac cd\notin\mathbb Q$

5) $ a\equal{}0,b\ne 0,c\equal{}0,d\equal{}0$ : $ f_r(0,0)\equal{}f_(1,0)$ and so $ f_r(x)$ not injective

6) $ a\equal{}0,b\ne 0,c\equal{}0,d\ne 0$ : $ f_r(0,0)\equal{}f_r(1,0)$ and so $ f_r(x)$ not injective

7) $ a\equal{}0,b\ne 0,c\ne 0,d\equal{}0$ : $ f_r(x)$ injective

8) $ a\equal{}0,b\ne 0,c\ne 0,d\ne 0$ : $ f_r(x)$ injective

9) $ a\ne 0,b\equal{}0,c\equal{}0,d\equal{}0$ : $ f_r(0,0)\equal{}f_r(0,1)$ and so $ f_r(x)$ not injective

10) $ a\ne 0,b\equal{}0,c\equal{}0,d\ne 0$ : $ f_r(x)$ injective

11) $ a\ne 0,b\equal{}0,c\ne 0,d\equal{}0$ : $ f_r(0,0)\equal{}f_r(0,1)$ and so $ f_r(x)$ not injective

12) $ a\ne 0,b\equal{}0,c\ne 0,d\ne 0$ : $ f_r(x)$ injective

13) $ a\ne 0,b\ne 0,c\equal{}0,d\equal{}0$ : $ f_r(x)$ injective $ \iff$ $ \frac ab\notin\mathbb Q$

14) $ a\ne 0,b\ne 0,c\equal{}0,d\ne 0$ : $ f_r(x)$ injective

15) $ a\ne 0,b\ne 0,c\ne 0,d\equal{}0$ : $ f_r(x)$ injective

16) $ a\ne 0,b\ne 0,c\ne 0,d\ne 0$

16.1) If $ ad\minus{}bc\ne 0$ : $ f_r(x)$ injective

16.2) If $ ad\minus{}bc\equal{}0$ and $ \frac ca\notin Q$, then $ f(A)\equal{}\{(0,0)\}$ and :

16.2.1) If $ \frac ab\notin Q$ : $ A\equal{}\{(0,0)\}$ and $ f_r(x)$ injective

16.2.2) If $ \frac ab\in Q$ $ \{(0,0)\}\subsetneq A$ and $ f_r(x)$ not injective

16.3) If $ ad\minus{}bc\equal{}0$ and $ \frac ca\in Q$ :

16.3.1) If $ \frac ab\notin Q$ : $ f_r(x)$ injective

16.3.2) If $ \frac ab\in Q$ $ f_r(x)$ not injective

[u][b]Hence the result [/b][/u]:

$ f_r(x,y)$ is an injective function $ \iff$

Either (see case 4) $ a\equal{}0,b\equal{}0,c\ne 0,d\ne 0$ and $ \frac cd\notin\mathbb Q$

Either (see case 7) $ a\equal{}0,b\ne 0,c\ne 0,d\equal{}0$

Either (see case 8) $ a\equal{}0,b\ne 0,c\ne 0,d\ne 0$

Either (see case 10) $ a\ne 0,b\equal{}0,c\equal{}0,d\ne 0$

Either (see case 12) $ a\ne 0,b\equal{}0,c\ne 0,d\ne 0$

Either (see case 13) $ a\ne 0,b\ne 0,c\equal{}0,d\equal{}0$ and $ \frac ab\notin\mathbb Q$

Either (see case 14) $ a\ne 0,b\ne 0,c\equal{}0,d\ne 0$

Either (see case 15) $ a\ne 0,b\ne 0,c\ne 0,d\equal{}0$

Either (see case 16.1)  $ a\ne 0,b\ne 0,c\ne 0,d\ne 0$ and $ \frac ab\ne\frac cd$

Either (see cases 16.2.1 and 16.3.1) $ a\ne 0,b\ne 0,c\ne 0,d\ne 0$ and $ \frac ab\equal{}\frac cd\notin Q$


Why do so many different phenomena match the normal distribution so well? I assume it has something to do with the central limit theorem, but still- what makes this particular curve so special, that so many different distributions in nearly every science match it so well?

You're right, it probably has something to do with the central limit thoerem.

I would go further, and claim that nothing in nature is [i]exactly[/i] normally distributed, but rather that [i]all[/i] of those approximately-normal distributions that you are talking about are consequences of the central limit theorem.

The way I understand it, the central limit theorem states that with a large enough sample size, the set of all possible sample means or proportions is normally distributed. But if this is true, then all we can expect is that the set of [i]means[/i] of samples of measurements (in any science you like) will roughly follow a Gaussian distribution, not necessarily that the measurements themselves will follow a normal distribution. But clearly, the measurements do vary normally.

Wikipedia states that
[i]The importance of the normal distribution as a model of quantitative phenomena in the natural and behavioral sciences is due to the central limit theorem. Many psychological measurements and physical phenomena (like noise) can be approximated well by the normal distribution. While the mechanisms underlying these phenomena are often unknown, the use of the normal model can be theoretically justified by assuming that many small, independent effects are additively contributing to each observation.[/i]

Can someone explain how these "small, independent" effects would tie individual observations to the central limit theorem? One example would be enough.

This is just something I've been really curious about, I'd be very appreciative if someone explained it to me.

The central limit theorem and the law of large numbers are maybe the two biggest results in probability, and they 'live' on different scales.

The law of large numbers just says that if you take the average of a bunch of iid random variables, then in the limit you get a constant, the expectation of any of the random variables.

The central limit theorem says that if you sum up a bunch of iid random variables, subtract off the expectation of the sum, and divide by an appropriate constant, then in the limit (a different kind of limit than above!) you get the familiar normal distribution with mean 0 and variance 1.  One way of looking at this is that the central limit theorem tells you what sort of fluctuations around the mean you can expect when taking the average of a bunch of iid random variables.

As a heuristic, you can write the central limit theorem as $ S_n/n \approx \mu \plus{} n^{\minus{}1/2}N(0,1) \equal{} \mu \plus{} N(0,1/n) \equal{} N(\mu,1/n)$, which should give you a quantitative idea of what is going on.  If $ n$ is very large, then you essentially just get the (weak) law of large numbers, but if $ n$ is a 'reasonable' number like $ 4$, this allows you to compute the probability that your sample mean lies in some range using the familiar normal distribution.

The rest is just noting that, as Dr. Merryfield said, a lot of processes in nature are the 'sum' of many small, independent effects, and hence produce phenomena which follow (closely) a normal distribution (by CLT)

One point about the central limit theorem: the theorem itself guarantees convergence, but only very slowly. In practice, the convergence is usually much faster and the normal approximation works even on very small scales. If you want to estimate the probability that exactly 513 of your 1000 coins come up heads, the value coming from the normal distribution (integrated between 512.5 and 513.5) is very close to the actual value.


Find $ a,b,c\in R$ if we have $ |4x^3\plus{}ax^2\plus{}bx\plus{}c|\leq 1$ ($ \forall x\in[\minus{}1,1]$)

It's the Tchebyshev polynomial.
$ a\equal{}c\equal{}0;b\equal{}\minus{}3$


Prove that $4^{545}+545^{4}$ is not prime.

Sophie Germaine's Identity

$x^4 + 4y^4$
$= x^4 + 4x^2 y^2 + 4y^4 - 4x^2 y^2$
$= (x^2 + 2y^2)^2 - 4x^2 y^2$
$= (x^2 + 2y^2 + 2xy)(x^2 + 2y^2 - 2xy)$

I hope there were no typos in that sequence.

Anyway, set 545 to be $x$ in the above identity, and $4^{136}$ to be $y$. Then just plug it in and you're done.

Basically any problem that involves primality or even contains anything remotely related to something to the fourth power plus some other junk (especially if there's a 4 involved) pretty much always has something to do with the Sophie Germaine identity.

Consider the following problem:

Prove $n^4 + 4^n$ is never prime for integers $n > 1$.

well here is my proof:

CASE 1: if $n$ is an even number, we know it is true, because even number plus even number would equal even number, which is never a prime.

CASE2: $n$ being an odd number, it really doesn't matter, because $4^n$ will alway end in 4, and $n^4$ will always end in either 5 or 1.

so case 2-1, if $n^4$ ends in a 1:
we don't need any proof, because then $n^4 + 4^n$ will always end in 5, meaning it is a multiple of 5.

case 2-2, if $n^4$ ends in a 5:
it only ends in 5 if $n$ ends in a 5, which means $n$ is amultiple of 5.
so lets say $n = 5a$

then $n^4 + 4^n = 625a^4 + 1024^a$
this seems really hard, but referring back to Sophie Germaine's theory: $x^4 + 4y^4$,

if we set $x = 625a$, $y^4 = 4^{\frac{5a-1}{4}}$ ----- (1)

since 5a ends in 5, 5a-1 ends in a 4, which makes $\frac{5a-1}{4}$ an integer. So, (1) is proven not prime,

and i have proven all the cases. QWD.

Case 1: If $n$ is even, then the number is clearly divisible by 2.

Case 2: If $n$ is odd then let $n=2k+1$ and we obtain the following:

\[n^4+4^n=n^4+4 \cdot 4^{2k}=n^4+4 \cdot (2^{k})^4\]

which has the form $m^4+4n^4$ and is true by Sophie Germain's indentity.

[quote="DPopov"]Prove that $4^{545}+545^{4}$ [b]is prime[/b].[/quote]  ;)  :lol:

[quote="MithsApprentice"][quote="DPopov"]Prove that $4^{545}+545^{4}$ is prime.[/quote]

Mathematica says it is (after running the PrimeQ function for a good half hour no prime factor was found so I gave up).[/quote]

darktreb's solution also shows that it's not prime.

[quote="MithsApprentice"][quote="DPopov"]Prove that $4^{545}+545^{4}$ is prime.[/quote]

Mathematica says it is (after running the PrimeQ function for a good half hour no prime factor was found so I gave up).[/quote]

I have heard that Mathematica before in the MathWorld. Is it free and how did u get it?

[quote="nr1337"][quote="MithsApprentice"][quote="DPopov"]Prove that $4^{545}+545^{4}$ is prime.[/quote]

Mathematica says it is (after running the PrimeQ function for a good half hour no prime factor was found so I gave up).[/quote]

?  It did?  What version do you have?  Mathematica 5.1 said  that it wasn't prime after about 5 seconds:

In[3]:=
PrimeQ[4^545 + 545^4]

Out[3]=
False[/quote]

Same. But I can't seem to type. Entered the wrong number...and so it was "Running..." forever.

Now that I actually got the number, Mathematica did indeed get it within a couple of seconds.

henry, Mathematica is not free. It is available for approximately 140 USD (Student Edition) at http://www.wolfram.com. It is also available for free at some math programs and competitions (MOP, etc.)

[quote="henryyangrui"]
I have heard that Mathematica before in the MathWorld. Is it free and how did u get it?[/quote]

By no means.  I think it's around $1000.

Depends on which edition you want, professional or student.

[quote="WindSlicer"]1000 or 140? ...[/quote]

For his purposes, I'd assume he wants the Student Edition...

Whats the difference between professional and student?

[quote="MithsApprentice"]Well, the professional edition is used by a lot of professionals (go figure) like engineers and whatnot. There's loads of features you don't and won't need.[/quote]
Actually, the student edition comes with all of the features of the Professional edition - it just adds a footnote at the bottom of every printed page so that you can't pass off documents printed by the Student edition in a professional environment.  I'm not sure if that's what you were implying with your post, but I figured I'd clarify anyway.  I agree, though - most of the features of Mathematica are so esoteric that very few people use them, but at least they're there if you want them.


