calculus
integration
function
factorial
real analysis
algebra
domain

Evaluate: \[ \int_0^\infty \frac{x^3}{e^x\minus{}1} dx\]
[hide="What is Polylog function?"]
In Mathematica it says that the integral of function is in Polylog Function which I don't really understand.  :|[/hide]

I think this does not have an elementary  antiderivative

We have,
$ I \equal{} \int_{0}^{\infty}\frac{x^{3}}{e^{x}\minus{}1}dx$

$ \equal{} \int_{0}^{\infty}\frac{x^{3}}{e^{x}\minus{}1} \cdot \frac{e^{\minus{}x}}{e^{\minus{}x}} dx$

$ \equal{} \int_{0}^{\infty} x^{3} e^{\minus{}x} \sum_{n\equal{}0}^{\infty} \frac{e^{\minus{}nx}}{n!} dx$

$ \equal{}3! \sum_{n\equal{}0}^{\infty} \frac{1}{n!} \int_{0}^{\infty} \frac{x^{3}}{3!} e^{\minus{}(n\plus{}1)x} dx$ [the switching of the sum with integral sign would need justification, here one can invoke Tonelli's]

$ \equal{} 3! \sum_{n\equal{}0}^{\infty} \frac{1}{n!} \cdot \boxed{\int_{0}^{\infty} \frac{x^{3}}{3!} e^{\minus{}(n\plus{}1)x} dx}$

$ \equal{} 3! \sum_{n\equal{}0}^{\infty} \frac{1}{n!} \cdot \boxed{\frac{1}{\left(n\plus{}1\right)^4}}$ [this is obtained by integration by parts or by appealing to Laplace transform]

$ \equal{} \sum_{n\equal{}0}^{\infty} \frac{3!}{\left(n\plus{}1\right)! \left(1 \plus{} n\right)^3}$

Ah, akech's calculation is wrong. In particular, $ \frac1{e^x\minus{}1}\equal{}\frac{e^{\minus{}x}}{1\minus{}e^{\minus{}x}}\equal{}e^{\minus{}x}\plus{}e^{\minus{}2x}\plus{}\cdots\plus{}e^{\minus{}nx}\plus{}\cdots$. There are no factorials in that expansion.
Now,
$ I\equal{}\int_0^{\infty}x^3\sum_{n\equal{}1}^{\infty}e^{\minus{}nx}\,dx$
$ \equal{}\sum_{n\equal{}1}^{\infty}\int_0^{\infty}x^3 e^{\minus{}nx}\,dx$ (Tonelli's theorem justifies the switch)
Substituting $ t\equal{}nx$, $ \int_0^{\infty}x^3 e^{\minus{}nx}\,dx\equal{}\int_0^{\infty} \frac{t^3}{n^3} e^{\minus{}t}\,\frac{dt}{n}\equal{}\frac1{n^4}\int_0^{\infty}t^3e^{\minus{}t}\,dt\equal{}\frac{6}{n^4}$
$ I\equal{}\sum_{n\equal{}1}^{\infty}\frac{6}{n^4}\equal{}6\zeta(4)\equal{}\frac{\pi^4}{15}$

If you believe akech, then $ \sum_{n \equal{} 0}^{\infty}\frac6{(n \plus{} 1)!(n \plus{} 1)^3}\approxeq 6.416386,$ while

$ \frac {\pi^4}{15}\approxeq 6.493939$

So between akech and transseries, at most one is right.

I just did a numerical integration and got

$ \int_0^{20}\frac {x^3}{e^x \minus{} 1}\,dx\approxeq 6.49392$

That doesn't look good for akech - see jmerry's post.

$ \Re x>1$, $ \zeta(z)\Gamma(z)\equal{}\int_0^\infty \frac{t^{z\minus{}1}}{e^t\minus{}1}dt$.
So, $ \int_0^\infty \frac{x^3}{e^x\minus{}1} dx \equal{} \zeta(4)\Gamma(4)\equal{}\frac{\pi^4}{90}\times 6\equal{}\boxed{\frac{\pi^4}{15}}$.

The error is acknowledged. Of course the geometric series expansion does not have a factorial in it.