Let $ a,b,c\in \mathbb {R}$  Prove that  $ max\{ \ a^2-b,b^2-c,c^2-a \}\leqq -\frac{1}{4}$ $ a,b,c\in \mathbb {R}$
 :rotfl:

____________________________________________
$ \begin{array}{c}
\mathfrak{Azerbaijan } \\
\mathfrak{Land } \\
\mathfrak{of} \\
\mathfrak{Fire} \end{array}$

[quote="Pirkuliyev Rovsen"]Let $ a,b,c\in \mathbb {R}$  Prove that  $ max\{\ a^2 \minus{} b,b^2 \minus{} c,c^2 \minus{} a \}\leqq \minus{} \frac {1}{4}$ $ a,b,c\in \mathbb {R}$
[/quote]
Please change that to minimum instead of maximum.

[quote="Pirkuliyev Rovsen & Arrange your tan"]Let $ a,b,c\in \mathbb {R}$.
 Prove that  $ max\{a^2 \minus{} b, b^2 \minus{} c, c^2 \minus{} a \}\le \minus{} \frac {1}{4}$ or it is supposed to be
$ min\{a^2 \minus{} b, b^2 \minus{} c, c^2 \minus{} a \}\le \minus{} \frac {1}{4}$ according to Rijul saini
[/quote]

Why would either of these versions claimed be true?

If $ a \equal{} b \equal{} c \equal{} k$, where $ k$ and $ t$ are real numbers, $ t \equal{} k^2 \minus{} k$,

and $ k < \frac12$ or $ k > \frac12$, then the $ max$ or $ min\{k^2 \minus{} k, k^2 \minus{} k, k^2 \minus{} k\} \equal{}$

$ max$ or $ min\{t, t, t\} \equal{} t$, where $ t > \frac { \minus{} 1}{4}$.

Only if $ k \equal{} \frac12$ does the $ max$ or $ min\{t, t, t\} \equal{} t$, where $ t \equal{} \frac { \minus{} 1}{4}$.

But suppose $ a, b, c$ are distinct.

Example:

Let $ a \equal{} \frac32, b \equal{} \frac43, c \equal{} \frac54$

The $ max$ or $ min\{a^2 \minus{} b, b^2 \minus{} c, c^2 \minus{} a\}$ gives a positive value,

regardless of the whether the maximum or the minimum is taken.

And it is not less than or equal to $ \frac{\minus{}1}{4}$.

____________________________________________
$ \mathfrak{Azerbaijan } \\
\mathfrak{Land } \\
\mathfrak{of} \\
\mathfrak{Fire}$

Hidden stuff should not be displayed in a forum like this .... you should be above that by now.


inequalities
logarithms
calculus
calculus computations

Find the value of $a$ such that $x-\frac{x^{2}}{2}+ax^{3}\geq \ln (1+x)$ always holds for $x>-1$.

Clearly, $ln(1+x) \leq x$
Then we are multiplying the right hand side by $(1-\frac{x}{2}+ax^{2})$ and in order for the inequality to hold, it must be true that $(1-\frac{x}{2}+ax^{2})$ > 0. This is how I thought about it.


algebra
polynomial
number theory unsolved
number theory

A polynomial $\Phi_{n}(x)$is defined as $\Pi_{i}(x-\epsilon_{i})$ where $\epsilon_{i}$are all roots of unity of degree exactly $n$
prove that if $p$ is a divisor of $\Phi_{n}(m)$ for some postive integers $m,n$ then either $p|n$ or $n|p-1$

See here!
http://www.mathlinks.ro/Forum/viewtopic.php?t=126566  :wink:


Here's a relatively simple set of simultaneous equations:

$\frac{w^2 + 1}{xy} = \frac{x^2 + 1}{yz} = \frac{y^2 + 1}{zw} = \frac{z^2 + 1}{wx} = 1$

Find all real solutions to these equations, or prove that none exist.

[hide]Since all four terms are 1, their product, $\frac{(w^2+1)(x^2+1)(y^2+1)(z^2+1)}{w^2x^2y^2z^2}=1$, so that $\frac{w^2+1}{w^2} \times \frac{x^2+1}{x^2} \times \frac{y^2+1}{y^2} \times \frac{z^2+1}{z^2}=1$.

But each of these terms are greater than 1. Therefore no real solutions.[/hide]


Coupons in cereal boxes are numbered 1 to 5, and a set of one of each is required for a prize. With one coupon per box, how many boxes on average are required to make a complete set?

heres a hint:
find the expected number of boxes needed to get 1 distinct type of coupon
then find the expected number of boxes needed to get 2 distinct types of coupons
...
then find the expected number of boxes needed to get 5 distinct types of coupons
for example the expected number of boxed to get 1 distinct type of coupon is 1 since any type of box will work
so basically you are considering the 1st, 2nd, 3rd, 4th, 5th time you get a unique coupon

... hope u understood

Take one box.

The probability that the next box will be different from the one you chose is $ \frac 45$. Therefore, the expected number of boxes you need to draw until you draw something different is $ \frac 54$.
The probability that the next box will be different from each of the other two is $ \frac 35$. Therefore, the expected number of boxes you need to draw until you draw something different is $ \frac 53$.
The probability that the next box will be different from each of the other three is $ \frac 25$. Therefore, the expected number of boxes you need to draw until you draw something different is $ \frac 52$.
The probability that the last box will be different from the previous four is $ \frac 15$. Therefore, the expected number of boxes you need to draw until you draw something different is $ 5$.

Adding the expected number of boxes for each stage, along with 1 for the first stage, yields $ 1 \plus{} \frac 54 \plus{} \frac 53 \plus{} \frac 52 \plus{} 5 \equal{} \frac {137}{12} \approx 11.4167$ boxes.


AMC USA(J)MO USAMO AIME Putnam Princeton college

How to learn Maths Olympic quickly? So that I could have a good problem solving ablity in a short time.

[quote="Myth"]I think, noone knows quick way. :)[/quote]Mikhail is right. There is no quick way to "learn math olympic". You might be lucky and have good results while you're still young (but I doubt you can be lucky in China - wpolly can tell us more about how tough is the olympiad) but you need lots of work to succede.

Also there is a great difference because of the different level of math olympiad. There's the local olympiads, there's the state olympiads, math contests, national olympiads (usamo, cmo, etc.), and finally the hardest most-effort requiring imo (international mathematical olympiad) which probably is the goal of every high-school student participating in olympiads :). So to which one are you reffering?

This topic reminds me of the following story about Euclid (or another greek mathematician, I don't remember exactly):
One day, the king asked Euclid to quickly teach him mathematics, but Euclid answered:
"There are no royal paths in mathematics" :)

[quote="Valentin Vornicu"][finally the hardest most-effort requiring imo (international mathematical olympiad) which probably is the goal of every high-school student participating in olympiads :). So to which one are you reffering?[/quote]

I think you're wrong about that.   (Or maybe you were just joking, given the smilie.)  I think there are plenty of high school students participating in local/national level contests/olympiads with no thought of making it to IMO.  There are lots of reasons to participate in math competitions even if you don't think you can make the IMO team -- fun, challenge, to see where you stand, to qualify for scholarships, etc.

hmm.. how about say doing well in USAMO?
 getting around 3~4 Q's rite

[quote="Valentin Vornicu"] There's the local olympiads, there's the state olympiads, math contests, national olympiads (usamo, cmo, etc.), and finally the hardest most-effort requiring imo (international mathematical olympiad) which probably is the goal of every high-school student participating in olympiads :). So to which one are you reffering?[/quote]

There are only two rounds in China, one is the Senior High School Combinative Competition, the ones who got the first 3 to 6 position in each Province could take part in CMO.(that's tooooooooooooooooooooooooooo hard, China is such a great country with 1300000000 people, each province has the population more than 50000000, like my hometown--Jiangsu, the population is 70000000, then the students take part in the Senior High School Combinative Competition should be more than 70000, choose 6 students from 70000 students, that's too hard to join CMO for students, so luck and ablity should be both great.) In fact, some of the schools in our Province have cheated in the Senior High School Combinative Competition this year, some was caught, some are still out of the law, why they cheated? Because the first 45 students can go to the best colleges without examination, the first 100 students can add 20 marks to the the achievements of college attending exam, these are glory for a school, so some school cheated. These should make it harder for students in schools which haven't cheated. Luckily, the province maths club decides to add a test before the Senior High School Combinative Competition next year, then the students with high marks can join the High School Combinative Competition in the capital city of the province but not in each school, so the cheatments shoud be less a lot.
    For me, I think to join the CMO maybe hard, so I think the High School Combinative Competition is more proper for me, if lucky enough, then the CMO.

Wow! that sounds extremely tought! In Romania there are numerous stages: first school olympiad (75% qualify further), then city (local) olympiad (20% qualify further), then state olympiad (10% qualify further), then national olympiad (first 10 in each class qualify for the first TST), then first TST for Balkan Olympiad (first 25 go forward, first 6 go to Balkan MO), then another 4 tests :)

Another thing, the Senior High School Combinative Competition is not easy, I think maybe it's harder than most national Olympiads. It includes 2 parts, the full mark of the first part is 150, it includes 6 choices, 6 fill-in-the-blanks like the AIME, and 3 big problems. the experts of this competition say that it is only a little harder than the college attending exam. But most students don't think so, the time for first part maybe short, only 1.5 hours. 90% students couldn't make their marks more than 80.
The second part includes 3 problems,its full marks is also 150. It' only a little easier than the CMO(I think CMO is a lot harder than IMO), most students could make 0 mark in the second part.

Should I do a lot of problems?

    Or if I am trying to work out a problem, but I can't think of the solution at the moment, after how much time thinking about the problem should I look the solution in the books?

The province-by-province qualification procedure
for the Chinese national olympiad is surprising.
The China IMO team would be stronger, I assume, if
people in the big cities had the same chances to
qualify as people in the remote provinces.

The IMO team from USSR was mostly people from
Leningrad and Moscow -- unfair in a regional sense,
but without any artificial geographic restrictions on
who could compete.

[quote="wpolly"]I just got 189 (119+70) in SHSCC (though I still got the 2nd place in Gansu Province and are going to CMO).
The differences between the Provinces are HUGE sometimes.
For example, In SHSCC 2003 I got 250 and ranked 1st in gansu while the 4th(my deskmate) got only 184. Meanwhile, My mark can rank only 10th in Jiangsu(and no other one in Gansu can make their way into Jiangsu top-50). Also, AFAIK, there've no onein Tibet who have ever get 100+ mark.
I just hope I could perform well on the CMO ......[/quote]

Well, I got 86+70(for the first time I take part in the SHSCC.)this year :blush: , while the First Prize(whole country) in Jiangsu is about 220, the First Prize(province) is 194. The least marks to go to CMO is 265 in Jiangsu. 45 stuhents got the First Prize(whole country), 108 students got the First Prize(province) .

can you guys post the questions on the site? (also there is a new [url=http://www.mathlinks.ro/Forum/index.php?f=252]Chinese Forum[/url] in which discussions in Chinese will be allowed, and the controls will soon also be in chinese there - you can also post in there if you want).

[quote="siuhochung"]wow... it is really sooooooooooooooooooooooooooo hard to qualify to the cmo....
in hong kong the story is totally different as you know that hong kong is much much weaker than china....[/quote]that's probably because you have to select 6 students out of less total students than in china.

If you look at the TSTs you will see that with little exceptions (Balkan countries especially) their difficulty is proportional with the population numbers. And so are their results at IMO.

[quote="Valentin Vornicu"]If you look at the TSTs you will see that with little exceptions (Balkan countries especially) their difficulty is proportional with the population numbers. And so are their results at IMO.[/quote]

Being a fan of the United States in math competitions, I hope that the United States learns from the Balkan countries how to exceed the expectations established by its high rank in national population.  ;)

It's not population per se so much as the distribution and use of
advanced (or competition-relevant) math books with problems.
What helped Russia, China and many of the socialist countries
was that good books were printed in large and dirt-cheap editions
(e.g. cheaper than food).   In the Balkan countries there was a huge
competition-specific literature with books of problems from various
national olympiads, collected problems from magazines, and so forth,
that many high school students could and did purchase.
That is a nice tradition but dying out with the old subsidies in most of
the ex (and de-facto ex) communist countries.

Bear in mind also that for purposes of the IMO, "Russia" meant
mostly the big cities and I suppose the same is true for China
and now India.  Most of the rural population is nowhere near competing
for spots at the IMO.

fleeting_guest, how come you know so much about the math policy in the balkan countries?

I think that there is a different perception in balkan countries, as to what the olympiads stand for, and this motivates the students more. Let me be more concrete:

In US, after Reid Barton in 2001 got a full-score (max) basically the first in the world, breaking a world record of 4 gold medals in a row at IMO, which was in Washington DC in that year!, what I could see was just a really really small column on the backside of Washington Post. I presume other papers didn't even treat the subject. Another american scored all max (Gabriel Caroll) along with Reid, making him also a 2 times gold medalist. Almost no mention of them in the media or the news papers (we don't even count Reid's medals in the IOI, Caroll's becoming Putnam Fellow in junior high, etc.!)

In Romania, Ana Caraini, which had no near performances at the time (in 2001/2002) with the two above, a huge media assault with tons of interviews, first page covers in news papers, tv shows, and much more was dedicated. I myself became the Minister of Educations' Counselor at the early age of 19, obviously because of my results in the IMO (along with other stuff). Andrei Negut (gold medalist in 2004) was also one of the targets of the Romanians news papers / media, and in 1999 when Hornet (now a Princeton student) won the first place at the IMO which was held in Romania, the prime-minister organized a press-conference just to congratulate him. All of the team members from 2002 (including myself :P) were decorated by the President of Romania with the Silver Excelence Insignia.

So, as you see, this very different perception of people regarding the olympiad subject has a huge impact in the balkan countries towards motivating the students into learning more, and compensates for the small number of population.

Valentin,

I didn't understand whether you are agreeing with, or criticizing,
my remarks about the Balkan countries. I'm well aware of the
national differences in "olympiad appreciation" that you mentioned,
such as the total indifference in the USA versus intense promotion
in Romania.

Yes, such differences are a partial explanation; if the group
of motivated students is 5 times higher per population that certainly
makes a difference.   However, there are many other countries
where an IMO gold medal would make national television and
newspapers, but who are not nearly as strong as Romania, Vietnam,
Hungary (until 1980's), etc.   Promoting olympiads means not
only giving nice treatment to the winners, but providing the
resources --- the "means of production", if you will --- for interested
students to develop their abilities.   Smaller countries such as
Romania and Vietnam had books with compiled olympiad problems,
TSTs, and other training materials available in cheap editions with
large printings (in Russia the printings sometimes reached the
tens or hundreds of thousands; on a per population basis I think
it was comparable for the smaller countries).
Other resources include specialized math/physics high
schools, math circles run by grad students or mathematicians,
high quality general (non-olympiad) math textbooks available
to the student population, olympiad (written proofs) competitions
starting in the earlier stages of school (ages 10-13, not only high
school).

The English-speaking countries always had large numbers
of motivated students, but before the era of Internet problem
databases, those people could never have dreamed of accessing olympiad
training materials to the extent available in the (ex)communist countries.

My point was something like this:  the critical parameter is not
population but "effective population", something like the number
of motivated people with access to relevant resources.  In that sense
in USSR and China one is dealing mainly with large cities with
their concentration of resources (i.e. a smaller effective population
than one might assume) and in Romania, Hungary, Vietnam, maybe
nowadays Iran, etc --- a larger effective population due to the
disproportionate availability of resources.

 [/b]

I was agreeing partially with what you said, and also supplied another strong reason for which Balkan countries' students perform better than most of the western countries.

Speaking only of the Chinese-language literature, I will agree that the amount of READABLE, ACCESSIBLE materials for young learners in Taiwan (some locally produced, mostly reprinted from China) and China about mathematical contests is ASTOUNDING. Although there are unquestionably people in some rural regions of China who are lucky even to learn to speak understandable Mandarin, most of the urban population in China would have access to materials such as these at affordable prices, and there are "key" schools in most of the cities that build their curriculum around such materials. There are tens of thousands of students in China (judging from the colophons of these books, which report the size of press runs) studying at a level that a few dozen young people reach at the same age in the United States.

Free public lending libraries and interlibrary loan networks are the "secret weapon" of American education, but too many kids in the United States aren't even aware that there is a better mathematical literature than the appallingly bad school textbooks to look for. It will take some time to build a market for more books of the nature of the AoPS books in the United States.

[quote="tokenadult"]It will take some time to build a market for more books of the nature of the AoPS books in the United States.[/quote]I personally think that the time is not going in the favor of the books (paper books that is). US is probably the largest internet-covered country, and at the rate things are going, I really doubt it that in 20 years there will be half the books of today selling. :) Same thing happens everywhere.

A huge beneficial effect in the socialist countries
was the disregard for Western "intellectual property"
rights.   In China there is still massive piracy of
university- and research-level texts such as Springer
graduate texts and lecture notes, both in translated
Chinese editions (authorized or not) and in
English editions that are basically bound photocopies.

People with mathematical, scientific or literary interests
in the ex-USSR would routinely accumulate huge
libraries of high quality books.  The books were
printed in large runs (e.g. volumes of Great Russian
Encyclopedia with printings on the order
of 10^5 - 10^6), often on toilet-quality paper,
at very cheap subsidized prices even by local
currency standards. To do this in the Western countries,
if the books even exist in the native languages one
would still need a lot of money.  It is no accident
that a lot of the top mathematicians (Weil, Selberg,
Bombieri, Wiener etc) were from wealthy or scholarly
families whose children had access to large home libraries.

To the extent that scholarship is a production process,
it relies on access to the means of production.
In the US, textbooks are not sold in bookstores and
are only loaned to the students, and a student would
be hard pressed to buy an adequate math textbook
in a typical (say 95-99 per cent of the market) book store.

Let me balance this screed/paean to socialist educational
and publishing practices, by mentioning that US libraries
and interlibrary loan that tokenadult referred to, are
the acme of civilization as far as I'm concerned.

I am sure that I wouldn't have come to where I am now (in terms of mathematical ability) without the help of the city and school libraries.

Contest-oriented books are viturally extinct in bookstores (in North America), they can be found only on Internet stores and they usually have a very significant price tag.

On the other hand, during my trip to China, I noticed that there is a wealth of academic and contest resources in book stores. And they are usually extremely cheap too (well, it may be because I was comparing the prices to that of Canada). For example, the four Tournament of Towns books published by AMT would have together costed $140, while I found a translation of the collection in Chinese that costed only 34 Yuan.

On the other hand, I feel that the quality of the the Chinese books (at least the ones I bought) were rather poor. Not just in terms of paper quality. Perhaps I was looking into too much of a specialty market, but I felt that the books were writen often just for the purpose earning money and filling paper-space. I bought an Olympiad problem book, in which most of the solutions were extremely unmotivated and poorly explained. I mean, it is one thing to have correct solutions, but trying to teach readers how to solve problems is a completely different matter.

In fact, the students in China, no matter he(she) is from urban areas or rural areas, they will have the same chance to take part in the maths competition. Tokenadult has said that there are "key" schools in urban areas of China, that's true. I am from rural areas of China, but now I study in a "key" school in a very large city, because when I graduated from the junior high school, I have a chance to take part in the exam which my present school held for all the graduted junior high school students in our province, I performed well in the exam, so I could attend this school. What I mean is that if the student has the ability, he or she could attend the "key" schools near his hometown or even in another city or province. So the students have the same chance.

hello Qxy,

Rural people who live away from home at their school
should (in my opinion) be classified as Urban population
for this discussion.  The people for whom this is not possible,
either for financial or family reasons, or because they don't
score high enough on the selection exams, have only the
Rural level of resources available.  Even if they would never
have made it to the IMO or CMO, they have a much lower chance
to contribute to the overall level of competition within the country
(or more precisely, their contribution be higher if they had the
same access to the materials that Urban people have).

In that sense, even with special away-from-home schools as in China,
Russia, etc,  basically it is the population in certain limited areas
(mostly big cities) that contributes to the "effective mathematical
population" of a country.

Best of luck with your studies.

[quote="fleeting_guest"]In that sense, even with special away-from-home schools as in China,
Russia, etc,  basically it is the population in certain limited areas
(mostly big cities) that contributes to the "effective mathematical
population" of a country. [/quote]I think this is going to change in the future years, along with the spreading of the internet, and rural population will gain more procents of the "effective mathematical population" of the developed or in course of development countries. Then again I might be too optimistic about this.

The Internet is superior already for databases of recent olympiads (and some old ones).
For most other purposes the Internet has several centuries of literature to catch up with.  It may not happen as fast as we'd like.

[quote="fleeting_guest"]For most other purposes the Internet has several centuries of literature to catch up with.  It may not happen as fast as we'd like.[/quote]Google is already on this: [url=http://www.google.com/press/pressrel/print_library.html]Google Indexes Books[/url]. ;)

[quote="fleeting_guest"]the Google book scan is several tens of thousands of
books.  I don't know how many Proj Gutenberg has done.[/quote]Yes, that will cover a great deal of common stuff that is not accessible through the net right now. In plus you can search within the books! It's not a simple scan-display picture scan :)

[quote="billzhao"]I'd consider Canada to be a rather technologically advanced country (or as far as Internet availability is concerned). Yet 5 out of 6 people of the IMO 2004 team came from the same city...

Being in a big city definitely is an advantage to competitors, in more ways than one.[/quote]

Canada definitely has a high degree of Internet use, which extends to the high Arctic. But, yes, there is still no substitute for occasional face-to-face interaction.

This AoPS-MathLinks site is leading the way in showing how Internet technology can reduce the isolation and increase the training resources of math-liking young people around the world, and it is invaluable to me in Minnesota, where there are always multiple USAMO qualifiers but usually not much interaction among those students except on the Minnesota ARML team. In a country like China that is still mostly unimaginably poor, it will be a long time indeed before Internet access--or even basic literacy--is pervasive in the rural population. Meanwhile, those of us who are lucky enough to live in wealthy parts of the world can share ideas and raise the level of math competition and research.

[quote="billzhao"]I'd consider Canada to be a rather technologically advanced country (or as far as Internet availability is concerned). Yet 5 out of 6 people of the IMO 2004 team came from the same city...

Being in a big city definitely is an advantage to competitors, in more ways than one.[/quote]

But not even that; the West Coast (read: BC) hasn't had one of the IMO qualifiers in the past few years.  There's little or no contact between high-achieving students here whereas I would expect that there is at least a modicum of interaction in Toronto.  I was in the top 5 in BC last year on both the Euclid and COMC, but yet I didn't meet any of the other high scorers (save for the ones who went to my school) except for at a completely unrelated International Baccalaureate function.

So it really is all about being in a big city (any city, really) with a pre-existing math tradition, I think.  For example, two of the US IMO'ers are from upstate New York, I think, which is hardly a 'big city'.


Solve for x, y, and z in this simultaneous equation
3x-y+2z=1
2x+3y+3z=4
x+y-4z=-9

[hide]Add the third equation and the first equation and get 4x - 2z = -8
Subtract the second equation from three times the third equation to get x - 15z = -31.