[quote="jmerry"]$ I \equal{} \sum_{n \equal{} 1}^{\infty}\frac {6}{n^4} \equal{} 6\zeta(4) \equal{} \frac {\pi^4}{15}$[/quote]
Yes, I think that $ \frac {\pi^4}{15}$ is the correct answer.
[quote="jmerry"]$ I \equal{} \int_0^{\infty}x^3\sum_{n \equal{} 1}^{\infty}e^{ \minus{} nx}\,dx$
$ \equal{} \sum_{n \equal{} 1}^{\infty}\int_0^{\infty}x^3 e^{ \minus{} nx}\,dx$ (Tonelli's theorem justifies the switch)
[/quote] I reckon that it is just integration by parts, isn't it? Please make me correct if I am wrong. I don't know what Tonelli's theorem is. :|

It's not integration by parts; it's the interchange of a limit (the infinite sum) and an integral.

When is this allowed? That is, when can we be confident that $ \int_E\sum_{k\equal{}1}^{\infty}f_k(x)\,dx\equal{}\sum_{k\equal{}1}^{\infty}\int_Ef_k(x)\,dx\ ?$

The two most general conditions are rooted in the Lebesgue theory of integration - indeed, the primary motivation for the Lebesgue theory is to justify things like this. What are the two conditions?

1. If $ f_k(x)\ge 0$ for all $ k$ and $ x$ (well, almost all $ x$), then this is justified. This includes the possibility of infinite results - if one side of the equation above is infinite, then both sides are infinite. jmerry called this Tonelli's Theorem: that normally refers to double integrals, and indicates that jmerry was adopting a point of view in which he considered this to be a double integral, over a product domain for which one factor has discrete measure. From another point of view, we'd say that this is justified by the Monotone Convergence Theorem.

2. If $ \sum_{k\equal{}1}^{\infty}\int_E|f_k(x)|\,dx<\infty,$ then this is justified. Note that by the previous point that this is the same as $ \int_E\sum_{k\equal{}1}^{\infty}|f_k(x)|\,dx.$ Most conventionally, this is said to be justified by the Dominated Convergence Theorem; to be consistent with his point of view, jmerry would call this Fubini's Theorem.

The same pair of possible justifications also appear for interchanging the order of summation or the order of integration in a double sum or a double integral.

There are stronger conditions (meaning weaker theorems) that could also be used, some of which are rooted in the Riemann theory of integration rather than the Lebesgue theory. For instance: if there exists numbers $ M_k$ such that for all $ x\in [a,b]$ (note that this needs to be a bounded set), $ |f_k(x)|\le M_k$  and $ \sum_{k\minus{}1}^{\infty}M_k<\infty.$ As long as each $ f_k$ is Riemann integrable on $ [a,b],$ we have $ \int_a^b\sum_{k\equal{}1}^{\infty}f_k(x)\,dx\equal{}\sum_{k\equal{}1}^{\infty}\int_a^bf_k(x)\,dx.$

This justification combines the Weierstrass $ M$ test for uniform convergence with the use of uniform convergence (on a bounded set) to justify interchanging limit and integral.

======

There is one place for integration by parts, and that is in deriving the basic properties of the Gamma function - in particular

$ \int_0^{\infty}t^3e^{\minus{}t}\,dt\equal{}3!\equal{}6.$

[quote="Kent Merryfield"]Most conventionally, this is said to be justified by the Dominated Convergence Theorem; to be consistent with his point of view, jmerry would call this Fubini's Theorem. [/quote]
One side point: the Dominated Convergence Theorem is strictly stronger than the mixed sum/integral case of Fubini's theorem, because the DCT can take into account the order structure of the integers. As an example, consider this calculation:
$ \ln(2)\equal{}\int_0^1 \frac1{1\plus{}x}\,dx\equal{}\int_0^1\sum_{n\equal{}0}^{\infty}(\minus{}1)^n x^n\,dx$
$ ?\equal{}\sum_{n\equal{}1}^{\infty}\int_0^1 (\minus{}1)^n x^n\,dx\equal{}1\minus{}\frac12\plus{}\frac13\minus{}\frac14\plus{}\cdots$

Fubini's theorem isn't strong enough to justify that interchange; if we put absolute values on the terms $ (\minus{}1)^n x^n$, it blows up to $ \int_0^1 \frac1{1\minus{}x}\,dx\equal{}1\plus{}\frac12\plus{}\frac13\plus{}\frac14\plus{}\cdots\equal{}\infty$.
On the other hand, the DCT looks at the partial sums $ \sum_{n\equal{}0}^N (\minus{}1)^n x^n$. By the alternating series estimate, $ 0\le \sum_{n\equal{}0}^N (\minus{}1)^n x^n\le 1$ for any $ N$ and for all $ x\in [0,1]$. $ 1$ is an integrable function on this interval, and the interchange $ \int_0^1\left(\lim_{N\to\infty}\sum_{n\equal{}0}^N (\minus{}1)^n x^n\right)\,dx\equal{}\lim_{N\to\infty}\int_0^1\left(\sum_{n\equal{}0}^N (\minus{}1)^n x^n\right)\,dx$ is justified.