Solve the system and substitute to get {x,y,z}={-1,0,2}[/hide]


function
AMC
AIME
algebra
domain
functional equation

The function $f$ is defined on the set of integers and satisfies \[ f(n)=\begin{cases} n-3 & \text{if } n\ge 1000 \\ f(f(n+5)) & \text{if } n<1000\end{cases}  \] Find $f(84)$.

How do you type the function in a better way?

The \{\limits^{a}_{b} was the only thing I could think of...  :(

\[ f(n)=\begin{cases} n-3 & \text{if } n\ge 3 \\ f(f(n+5)) & \text{if } n<1000\end{cases} \]

[code]f(n)=\begin{cases} n-3 & \text{if } n\ge 3 \\ f(f(n+5)) & \text{if } n<1000\end{cases}[/code]

[hide="Answer"]$f(84)=f(f(89))=...=f_{185}(1004)=...=f_{183}(998)=f_{184}(1003)...$
Following the pattern, we notice that we move down two numbers in the $f$ and return to the same value of $n$. Therefore, we reduce down to $f_3(1004)=f_2(1001)=f(998)=f_2(1003)=f(1000)=\boxed{997}$.[/hide]

[quote="1234567890"][quote="thebrokenlight"][hide]
$f(84)=f(89)=f(94)=...=f(1004)=1004-3=\boxed{1001}$
[/hide]

Is that right?

-BLt[/quote]

Probably not, as AIME answers are in between 0 and 999. :)[/quote]

Grr. I knew the problem seemed suspiciously easy...

-BLt

It's because you're supposed to add another $f(...)$ every time you use the second function. ;)

I got little different solution.
Don't know if its right but it gives right solution :)

$ \line{f(84) = f_{m - 1}(84 + 5m) = f(84 + 5m - 3(m - 2)) = f(90 + 2m) = f(1000) = 997}$

EDIT:

This should be like
$ f(84) = f_{m}(84 + 5(m - 1)) = f(84 + 5m - 5 - 3(m - 1)) = f(82 + 2m) = f(1000) = 997$

Could you define $ m$? As I'm reading it now, it doesn't seem to make sense.

$ f_{m}(x) = %Error. "underhat" is a bad command.
{f(f(f(f....f(x)))}$ m times.
This is based on $ f(84) = f((84 + 5)) = f(((84 + 5 + 5))).....$

And $ m\in N$

I mean, where you got the two statements

[quote="NapoleonXIV"]
$ f_{m \minus{} 1}(84 \plus{} 5m) \equal{} f(84 \plus{} 5m \minus{} 3(m \minus{} 2))$

...

$ f(90 \plus{} 2m) \equal{} f(1000)$
[/quote]

from. Maybe I'm just missing something, but it doesn't seem that $ m$ is consistently defined. Or are you just temporarily dropping the domain restrictions? ..

[quote="JesusFreak197"][hide="Answer"]$ f(84) \equal{} f(f(89)) \equal{} ... \equal{} f_{185}(1004) \equal{} ... \equal{} f_{183}(998) \equal{} f_{184}(1003)...$
Following the pattern, we notice that we move down two numbers in the $ f$ and return to the same value of $ n$. Therefore, we reduce down to $ f_3(1004) \equal{} f_2(1001) \equal{} f(998) \equal{} f_2(1003) \equal{} f(1000) \equal{} \boxed{997}$.[/hide][/quote]
I don't think I understand your pattern... any elaboration would be great...

The pattern I observed is
$ f_{184}(999) \equal{} f_{185}(1004)... \equal{} f_{183}(1000) \equal{} f_{182}(997) \equal{} ... \equal{} f_{183}(1004)$

So we have $ f_{n \plus{} 2}(1004) \equal{} f_{n \minus{} 1}(997) \equal{} f_n(1004)$.
Replacing $ n \equal{} 1$ gives $ f_3(1004) \equal{} f(1000) \equal{} 997$.

Firstly,
$f(1000)=997$
Let $n=995$
$f(n+5)=1000$
$n=998$
$f(f(n+5))=997$
$f(998)=997$
We now start to realize that we are simply substituting the same value over and over again to get 997. Checking one more value, we find that
$f(n+5)=998$
So $n=996$
and then $f(f(n+5))=997$
And we are done because thusly $f(996)=997$, which means this value repeats, so $f(84)=\fbox{997}$


linear algebra
matrix
vector
linear algebra unsolved

If $ n$ is even, it's false:
$ A\equal{}\begin{pmatrix}0&1\\\minus{}1&0\end{pmatrix}$.

[quote="loup blanc"]If $ n$ is even, it's false:
$ A \equal{} \begin{pmatrix}0 & 1 \\
\minus{} 1 & 0\end{pmatrix}$.[/quote]

I think it's false for $ n\equal{}2$ only. Unless $ \text{char }K\equal{}3$ is allowed (in which case I don't know whether it is true or not).

  darij

Setting $ i\equal{}j\equal{}k$ and $ i\equal{}j$ shows, that all matrices with the given property are skew-symmetric. Therefore, for $ n$ odd the determinant vanishes. For even $ n$ we can generalize [b]loup blanc[/b]'s counterexample by setting $ A\equal{}\left(\begin{array}{cc}0 & E\\\minus{}E&0\end{array}\right)$, where $ E\in M_{\frac{n}{2}}(\mathbb{R})$ is the matrix consisting only of 1-entries

[quote="hjbrasch"]For even $ n$ we can generalize [b]loup blanc[/b]'s counterexample by setting $ A \equal{} \left(\begin{array}{cc}0 & E \\
\minus{} E & 0\end{array}\right)$, where $ E\in M_{\frac {n}{2}}(\mathbb{R})$ is the matrix consisting only of 1-entries[/quote]

But this $ A$ trivially satisfies $ \det A\equal{}0$ for all $ n\geq 2$...

  darij

I stated complete nonsense with regard to the counterexample, since its determinant is zero.  :blush: P.S. I assume we work over the real number field

Doing a block decomposition $ A=\left(\begin{array}{cc}U& -V^T\\V & W\end{array}\right)$ for the even case $ n=2m$ with $ m>1$, one can immediately verify the  following two identities (e.g., for the first identity choose $ 1\leq i,j\leq m$ and $ k=m+i$):

$ U=V-diag(V)\cdot E$ and $ W=V-E\cdot diag(V)$

where $ E$ is the matrix of my previous post and $ diag(V)$ is generated from $ V$ by setting all off-diagonal entries to zero and keeping the diagonal elements. Since $ U,W$ are both skew-symmetric, we must have the relation $ V+V^T=E\cdot diag(V)+diag(V)\cdot E$ and we compute:

$ \begin{array}{lcl}\det A & = &\det\left(\begin{array}{cc}V-diag(V)\cdot E& -V^T\\V & V-E\cdot diag(V)\end{array}\right)\\[0.4cm]
 & = & \det\left(\begin{array}{cc}-diag(V)\cdot E& diag(V)\cdot E\\V & V-E\cdot diag(V)\end{array}\right)\end{array}$

by subtracting the lower blocks from the upper blocks. The determinant is clearly zero if $ m>1$ holds, since the first $ m$ rows are multiples of the vector $ (\underbrace{-1,\ldots,-1}_{m\mbox{ entries}},\underbrace{1,\ldots,1}_{m\mbox{ entries}})$, i.e. they are linearly dependent.


function
search
geometry
circumcircle
calculus
integration
combinatorics unsolved

Two triangle are said to be of the same type if they are both acute triangles, both right triangles, or both obtuse triangles.Let n be a positive integer and let P be

 a  n-sided regular polygon.Exactly one magpie sits at each vertex of P  . A hunter passes by, and the magpies fly away. When they return, exactly one magpie

 lands on each vertex of P, not necessarily in its original position. Find all n for which there must exist three magpies with the following property:the triangle

 formed by the vertices the magpies originally sit at, and the triangle formed by the vertices they return to after the hunter passes by, are of the same type.

Very nice problem.

First, we want to understand how many triangles of each type there are in the polygon. We can be optimistic and hope that in many cases, just counting those triangles will impose restrictions on how they can be permuted among themselves, restrictions which would suffice to rule out the possibility of every triangle changing class. This, indeed, turns out to be the case. So our strategy is first to determine those numbers, then check if any "scrambling" permutation is at all possible based on the combinatorics of those numbers, and finally check if we can realize the scrambling of the triangles with an actual permutation of vertices.

[hide="Counting Triangles"]
Let $ P$ be a regular $ n$-gon, and let $ T$ be the set of triangles whose vertices are vertices of $ P$. We write $ T \equal{} T_1 \cup T_2 \cup T_3$ where $ T_1$ are the acute triangles, $ T_2$ are the right ones, and $ T_3$ the obtuse triangles. Let $ t_i \equal{} |T_i|$ be the size of those sets. We want to calculate the numbers $ t_i$ as a function of $ n$. Obviously $ t_1\plus{}t_2\plus{}t_3 \equal{} \binom{n}{3}$.

The trick to doing so is similar to that used in the solution of [url=http://www.mathlinks.ro/viewtopic.php?search_id=257584860&t=226964]this very famous problem[/url]. Choosing an obtuse triangle is the same as choosing three vertices on the same (open) half-circle, and choosing a right triangle is the same as choosing three vertices of which two span a diameter of the circumcircle of $ P$. To do the precise calculation we separate the cases of $ n$ odd and $ n$ even.

If $ n$ is odd, $ n\equal{}2m\plus{}1$, then there are no right triangles (since no two vertices of $ P$ are diametrically opposed). To choose an obtuse triangle we choose any initial vertex in $ 2m\plus{}1$ ways, then look at the $ m$ vertices which follow that initial vertex in a counterclockwise sense and choose two of these in $ \binom{m}{2}$ ways. All obtuse triangles are obtained in this way precisely once: each obtuse triangle may lie on several arcs of length $ m$, but only one of them starts at a vertex and proceeds counterclockwise. Thus in this case:

1. $ n \equal{} 2m\plus{}1$
2. $ t_1 \equal{} \frac{(2m\plus{}1)(m\plus{}1)m}{6}$ (acute)
3. $ t_2 \equal{} 0$ (right)
4. $ t_3 \equal{} \frac{(2m\plus{}1)m(m\minus{}1)}{2}$ (obtuse)

If $ n\equal{}2m$ is even, we get using the same method:

1. $ n \equal{} 2m$
2. $ t_1 \equal{} \frac{m(m\minus{}1)(m\minus{}2)}{3}$ (acute)
3. $ t_2 \equal{} 2m(m\minus{}1)$ (right)
4. $ t_3 \equal{} m(m\minus{}1)(m\minus{}2)$ (obtuse)

The main thing to note here is that $ t_2$ grows much more slowly than the other ones (it's $ O(m^2)$ compared with $ O(m^3)$). This isn't at all surprising since right triangles are much more constrained than obtuse and acute ones.
[/hide]

[hide="Ruling out most cases"]
We now consider what these numbers say about the possibility of permuting the vertices such that each triangle changes class. The odd case is easy: since there are no right triangles ($ t_2\equal{}0$), we must have $ t_1\equal{}t_3$ to even have a chance of doing so. Looking at the formulas above we see that this happens only when $ m\equal{}2, n\equal{}5$. Therefore, any odd polygon except possibly the pentagon forces at least one triangle to keep its class under a permutation of the vertices.

In the even case, we reason as follows. Assume there is a permutation of the vertices which makes all triangles change class. Assume that $ x$ acute triangles became right after the permutation. Then the other acute triangles (of which there are $ t_1 \minus{} x$) all became obtuse. Of the $ t_2$ right triangles we got after the permutation, $ x$ were acute, so $ t_2\minus{}x$ were obtuse. Of the $ t_3$ obtuse triangles we got after the permutation, $ t_1\minus{}x$ were acute, so the $ t_3\minus{}t_1\plus{}x$ other ones were right. The other $ t_1\plus{}t_2\minus{}t_3\minus{}x$ right triangles became acute, so $ t_3\minus{}t_2\plus{}x$ obtuse triangles became acute.

All of those numbers: $ x, t_1\minus{}x, t_2\minus{}x, t_3\minus{}t_1\plus{}x, t_3\minus{}t_2\plus{}x, t_1\plus{}t_2\minus{}t_3\minus{}x$ must be integral and non-negative if we are to succeed in finding a permutation that causes that number of triangles to move to the prescribed classes. Looking at the formulas again, we see that $ t_3 \equal{} 3t_1$, so the last quantity is actually $ t_2\minus{}2t_1\minus{}x \geq 0$, so $ x \leq t_2\minus{}2t_1$. But a quick look at the formulas reveals that $ t_2 < 2t_1$ as soon as $ m \geq 6$, so all even polygons with $ 12$ or more vertices are safe (all permutations keep one triangle in its class). The remaining cases are easily checked by hand, and we find that the only candidate for a "scrambling" permutation is $ m\equal{}4, n\equal{}8$: the octagon. Indeed, the regular octagon contains $ t_1\equal{}8$ acute triangles, $ t_2\equal{}24$ right ones and $ t_3\equal{}24$ obtuse ones. Any value of $ x$ with $ 0 \leq x \leq 8$ works, in this case.
[/hide]

[hide="Verifying the two remaining cases"]
So, we are left with only two values of $ n$ for which we haven't yet ruled out the possibility of a "scrambling" permutation. Look at $ n\equal{}5$ first. The pentagon contains $ 5$ acute triangles and $ 5$ obtuse ones, and we can map them into each other by the permutation which takes $ 12345$ into $ 14253$. As you can easily check this permutation of the vertices makes each acute triangle into an obtuse one and each obtuse triangle into an acute one. So $ n\equal{}5$ is one instance where the claim "there must exist a triangle which stayed in its class" is false.

The case $ n\equal{}8$ is more interesting. I believe it is impossible to permute the vertices of an octagon and make all triangles change classes, but the only proof I have right now is computer-based. I'm pretty sure there's a simple argument to show this and I'll update the post if I find it.
[/hide]

[hide="Conclusion"]
So, overall, the result is that for every $ n \neq 5$, all permutations must keep at least one triangle in its class, while for $ n\equal{}5$ there is a permutation which changes all classes.
[/hide]


group theory
abstract algebra
superior algebra
superior algebra solved

If $G$ is a finite group such that $\forall a,b \in G-\{e\}, \exists f \in Aut(G), f(a)=b$, then $G$ is abelian.

First of all, we notice that all the non-trivial elements of the group have the same prime order $p$, because they clearly have the same order, and if it were $uv,\ u,v>1$, then the $v$'th power of an element would be a non-trivial element with order $u\ne uv$. This means that our group is a $p$-group, which must then have a non-trivial center. Since the center is a characteristic subgroup, the entire group must be abelian.


Find $(a,b,c)$ where both $a^{2}+1$ and $b^{2}+1$ are primes and also satisfying $(a^{2}+1)(b^{2}+1) = c^{2}+1$.

I found this hard for me, can anyone help ?

I have question bind with the thema. It's true that exist infinity many primes numbers form $a^{2}+1$ ?

we have that $a^{2}=\frac{c^{2}-b^{2}}{b^{2}+1}$. So $b^{2}+1|c-b$ or $b^{2}+1|c+b$. If we have $b^{2}+1=c-b$ or$b^{2}+1=c-b$ exercise is easy. So we can suppose that $c+b \geq 2(b^{2}+1)$. Because $b^{2}-b+1>0$ so we must to consider posibility only when $c=b^{2}+1+b$ or $c=b^{2}+1-b$. After substitute and dividing by $b^{2}+1$. We have that $a=b+1$ or $a=b-1$ It's gives us that$b=2$and $a=1$or$a=3$. We must to test this solution.

[EDIT] Not that question. This proof is when a,b are primes :blush:
Let $a \leq b$
1. if $a=2,b=2$ then we have $c^{2}=8$, which has no integers solutions.
2. if $a=2, b \geq 3$, then
$5(b^{2}+1) = c^{2}+1$
$5b^{2}= c^{2}-4 = (c+2)(c-2) = \lambda (\lambda+4)$
So $5=\lambda$ and $b^{2}| (\lambda+4)$ (I won't prove this. It's trivial.) This gives the solutions (2,3,7), (3,2,7)
2. if $a,b \geq 3$ then a,b are odd, so
$a^{2}+1 \equiv 2 \mod 4$
$b^{2}+1 \equiv 2 \mod 4$
$c^{2}+1=(a^{2}+1)(b^{2}+1) \equiv 0 \mod 4$
$c^{2}\equiv-1 \mod 4$
So in this case there are no solutions.

@wojto111 : I don't think so.. However, I can't prove that..  :(

@krassi_holmz : Where's the case if $a^{2}+1 = 2$ ? And why if $5b^{2}= x(x+4)$, then $x = 5$ ? Thanks..  :)


si asta e interesanta...calculati $\tan 1+\tan 2+...+\tan 89$.


[quote="bboypa"]Let a,b,c non negative reals with sum 1. Show that 4a³+4b³+4c³+24abc-1 is non negative.[/quote]
Well, we have

$ 4 a^3 \plus{} 4b^3 \plus{} 4c^3 \plus{} 24abc \minus{} (a \plus{} b \plus{} c)^3$
=$ 3 \left(a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc \minus{} \sum_{cyc}{ab(a \plus{} b)}\right)$
From Schur inequality for 3 variable , we have the answer

Very nice Gibbernergy!

My solution: a³+b³+c³+6abc = (a+b+c)((a+b+c)²-3(ab+bc+ca))+9abc= 1-3(ab+bc+ca)+9abc; we need to show that 36abc+3 ≥ 12(a(b+c)+bc), that is: 12(1-b-c)bc+1 ≥ 4(b+c-(b+c)²+bc) iff 16bc+1+4b²+4c² ≥ 4b+4c+12b²c+12bc². Define S=b+c ≤ 1 and P=bc ≤ (S/2)² ≤ 1/4, so: 4S²+8P+1≥4S+12SP iff (2S-1)²≥4P(3S-2) that is true in fact if 0≤S≤2/3 then (2S-1)²≥0≥4P(3S-2); otherwise if 2/3<S≤1 we have 3S-2≥2S-1 so it'd be sufficient to show that 2S-1≥4P, that is 2S-4P≥2S-4(S/2)²=S(2-S)≥1 iff (S-1)²≥0: we have also found all and unique cases of equality that is (0,1/2,1/2) and their permutations.

Edit: the use of lagrange multiplier is like cheating in olimpiad problems! However i realized now that i have showed a stronger statement: 4a³+4b³+4c³+12abc-1 is non negative.  :wink:

I think that this is very simple if we use the method of lagrange multipliers...right??? :wink:


analytic geometry
real analysis
real analysis unsolved

Let $(\mathbb{R}^{n},|\,\,| )$ be an euclidean space with an ordinary norm. Prove that if $(x_{i})$ converges weakly in $\mathbb{R}^{n}$ to the point $x$, then $|x_{i}-x|\to 0$.

The $n$ coordinates define linear functionals $(x^{1},x^{2},\dots,x^{n})\mapsto x^{k}$, which are of course continuous. By the definition of weak convergence $x_{i}^{k}\to x^{k}$ for $k=1,\dots,n$.


calculus
derivative
analytic geometry
function
graphing lines
slope
calculus computations

Consider the curve given by x^2 y - x^3 y = 6

a) Show that dy/dx= (3x^2 y - y^2) / (2xy - x^3)

b) Find all these points ont he curve whose x-coordinate is 1, and write an equation for the tangent line at each of these points.

c) Find the x-coordinate of each point on the curve where the tangent line is vertical.

I am completely lost with this and don't know where to go with it.  Thanks if anyone else knows.

This is implicit differentiation; have you heard that name?

Here's an explanation, at the beginning calculus level (the subject should be revisited in multivariable calculus):

Suppose we have a function defined by an equation in $x,y$ such as $x^3+5x^2y+2y^3=0$. Rewrite this as $x^3+5x^2\cdot Y(x)+2(Y(x))^3=0$, and assume that $Y$ is a differentiable function. In order to find $Y'$, we differentiate the above equation, treating $Y(x)$ as an unknown function of $x$.
In this case, that's $3x^2+(10x\cdot Y(x)+5x^2\cdot Y'(x))+2\cdot 3\cdot (Y(x))^2\cdot Y'(x)=0$,
which can be rearranged to $(3x^2+10xY(x))=(-5x^2-6(Y(x))^2)Y'(x))$, so
$Y'(x)=-\frac{3x^2+10xY(x)}{5x^2+6(Y(x))^2)Y'(x)}$. We would usually just write $y$ instead of $Y(x)$, so then that becomes $\frac{dy}{dx}=-\frac{3x^2+10xy}{5x^2+6y^2}$.

All of the above can be justified at this level except the assumption that $Y$ is differentiable.

Okay, so I looked at it a bit more and got done with (a) and (b).

Any help with (c)...really lost on that.  Would it work since the vertical would mean no slope, to set the bottom of the equation = to 0, then solve for Y, then plug in those values to the original equation...

Or I may be completely wrong and need help.

That's right. Another way to think of it, and get the same thing, is simply to reverse the roles of $x$ and $y.$ We have (by the very same calculation) that

$\frac{dx}{dy}=-\frac{5x^2+6y^2}{3x^2+10xy}$

(This is jmerry's example, not your original.)

Vertical tangents happen when $\frac{dx}{dy}=0.$


advanced fields
advanced fields unsolved

{x} denote the fractional part of x.
1.(easy) any a$ \neq$b  a,b$ \notin$Z  there exist infinit interger n that $ a^n \minus{} b^n$ is not interger.
2.(hard) a is rational, and a>1. are {$ a^n$} $ n\in N$dense on [0,1]? further more are they equably distributed?

2. Did you mean a<1 ?

edit: i'm sorry, i didn't notice { }


Does anyone know of any thorough, well-explaining probability books that are solid and good enough to prepare you for Mu Alpha Theta Probability (at the National Topic Test level) and the AMCs?

The AoPS probability and counting books are really good.

http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3

I saw that but I'm wondering... does it actually teach you how to "think" in probability. For the most part, I know almost all the Algebra 2 Probability formulas off the top of my head (except for fractional and negative expansions), but when they stick probability into a word problem, I can't see how to put everything into the formulas. The Algebra 2 books I have looked at (Foerster included) have fair-good prob. sections, however they still haven't enabled me to fully comprehend the probability problems on some of the tests I take.

The Art of Problem Solving books will teach you how to "think" (in probability or otherwise) much better than most other math books.  If you're looking for books that teach you to think, then those are for you.


quadratics
calculus
integration
algebra
algebra proposed

If positive integers $p,q,r$ are such that the quadratic equation $px^2-qx+r=0$ has two distinct real roots in the open interval $(0,1)$, find the minimum value of $p$.

Must be fixed q or r, because (p,q,r) and (pc,qr,cr) both satisfyed your property.

I don't think there's a problem. Note that what we want is the minimal integral value of $p$ (which must exist).