Hey, if you'd asked me to drove the alternating harmonic series converges to $ \ln 2$ using the DCT, I would have looked at you funny. But I was just looking for a nice simple example of something with bounded partial sums but not absolute convergence... which of course leads to a conditionally convergent series...

[quote=pezzeepenguinstalker]Why the ten year revive?[/quote]

I personally don't mind.  Someone had a genuine question, and the original problem is interesting.

[quote=winnertakeover][quote=Kent Merryfield]Hmm... I would have quoted Abel's Theorem. I hadn't thought about the DCT proof.[/quote]

Can someone explain what DCT states without using measure theory terms? I've looked it up online, but every site I see seems to mention something about being Lebesgue Measurable and I don't really know what this means.[/quote]

I do not know much of measure theory, but from what I can tell, the Lebesgue Dombinated Convergence Theorem would be difficult to describe outside any measure theoretic language.  I first came across the Lebesgue DCT in a formal probability course, though I am unsure how I would translate the probabilistic context into the more general measure theory context.

I think, though, that for many applications, the following DCT version for [i]Riemann integration[/i] is enough (i.e. NOT Lebesgue integrals):

Let $I \subseteq \overline{\mathbb{R}}$ be an interval (not necessarily a bounded interval).  Let $f:I \to \mathbb{R}$ be a Riemann integrable function.  Let $f_1,f_2,\cdots : I \to \mathbb{R}$ be a sequence of functions that converges [i]pointwise[/i] to $f$.  This means that
$$\lim_{n \to \infty} f_n(x) = f(x)$$
for all $x \in I$.

Now suppose that the functions $f_1,f_2,\cdots$ and $f$ are all [i]dominated[/i] by a function $g:I \to \mathbb{R}$.  That is,
$$|f(x)| \leq g(x)$$
$$|f_n(x)| \leq g(x)$$
for all $x \in I$, for all $n \geq 1$.

If all these conditions holds, then it follows that:
$$\lim_{n \to \infty} \int_I |f_n(x) - f(x)| \,dx = 0$$

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

It may help to rewrite the conclusion of the theorem to better ascertain the importance of this theorem.  Note that:
$$0 \leq \left|\int_I f_n(x)-f(x)\,dx\right| \leq \int_I |f_n(x)-f(x)|\,dx$$
Sending $n \to \infty$ and applying squeeze rule, we obtain:
$$\lim_{n \to \infty} \int_I f_n(x)\,dx = \int_I f(x)$$
It may be more pleasing to view this as:
$$\lim_{n \to \infty} \int_I f_n(x)\,dx = \int_I \lim_{n \to \infty}f_n(x)$$
With this view, the theorem states that we may exchange the limit and the integral.

So what?  Remember that derivatives are, in essence, limits.  So with certain conditions, we may be able to switch the derivative and an integral, and hence justify differentiation under the integral.  Semi-rigorously, consider:
$$\frac{d}{dx} \int_I f(x,y)\,dy$$
Write this as:
$$\lim_{t \to 0} \frac{\int_I f(x+t,y)\,dy - \int_I f(x,y)\,dy}{t}=\lim_{t \to 0} \frac{\int_I f(x+t,y)-f(x,y)\,dy}{t} = \lim_{t \to 0} \int_I \frac{f(x+t,y)-f(x,y)}{t}\,dy$$
Can we swap the limit and the integral by DCT?  I have to go do more productive things, but see if you can find valid conditions for applying the theorem of DCT.