Let $0<a<b<1, q=p(a+b),r=pab,$ p any positive satisfyed your condition. Has not minimum. inf(p)=0.

To Rust: $p$, $q$, $r$ must be positive integers!!

Possibly a more rigorous solution?

Let the quadratic be $ax^2-bx+c$ where $a,b,c$ are positive integers.  It suffices to find $a,b,c$ such that

$\frac{b-\sqrt{b^2-4ac}}{2a} > 0$
$\frac{b+\sqrt{b^2-4ac}}{2a} < 1$

and $a$ is as small as possible.  The first equation always holds so rewrite the second as

$b+ \sqrt{b^2-4ac} < 2a$

We note that $b^2 > 4ac \geq 4a$ since $c \neq 0$ (roots are distinct) so ...
For $a = 1$ note $b \geq 3$ so the LHS is at least $3+\sqrt{3^2-4} > 2$
For $a = 2$ note $b \geq 3$ so the LHS is at least $3+\sqrt{3^2-8} = 4$
For $a = 3$ note $b \geq 4$ so the LHS is at least $4+\sqrt{4^2-12} = 6$
For $a = 4$ note $b \geq 5$ so the LHS is at least $5+\sqrt{5^2-16} = 8$
For $a = 5$ we may take $b = 5$ and $c = 1$ and verify that $5 + \sqrt{5^2-20} < 10$.

So minimal $a = 5$ as shown by the quadratic $5x^2 - 5x + 1 = 0$.

[quote="ThAzN1"]It suffices to find $a,b,c$ such that
$\frac{b-\sqrt{b^2-4ac}}{2a} > 0$
$\frac{b+\sqrt{b^2-4ac}}{2a} < 1$
and $a$ is as small as possible.[/quote]
You forgot the end braces for \frac, by the way.

letting $f(x)=px^2-qx+r$ then $f(1)>0$ so $p+r>q$ , $q^2>4pr$ also the minimum point is $\frac{q}{2p}<1$

then we can find $p>r$ so $2p>p+r>q$ and we get $2p\geq q+2$

also $p>r\geq 1$ so $p\geq 2$

when $p=2$ , $q^2>8$ and so $q\geq 3$  ( does satisfy $q<2p$)

when $p=3$ , $q^2>12$ and $q\geq 4$ (doesnt satify $p+r>q$)

when $p=4$ , $q\geq 5$ (doesnt satisfy $p+r>q$)

when $p=5$ , $q\geq 5$ and this one satisfy the condition . So the min for $p=5$


Points $ A,B,C,D$, and $ E$ are located in 3-dimensional space with $ AB\equal{}BC\equal{}CD\equal{}DE\equal{}EA\equal{}2$ and $ \angle ABC \equal{} \angle CDE \equal{} \angle DEA \equal{} 90^{\circ}$. The plane of $ \Delta ABC$ is parallel to $ DE$. What is the area of $ \Delta BDE$.

[hide] My solution makes the problem trivially easy for a #25, which is why I'm suspicious and posting it here. The area of $ \Delta BDE$ is the length of $ DE$ times the height from $ B$ to $ DE$. Since $ BA$ is parallel/skew to $ DE$, the height from $ B$ to $ DE$ is the same as the height from $ A$ to $ DE$. Since we're given that $ AE$ is perpendicular to $ DE$ and that $ AE\equal{}2$, that height is clearly $ 2$. So our answer is $ \frac{1}{2} (2*2) \equal{} 2$ (which is the correct answer). This seems far too easy for a #25, so I suspect I made an error in reasoning and just ended up with the right answer by luck. Thoughts?[/hide]

[hide]yeah when i did it also turned out to be a prism with equilateral triangle base and same height as base, all 2s. I guess they put it as 25th question either as a joke or trying to pick on the 3D-challenged. lol[/hide]


A vector $ (x_1,x_2,\dots,x_n)\in%Error. "mathdd" is a bad command.
{R}^n$ is called [b]$ k$-alternating [/b] (where $ k \leq n$) if it satisfies:
 $ a$) $ sup_{i}|x_i|=1$
 $ b$) there exists a subsequence $ (n_i)_{i=1}^{k}$ of $ \{1,2,\dots,n\}$ such that $ x_{n_i}=(-1)^{i}$ for all $ i\leq k$.

[u]Problem:[/u] Show that for any $ k\leq n$, any $ k$-dimensional subspace of $ %Error. "mathdd" is a bad command.
{R}^n$ contains a $ k$-alternating vector.

[u]Generalization:[/u] Fix $ \varepsilon=(\varepsilon_i)_{i=1}^{n}$, where $ \varepsilon_i=\pm 1$ for any $ i\leq n$. A vector $ (x_1,x_2,\dots,x_n)\in%Error. "mathdd" is a bad command.
{R}^n$ is called [b]$ (k,\varepsilon)$-sample [/b] (where $ k \leq n$) if it satisfies:
 $ a$) $ sup_{i}|x_i|=1$
 $ b$) there exists a subsequence $ (n_i)_{i=1}^{k}$ of $ \{1,2,\dots,n\}$ such that $ x_{n_i}=\varepsilon_i$ for all $ i\leq k$.

Give necessary and sufficient condition on $ \varepsilon$ such that for any $ k$, any $ k$-dimensional subspace of $ %Error. "mathdd" is a bad command.
{R}^n$ contains a  $ (k,\varepsilon)$-sample vector.


Treething gets banned once every five days. If he is first banned on January 3rd, 2007, how many times will he be banned in total throughout 2007?

[hide]There are 365 days in 2007, and the first two days have already passed.  365-2=363.  So you will get banned 363/5 or 73 times (after rounding).[/hide]


inequalities
function
inequalities proposed

Prove that if a,b,c>0 have product 1, then sum sqrt(3x^2+1)<=2 sum x.

[quote="harazi"]Prove that if $a,b,c>0$ have product $1$, then $\sum \sqrt{3x^2+1}\leq 2 \sum x$.[/quote]

 :)

[quote="zhaobin"][quote="Soarer"]but the trick applied to this inequality doesn't seem to work with harazi's one.[/quote]
what do you mean?[/quote]

sorry i was speaking nonsense... :blush:

[quote="zhaobin"]$Soarer$,I think you are right,because the inequality posted by $nthd$ is different to harazi's...

$nthd$,can you post your solution?[/quote]

no, nthd's ineq is a generalization. Put k = 4/3 in his ineq

[quote="zhaobin"]$Soarer$,I think you are right,because the inequality posted by $nthd$ is different to harazi's...

$nthd$,can you post your solution?[/quote]

no, nthd's ineq is a generalization. Put k = 1/3 in his ineq

Soarer, what are you doing? :?
Isn't it $n=3$ and $k=3$?

[quote="zhaobin"]ok,you are right leepakhin.I and Soarer are both have a mistake.
And nthd can you post your solution?[/quote]yes,wait me a day.I must translate it to English :)

I posted my solution ,but I don't know whyI cann't posting it.

why?do you want tu upload a file,then you can't?

[quote="zhaobin"][quote="nthd"]I don't know.when I click Submit,my post is deleted![/quote]
ok,can you send you solution to me?Then I will post it here for you?[/quote]
I sent my solution toyou,zhaobin! :) Thanks.

ok,now let me post nthd's solution,(Thanks to nthd)
PROBLEM::with  all $a_1,a_2,...,a_n>0: a_1a_2...a_n=1$ and $k\ge 0$  $(1)$
then:

$\sum_{i=1}^n \sqrt{1+ka_i^2} \le \sqrt{1+k} \sum_{i=1}^n a_i$
SOLUTION:let $f(x)=\sqrt{1+k}x -\sqrt{1+kx^2}$
LEMMA1:IF $a \ge 1 \ge b>0$ and $ab \le 1$,then we have:(zhaobin's PS:I don't why should we restrict that $ab \le 1$:))
$f(a)+f(b) \ge f(1)+f(ab)$
LEMMA2:IF $a\ge b > 0$ then:$f(a)+f(b) \ge 2f(\sqrt{ab})$
PROOF of LEMMA1:
fixed $ab=x \in (0,1],a \in [1,+\infty)$
$b=b(a)=\frac x a$,then $\frac{d b(a)}{d a}=-\frac{x}{a^2}=-\frac{b}a$
let $F(a)=f(a)+f(b)=f(a)+f(\frac x b),a \in [1,+\infty)$
then:$F'(a)=\frac{a-b}{a} \sqrt{1+k}-\frac k a(g(a)-g(b))$
where $g(y)=\frac{y^2}{\sqrt{1+ky^2}}$,
use largrange theorem:$g(a)-g(b)=(a-b)g'(t),t \in (a,b)$
notice:$g'(t)=\frac{2t}{\sqrt{(1+kt^2)^3}}$
then we get:
$F'(a)=\frac{a-b}{a}(\sqrt{k+1}-\frac{2kt}{\sqrt{(1+kt^2)^3}}) \ge 0$ (it is easy to check)
so $F(a) \ge F(1)$
which means lemma 1 is proved.

PROOF OF LEMMA 2:
$F'(a)=\frac{a-b}{a}(\sqrt{k+1}-\frac{2kt}{\sqrt{(1+kt^2)^3}}) =\frac{a^2-x}{a^2}(\sqrt{k+1}-\frac{2kt}{\sqrt{(1+kt^2)^3}}) \ge 0,\forall a \ge \sqrt x$
thus $F(a) \ge F( \sqrt x)$
lemma 2 is proved.

now for the original problem:
I think use the lemma2 is enough :)
$(1) \iff f(a_1)+f(a_2)+\cdots+f(a_n) \ge 0$
it is easy to now by lemma 2,and the continuausty of the function :)

by the hint of nthd.I think we can prove by
 $f(x)=\sqrt{1+k}e^x -\sqrt{1+ke^{2x}}$
 is convex on $R$ :)

[quote="zhaobin"] by the hint of nthd.I think we can prove by
 $f(x)=\sqrt{1+k}e^x -\sqrt{1+ke^{2x}}$
 is convex on $R$ :)[/quote] But $f(x)$ is not convex on $R$. Therefore, the hint of nthd is wrong. I thing there is a mistake for $g'(x)$.
Moreover, the inequality is not valid for any positive k.  ;)


Here are two problems that are bothering me.  As always, any help would be much appreciated:

5. The age of a wooden artifact was determined by carbon-14 dating. Three different measurements gave values of 6925, 4815, and 5485 years. Which is the best value to report for the age of this artifact?

(A) 5700 years
(B) 5740 years
(C) 5741 years
(D) 6000 years

[hide]Obviously, you need to average.  Averaging gives you 5741.  How do you decide the correct number of significant figures?  The correct answer is D.[/hide]

36. When these acids are arranged in order of increasing strength (weakest acid first), what is the correct order?

(A) HOCl(aq) < H2SeO4(aq) < H3PO4(aq)
(B) H3PO4(aq) < H2SeO4(aq) < HOCl(aq)
(C) H2SeO4(aq) < HOCl(aq) < H3PO4(aq)
(D) HOCl(aq) < H3PO4(aq) < H2SeO4(aq)

[hide]The correct answer is D.  I have no clue as to what kind of trend this is following.[/hide]

Thanks![/hide]

36. You compare the number of O's to H's in the oxyacids.

An age of "6000 years" implies an error of +/- 500 years.  An age of 5700 years implies a possible error of +/- 50 years,  which seems pretty optimistic in this case.


Prove that for every n,n is natural number,the numbers 18^n and 18^n+2^n have the same number of digits.(For example if n=1,then 18^1=18 and has 2 digits;18^1+2^1=20 again 2 digits.)

This problem doesn't belong here, way too difficult for Middle School Classroom Math. Pre-Olympiad is the right forum for this problem.