Using weighted AM-GM inequality only.
Note that $(\frac{1}{x})^{y}\leq  \frac{x+y-xy}{x}$.


inequalities
trigonometry
inequalities unsolved

Let $ \Delta ABC,max (A,B,C)<90$.Prove that
$ sin A\plus{}sinB\plus{}sinC\ge sin^3 A\plus{}sin^3 B\plus{}sin^3 C \plus{} sinAsinBsinC$
P/s.To have a nicer proof than arqady 's  http://www.mathlinks.ro/viewtopic.php?t=231077 ,I must prove this ineq
I proved it by schur,And you?

[quote="Allnames"]Let $ \Delta ABC,max (A,B,C) < 90$.Prove that
$ sin A \plus{} sinB \plus{} sinC\ge sin^3 A \plus{} sin^3 B \plus{} sin^3 C \plus{} sinAsinBsinC$
P/s.To have a nicer proof than arqady 's  http://www.mathlinks.ro/viewtopic.php?t=231077 ,I must prove this ineq
I proved it by schur,And you?[/quote]
I think, your inequality we can prove by Cauchy-Schwartz too and it's  much easier than
 http://www.mathlinks.ro/viewtopic.php?t=231077
 :P

[quote="ducpbc"]I think we need move it.It isn't unsolved  :lol:[/quote]
Yes,It is my error,I must posted in other box
To arqady:No,I dont think it is easier than that ineq,because I use this ineq to prove conan naruto's one,so I think it is stronger :maybe:

[quote="Allnames"]
To arqady:No,I dont think it is easier than that ineq,because I use this ineq to prove conan naruto's one,so I think it is stronger :maybe:[/quote]
Here is my proof.
Let $ b^2 \plus{} c^2 \minus{} a^2 \equal{} x,$ $ a^2 \plus{} c^2 \minus{} b^2 \equal{} y$ and $ a^2 \plus{} b^2 \minus{} c^2 \equal{} z.$ Then, $ x,$ $ y$ and $ z$ are positive and
$ \sin A \plus{} \sin B \plus{} \sin C\ge \sin^3 A \plus{} \sin^3 B \plus{} \sin^3 C \plus{} \sin A\sin B\sin C \Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {\cos^2A}{\sin B\sin C}\geq1\Leftrightarrow\sum_{cyc}\frac {\frac {(b^2 \plus{} c^2 \minus{} a^2)^2}{4b^2c^2}}{\frac {2S}{ac}\cdot\frac {2S}{ab}}\geq1\Leftrightarrow\sum_{cyc}\frac {a^2(b^2 \plus{} c^2 \minus{} a^2)^2}{16S^2bc}\geq1\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}a^3(b^2 \plus{} c^2 \minus{} a^2)^2\geq abc\sum_{cyc}(2a^2b^2 \minus{} a^4)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}x^2(y \plus{} z)^{1.5}\geq(xy \plus{} xz \plus{} yz)\sqrt {(x \plus{} y)(x \plus{} z)(y \plus{} z)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}x^4(y \plus{} z)^3 \plus{} 2\sum_{cyc}x^2y^2\left((x \plus{} z)(y \plus{} z)\right)^{1.5}\geq\left(\sum_{cyc}xy\right)^2\prod_{cyc}(x \plus{} y)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}x^2y^2(x \plus{} z)(y \plus{} z)\sqrt {(x \plus{} z)(y \plus{} z)}\geq\sum_{cyc}(3x^3y^3z \plus{} 5x^3y^2z^2).$
By Cauchy-Schwartz we obtain: $ \sqrt {(x \plus{} z)(y \plus{} z)}\geq\sqrt {xy} \plus{} z.$
Hence, it remains to prove that
$ \sum_{cyc}x^2y^2(x \plus{} z)(y \plus{} z)(\sqrt {xy} \plus{} z)\geq\sum_{cyc}(3x^3y^3z \plus{} 5x^3y^2z^2).$
Let $ x \equal{} \frac {1}{p^2},$ $ y \equal{} \frac {1}{q^2}$ and $ z \equal{} \frac {1}{r^2}.$
Hence, $ \sum_{cyc}x^2y^2(x \plus{} z)(y \plus{} z)(\sqrt {xy} \plus{} z)\geq\sum_{cyc}(3x^3y^3z \plus{} 5x^3y^2z^2)\Leftrightarrow$
$ \Leftrightarrow \sum_{cyc}(p^7 \plus{} p^3q^2r^2 \minus{} 2p^5qr \plus{} p^5q^2 \plus{} p^5r^2 \minus{} 2p^3q^3r)\geq0,$ which is obvious.