[hide]
Let $ d$ be the number of digits in $ 18^n$. Then $ 10^{d\minus{}1}\leq 18^n<10^d$, and we must show $ 10^{d\minus{}1}\leq 18^n\plus{}2^n<10^d$.

The left hand inequality is obvious. For the right hand inequality, as $ 18^n<10^d<18^d$, $ d>n$. As a result, $ 18^n$ and $ 10^d$ are both multiples of $ 2^n$, so their difference is. We get $ 10^d\minus{}18^n\geq 2^n$, so $ 18^n\plus{}2^n\leq 10^d$.

It remains to show equality can't happen. Suppose it did, so $ 2^n(9^n\plus{}1)\equal{}10^d$. As $ 9^n\minus{}1$ is divisible by 4 for all $ n$, $ 9^n\plus{}1$ is not. So the left hand side is not divisible by $ 2^{n\plus{}2}$, giving $ d<n\plus{}2$.

Now as $ n<d<n\plus{}2$, we must have $ d\equal{}n\plus{}1$ and so $ 9^n\plus{}1\equal{}2\cdot 5^{n\plus{}1}$. Just to have $ 9^n\plus{}1$ divisible by 5, $ n$ is odd.

So $ n\equal{}1\plus{}2m$ for some nonnegative $ m$. As a result, $ 9^n\plus{}1$ is 10 more than a multiple of 16 (because $ 81^m$ is 1 more than a multiple of 16). This must be true of $ 2\cdot 5^{n\plus{}1}$ as well, so $ 5^{n\plus{}1}$ must be 5 more than a multiple of 8. But for odd $ n$, $ 5^{n\plus{}1}$ is 1 more than a multiple of 8, a contradiction.
[/hide]

$ 18^n$ has $ n \plus{} 1$ digits.

The proof that $ \displaystyle \lfloor n\log_{10}{2} \plus{} \log_{10}{(3^{2n} \plus{} 1)}\rfloor \equal{} n$ seems to have been shown by scorpius119...

EDIT: sorry that alone probably makes no sense at all. The digits of $ 18^n$ is found by $ 1\plus{}\lfloor \log_{10}(18^n)\rfloor$.


function
algebra
domain
limit
topology
real analysis
real analysis unsolved

find f(x) defined on the interval [0,n] whose range on any interval (a,b) is [0,n] whenever $ 0 \leq a < b \leq n$, where $ n$ is a fixed natural number.

It has been solved I think by t0rajiru a long time ago using the concept of equivalence classes (and axiom of choice).

Of course, it's sufficient to show for n=1 ;)

A stronger statement: there exists a function $ g: \mathbb{R}\to \mathbb{R}$ such that for each interval $ (a,b)$, and each $ \alpha\in \mathbb{R}$, the set $ \{x\in (a,b): g(x) \equal{} \alpha \}$ is uncountable (e.g. one constructed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228437]here[/url]).  Constructing the function you're looking for from this $ g$ is straightforward.  For example:
\[ f(x) \equal{}\min \left \{\max\{g(x),0\},n \right\}\]

I prefer [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=195618#195618]this version[/url].

Perhaps next week I will post a more constructive solution. In the meantime maybe you can understand that the fibers of f must be dense in [0,1]... ;)

JoeBlow's claim is hardly difficult if you can exhibit even countable dense fibers of f in your case since many ideas carry over.

I have been looking for this thread. I forgot about my promise.

It is sufficient to study the case n = 1.

Simply study the function $ f: [0,1] \to [0,1]$ for which if x is in the domain and we have the binary representation $ x \equal{} 0.a_1 a_2 ...a_n ...$, then $ f(x) \equal{} \overline {\mathop {\lim }\limits_{n \to \infty } } \frac {{\sum\limits_{k \equal{} 1}^n {a_k } }}{n}$.  You can show that a sequence of the form shown here converges to every positive x in the domain, which implies that f is surjective. Assume in your analysis that there is no real number whose binary representation has a tail of '1's. Show that one can modify finitely many digits of the binary representation of x without changing f(x) and observe that in any open set in the domain there is a number which has the digits of x as a tail end of its binary sequence. This gives the density of the fibers in the domain.


A game is played in which the player flips a fair coin. If a heads shows on the first flip the player gets nothing and the game ends. If a tail shows on the first flip, the payout is $ 2 and the player continues to flip the coin until a head shows, or she has flipped the coin ten times at which time the game ends. The total payout doubles for each tail that shows in succession. What is the probability that a player earns at least$10? Express your answer as a common fraction.

I don't exactly understand what the question is asking I guess, because I've gotten it wrong 4 times.

Oh. Thanks!

I have another probability problem. (It's my weak spot)

Andy, Barbara and Christy play a game. Their probablilities of winning are $ \frac{1}{2}$, $ \frac{1}{3}$ and $ \frac{1}{6}$, respectively. What is the probability that each of them wins exactly one of the first three games played? Express your answer as a common fraction.

It's simply 1/2 times 1/3 times 1/6 times 3! to arrange the order they can win the games, which equals 1/6.

EDIT: Beaten


So here's how this game goes.  Anyone can post a logo of a [b]useful[/b] Web 2.0 site (or any useful website) and you PM the person who made the post with the name and URL of the website.  If the name is in the logo, post only part of the logo.  The answer will be posted the next day.

PMs should be submitted in the following form:
[b]Subject:[/b] Web logo answer.
[b]Body:[/b][code]For the logo posted on -Month- -Date-, -Year-,
[b]URL:[/b] http://...
[b]Name:[/b] ...[/code]

Simple, right?  If it is not in this form (or something very similar), it will not be counted.  a.k.a. Don't comment in your PMs.

I'll start easy.  It is part of the site's full logo.

Ok, time's up for the first one.
[hide="Click to view the answer."][b]URL:[/b] http://www.meebo.com/
[b]Name:[/b] Meebo[/hide][hide="Click to view submissions."][color=green][b]Correct answers (4):[/b][/color]
emannes
7h3.D3m0n.117
kimmystar94
Klebian
[color=red][b]Wrong answers (0):[/b][/color]
(none)[/hide]
Good job.

(probably) Not web 2.0 (I don't really know what that means lol), but here's a good one.

[img]http://img89.imageshack.us/img89/2861/logoaw2.png[/img]

Quick note: Please do not link images directly from their site. Participants will easily quote the post to find the original site and there would not be any point to this. None of the above 2 images have anything; again, just a warning.

The answer was [url]http://www.gentoo.org[/url] - gentoo linux

Two correct, no incorrects

[hide="corrects"]
emannes
MustItAlwaysBeMe
[/hide]

I'll post one! Sorry it's kinda blurry...

Well, the answer was[url=http://www.stumbleupon.com]StumbleUpon[/url]
2 corrects, 0 incorrects
[hide="correct"]
Klebian
MustItAlwaysBeMe
[/hide]


Let a_1, a_2, ..., a_n be positive real numbers and $ \sum_{i \equal{} 1}^{n} a_i > n\minus{}1$, prove that for $ n \in N$,

$ n^4 \sqrt[n]{\sum_{i \equal{} 1}^{n} a_i \minus{}n\plus{}1} (\sqrt[n]{\sum_{i \equal{} 1}^{n} a_i \minus{}n\plus{}1}\plus{}\frac{1}{n})(\sqrt[n]{\sum_{i \equal{} 1}^{n} a_i \minus{}n\plus{}1}\plus{}\frac{1}{n\plus{}1})(\sqrt[n]{\sum_{i \equal{} 1}^{n} a_i \minus{}n\plus{}1}\plus{}\frac{1}{n(n\plus{}1)}) \leq
\newline
 \sum_{i \equal{} 1}^{n} a_i(\sum_{i \equal{} 1}^{n} a_i\plus{} \frac{1}{n\plus{}1})(\sum_{i \equal{} 1}^{n} a_i\plus{}1\minus{}\frac{1}{n\plus{}1})(\sum_{i \equal{} 1}^{n} a_i\plus{}1)
\newline
\leq (n\sum_{i \equal{} 1}^{n} a_i ^2 \plus{}\sum_{i \equal{} 1}^{n} a_i \plus{}\frac{n}{2(n\plus{}1)^2})^2 \minus{} \frac{n^2}{4(n\plus{}1)^4}$

[quote="4865550150"]Let a_1, a_2, ..., a_n be positive real numbers and $ \sum_{i \equal{} 1}^{n} a_i > n \minus{} 1$, prove that for $ n \in N$,
$ \newline \sum_{i \equal{} 1}^{n} a_i(\sum_{i \equal{} 1}^{n} a_i \plus{} \frac {1}{n \plus{} 1})(\sum_{i \equal{} 1}^{n} a_i \plus{} 1 \minus{} \frac {1}{n \plus{} 1})(\sum_{i \equal{} 1}^{n} a_i \plus{} 1) \newline \leq (n\sum_{i \equal{} 1}^{n} a_i ^2 \plus{} \sum_{i \equal{} 1}^{n} a_i \plus{} \frac {n}{2(n \plus{} 1)^2})^2 \minus{} \frac {n^2}{4(n \plus{} 1)^4}$[/quote]

If we denote by $ \sum a_i\equal{}x$ and use CS for $ n\sum a^{2}_{i}\geq x^2$ then the inequality is obviously true after expanding :lol: . Ill check the other part tomorrow


Let $ \triangle ABC$ be right and $ AH$ be the altitude to the hypotenuse. Prove that $ A$ is the orthocenter of the triangle formed by the excenters of $ \triangle AHB$, $ \triangle AHC$, and $ \triangle BAC$.

Edit: Assume excenter of $ \triangle AXB$ is taken according to $ \angle AXB$. Sorry for this.

Let $ D$ be the excenter of $ \angle AHB$ of $ \triangle AHB$.  Let $ E$ be the excenter of $ \angle AHC$ of $ \triangle AHC$. Let $ F$ be the excenter of $ \angle BAC$ of $ \triangle BAC$.
It is obvious that $ D,B,F$ are collinear and so $ C,E,F$. With some simple angle algebra you can get $ \angle BFC \equal{} 90^\circ \minus{} \frac {\angle BAC}2 \equal{} 45^\circ$. Similarly $ \angle BDA \equal{} \angle AEC \equal{} 45^\circ$. So $ DA \perp FE$ and $ EA \perp FD$. Thus $ A$ is the orthocenter of $ \triangle DFE$.

It is obvious that $ (A$ is the orthocenter of $ \triangle DEF ) \Leftrightarrow (\angle BAC \equal{} 90^\circ$  and $ [AH]$ is the altitude $ )$


inequalities function calculus derivative

Let $ a,b,c,d$ be positive reals with sum 4 prove that -
$ \sum_{cyclic}\frac {1}{a^2 \plus{} 1} \ge 2$  :)

[hide="Solution?"]First, by the Trivial Inequality, $ a^2\plus{}1\plus{}b^2\plus{}1\plus{}c^2\plus{}1\plus{}d^2\plus{}1 \ge 2(a\plus{}b\plus{}c\plus{}d)\equal{}8$. Then we multiply $ sum_{\text{cyc}}\frac{1}{a^2\plus{}1}$ by $ 8\equal{}a^2\plus{}1\plus{}b^2\plus{}1\plus{}c^2\plus{}1\plus{}d^2\plus{}1$. By C-S this is $ \ge 16$, so dividing by $ 8\equal{}a^2\plus{}1\plus{}b^2\plus{}1\plus{}c^2\plus{}1\plus{}d^2\plus{}1$ we have the desired result.[/hide]

[quote="ThinkFlow"][hide="Solution?"]First, by the Trivial Inequality, $ a^2 \plus{} 1 \plus{} b^2 \plus{} 1 \plus{} c^2 \plus{} 1 \plus{} d^2 \plus{} 1 \ge 2(a \plus{} b \plus{} c \plus{} d) \equal{} 8$. Then we multiply $ sum_{\text{cyc}}\frac {1}{a^2 \plus{} 1}$ by $ 8 \equal{} a^2 \plus{} 1 \plus{} b^2 \plus{} 1 \plus{} c^2 \plus{} 1 \plus{} d^2 \plus{} 1$. By C-S this is $ \ge 16$, so dividing by $ 8 \equal{} a^2 \plus{} 1 \plus{} b^2 \plus{} 1 \plus{} c^2 \plus{} 1 \plus{} d^2 \plus{} 1$ we have the desired result.[/hide][/quote]

Your solution is not correct .

I see where I went wrong.  :oops:

What about:

The function $ f(x)\equal{}\frac{1}{x^2\plus{}1}$ is concave, so we have, by Jensen's Inequality, $ \frac{1}{4} \left( \sum_{\text{cyc}}\frac{1}{a^2\plus{}1} \right) \ge \frac{1}{\frac{a\plus{}b\plus{}c\plus{}d}{4} \plus{} 1} \equal{} \frac{1}{2}$, then multiply by $ 4$.

Power Mean rules.

$ \frac{1}{a^2\plus{}1}\equal{}a\minus{}\frac{a^3\plus{}a\minus{}1}{a^2\plus{}1}\ge a\minus{}\frac{a^3\plus{}a\minus{}1}{2a}\equal{}a\minus{}\frac{a^2}{2}\minus{}\frac12\plus{}\frac{1}{2a}$

Let $ x\equal{}\sum a^2,y\equal{}\sum \frac1a$

Then, by the above we have to prove that $ 4\minus{}\frac x2\minus{}2\plus{}\frac y2\ge 2$, or $ y\ge x$

It suffices to prove $ z\equal{}a^3\plus{}b^3\plus{}c^3\plus{}d^3\ge a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}x$, because $ (\sum \frac1a)(\sum a^3)\ge (\sum a^2)^2$ by Cauchy Schwarz

But, by Power Mean Inequality, we have

$ \sqrt[3]{\frac{z}{4}}\ge \sqrt{\frac{x}{4}}$, so

$ \frac{z^2}{16}\ge \frac{x^3}{64}$, or

$ 4z^2\ge x^3$.

Again use Power Mean:

$ \sqrt{\frac{x}{4}}\ge \frac{a\plus{}b\plus{}c\plus{}d}{4}\equal{}1$, so $ x\ge 4$. Similarly $ z\ge 4$ (we need this because when numbers are less than 1, we can't use the last inequality below).

$ 4z^2\ge x^3\ge 4x^2$, so $ z^2\ge x^2$. We have $ z\ge x$.

hmm..  bugi in your solution you have to prove $ 1 \ge \frac {y}{x}$ but u have proved that $ k \ge \frac {x}{y}$ where $ k \ge 1$ instead you are done only if you proved it the other way round ,so your solution is wrong i think.
[hide="my solution"]
write the expression as $ 4 \minus{} \sum_{cyclic} \frac {a^2}{a^2 \plus{} 1} \ge 4 \minus{} \sum_{cyclic} \frac {a^2}{2a} \equal{} 4 \minus{} (\frac {a \plus{} b \plus{} c \plus{} d}{2}) \equal{} 4 \minus{} 2 \equal{} 2$
this is much simpler i think[/hide]

Oh, yes it's wrong  :blush:

The error is in the $ a^3\plus{}b^3\plus{}c^3\plus{}d^3$, I shouldn't have brought it in.

$ \frac {1}{a^2 \plus{} 1} \ge \minus{} \frac {1}{2}a \plus{} 1 \iff a(a \minus{} 1)^2 \ge 0$, so $ LHS \ge \minus{} \frac {1}{2}(a \plus{} b \plus{} c \plus{} d) \plus{} 4 \equal{} 2$
How did I come up with this? I dunno, it was just magic, I pulled this identity out of thin air :D

Just kidding, I learned this trick from the poster [u]onlylovemath[/u], you suppose $ \frac {1}{a^2 \plus{} 1} \ge ma \plus{} n$, and equality holds when $ a \equal{} 1$, so let $ f(a) \equal{} \frac {1}{a^2 \plus{} 1} \minus{} ma \minus{} n$, then $ f(1) \equal{} 0$ since equality holds when $ a \equal{} 1$ and $ f'(1) \equal{} 0$ because that is the minimal point of $ f(a)$, solving, $ m \equal{} \minus{} \frac {1}{2}, n \equal{} 1$.

This is actually just geniusbliss' solution, but I wanted to show an interesting method that can be applied to more problems.

one doubt when we use functions,calculus do we always have to take second derivative and all to see if it is convex?


Another one I had trouble with.
A cube is sliced with one straight slice which passes through
two opposite edges. The result is two solids, as shown. The
area of the largest face on one of these two solids is
242 2 square units. What was the exact surface area of the
original cube? Express your answer in simplest form.
9. see [url]http://www.mnspe.org/mathcounts/2007tests/510-07_C_Team.pdf[/url] for picture although it just shows the cube being split diagonailly.

[color=darkred][size=75]Extra topic removed.
-Neal (nebula42)[/size][/color]

[hide="Slow Way"]To solve this, we can make an equation to find the side length of the cube using the area of the slant.

$ x\sqrt {2x^2} \equal{} 242\sqrt {2}$<--Equation
$ \sqrt {2x^4} \equal{} (242\sqrt {2})^2$<--Square both sides
$ x^2\sqrt {2} \equal{} 242\sqrt {2}$<--Square Root both sides
$ \sqrt {2} \equal{} \frac {242\sqrt {2}}{x^2}$<--Divide by $ x^2$
$ \frac {242\sqrt {x^2}}{\sqrt {2}} \equal{} x^2$<--Make $ x^2$ on its own
$ 242 \equal{} x^2$<--Cancel out square roots.
$ \sqrt {242} \equal{} x$

To to find the surface area, we square the side length and multiply by 6.

$ \sqrt {242}^2 \equal{} 242\times6 \equal{} \boxed{1452}$[/hide]

[hide="Faster Way"]Same as above, but way faster.

$ x^2\sqrt {2} \equal{} 242\sqrt {2}$<--We know the diagnal is $ x\sqrt {2}$ so it makes it faster.
$ x^2 \equal{} 242$<--Divide$ \sqrt {2}$
$ x \equal{} \sqrt {242}$
$ (\sqrt {242}^2)6 \equal{} \boxed{1452}$[/hide]

Hope that helps.

the fastest way:

$ x^2\sqrt{2}\equal{}242\sqrt{2}$
$ x^2\equal{}242$
$ 6x^2\equal{}242*6\equal{}1452$ <--- we don't need to find x, so square root of 242 is needless.

wow. a solution off. Theres an idea for my blog...

[quote="Bachukas"]the fastest way:

$ x^2\sqrt {2} \equal{} 242\sqrt {2}$
$ x^2 \equal{} 242$
$ 6x^2 \equal{} 242*6 \equal{} 1452$ <--- we don't need to find x, so square root of 242 is needless.[/quote]

eh....Bachukas' solution is faster than mine by one step....

Did you read and understand the solution?


Can somone please help me symbolize this difficult equation

http://i.imagehost.org/0332/Kopia_av_3095.jpg

I love you all generous helpfull people!

[hide="Solution"]Let the first number be $ x$ and the second number be $ y$.

First: $ x\plus{}7 \equal{} 2y$.

Second: $ x\minus{}14 \equal{} \displaystyle\frac{y}{2}$.[/hide]

It is useful to learn how to identify what are the variables, and to be able to convert verbal expressions to mathematical expressions. This is generally the first thing they teach you in Algebra.

To solve the equation now:

$ x\plus{}7\equal{}2y$
$ x\minus{}14\equal{}\frac{y}{2}$

$ x\equal{}2y\minus{}7$

$ 2y\minus{}7\minus{}14\equal{}\frac{y}{2}$

$ 4y\minus{}42\equal{}y$

$ 3y\equal{}42$

$ y\equal{}14$

$ x\equal{}2(14)\minus{}7\equal{}28\minus{}7\equal{}21$

$ (21,14)$ is the solution. Looking back at the sentences, we see that

21 is the first number and 14 is the second.


Prove that the product of two imaginary numbers in form $ a\plus{}bi$ cannot equal zero.

[hide]

There are three cases:
I. The complex numbers are identical
II. They are conjugates.
III. They are neither.

I.

$ (a\plus{}bi)(a\plus{}bi) \equal{} a^{2}\plus{}2abi\minus{}b^{2}$.  For the product to be zero, both the real parts and imaginary parts have to be zero.  The imaginary part is $ 2ab$, and it is zero if $ a \equal{} 0$ or $ b \equal{} 0$ or if both $ a,b \equal{} 0$.  The second and third cannot occur because $ b$ must be nonzero for the numbers to be imaginary.  For $ a \equal{} 0$, the imaginary part is ok, now let's look at the real part.  The real part of the product is $ a^{2}\minus{}b^{2}$, and since $ a \equal{} 0$, the real part becomes $ \minus{}b^{2}$.  To make the real part zero, $ b$ must equal zero.  However, that contradicts the fact that $ b$ cannot be zero.

[/hide]

[hide="One-liner"]
Zero product property (if the product of numbers is 0, at least one of the numbers must be 0)
[/hide]
[hide="Full solution"]
Let our numbers be $ a\plus{}bi$ and $ c\plus{}di$ with $ a,b,c,d$ real and neither of the complex numbers equal to zero. Otherwise, it's trivial since zero times anything is still zero. Assume the product is zero.

$ 0\equal{}(a\plus{}bi)(c\plus{}di)\equal{}ac\minus{}bd\plus{}(ad\plus{}bc)i$

If this equals zero, both the imaginary and real parts of the product must equal zero.

$ ac\minus{}bd\equal{}0$

$ ac\equal{}bd$

$ \frac{a}{b}\equal{}\frac{d}{c}$

$ ad\plus{}bc\equal{}0$

$ ad\equal{}\minus{}bc$

$ \frac{a}{b}\equal{}\minus{}\frac{c}{d}$

By the transitive property, we have
$ \frac{d}{c}\equal{}\minus{}\frac{c}{d}$

$ d^{2}\equal{}\minus{}c^{2}$
This is obviously impossible since $ c$ and $ d$ are real and they cannot both be zero. Thus we have a contradiction, and we know that it is impossible for the product of 2 nonzero complex numbers to equal zero.[/hide]
[hide="Remark"]
The above solution serves as a base case for a simple inductive proof of the zero product property for complex numbers. You can now prove that the product of $ n$ complex numbers never equals zero unless at least one of the numbers is zero.[/hide]

[quote="FightTheTide"][hide="One-liner"]
Zero product property (if the product of numbers is 0, at least one of the numbers must be 0)
[/hide][/quote]

As low as my standards for one-liners are, the problem asks to prove the zero product property for complex numbers, so implicitly this is not given.  By the way, the problem should state that the product of two [b]nonzero[/b] imaginary numbers is not zero.

[hide="A more acceptable one-liner"] $ (a\plus{}bi)(c\plus{}di) \equal{} 0\implies (a^{2}\plus{}b^{2})(c^{2}\plus{}d^{2}) \equal{} 0\implies a \equal{} b \equal{} c \equal{} d \equal{} 0$ [/hide]

$ N(w)*N(z)\equal{}N(zw)\equal{}N(0)\equal{}0$ but then $ N(z)\equal{}0\iff z\equal{}0$...