I think, the estimation $ \sqrt {(x \plus{} z)(y \plus{} z)}\geq\sqrt {xy} \plus{} z$ much easier than
$ \sum_{cyc}\frac {x^2y^2\sqrt {(x \plus{} z)(y \plus{} z)}}{x \plus{} y}\cdot\sum_{cyc}\frac {x \plus{} y}{\sqrt {(x \plus{} z)(y \plus{} z)}}\geq(xy \plus{} xz \plus{} yz)^2$ in
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231077
 :wink:
What is your proofs of  both inequalities, Allnames? Thank you!

perhaps we can use Klamkin Ine:aPA^2+bPB^2+cPC^2>=abc
R^2>=(a^3+b^3+c^3+abc)/(a+b+c)

[quote=Allnames]Let $ \Delta ABC,max (A,B,C)<90$.Prove that
$ sin A\plus{}sinB\plus{}sinC\ge sin^3 A\plus{}sin^3 B\plus{}sin^3 C \plus{} sinAsinBsinC$
[/quote]
This inequality is true for all triangle.


induction
inequalities
inequalities unsolved

[quote="Mathx"]let $a_1,a_2,\dots,a_n>0$ prove:
$\sqrt{a_n}+\sqrt{a_n+a_{n-1}}+\dots+\sqrt{a_n+a_{n-1}+\dots+a_1}\ge \sqrt{a_1+4a_2+\dots+n^2a_n}$[/quote]
Induction  :)
Suppose that the inequality is true for $n$, then put $a_{n+1}+a_n=b_n$ hence
$\sqrt{a_{n+1}}+\sqrt{b_n}+\sqrt{b_n+a_{n-1}}+\dots+\sqrt{b_n+a_{n-1}+\dots+a_1}$
 $\ge \sqrt{a_{n+1}}+\sqrt{a_1+4a_2+\dots+n^2b_n}$

by induction hypothesis.
Now $\sqrt{a_{n+1}}+\sqrt{a_1+4a_2+\dots+n^2b_n}$
$\ge \sqrt{a_1+4a_2+\dots+n^2a_n+(n+1)^2a_{n+1}}$
$\Leftrightarrow a_{n+1}+n^2a_{n+1}+2\sqrt{a_{n+1}}\cdot\sqrt{a_1+4a_2+\dots+n^2b_n}\ge$
$(n+1)^2a_{n+1}$

the last inequality is obvious  $\blacksquare$
The proof is complete...


A rectangular Plot requires 200 m of fencing to enclose it. If one of the sides is x meters, express its area, in square metres, as a function of x.?

may be this should be in intermediate if not in basic

[hide="This simple?"]
$ A=f(x)=x (100-x)$.
[/hide]


Guess what. Answer this question. It shouldn't take more than 10 seconds (if you know this):

[quote="anirudh"]What is the answer to the following sequence:
Visit my forum at http://dasarathy0.tripod.com
$\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}...\frac{9}{10}$ [/quote]

did you mean multiplication? This isn't so easy with addition.

whoops.i meant multiplication. sorry. typo. almost fixed. fixed. http://dasarathy0.tripod.com

okay,
[hide]cancel, and cancel, you are left with $\frac{1}{10}$[/hide]

[hide]$\frac{1}{10}$ because all of the numbers simpllify except for the 1 and the 10.[/hide]

That was easy.. al the numbers cancel out except for teh one and the ten so it is $1/10$

all numbers cancel out exept  and you are left with 1 and 10 so the answer is 1/10


[quote="benimath"]Suppose $ p_1,...,p_n$ are prime numbers. Are there any integers $ a,b,c$ having prime factors among the given primes such that $ a \equal{} b \plus{} c$?[/quote]

One of the primes must clearly be $ 2$ and then $ 2^2\equal{}2\plus{}2$

[quote="benimath"]Any proofs for that?? I'm not convinced about it.[/quote]

Huh !

If all primes are odd, $ a$, $ b$ and $ c$ are odd, but this is impossible since $ b,c$ odd imply $ a$ even.

So $ 2$ is in the set of primes.

And $ 2^2\equal{}2\plus{}2$ seems rather elementary to prove ... :)


calculus
integration
trigonometry
calculus computations

$ \int_{0}^{L}\frac {1}{\sqrt {Lx \minus{} x^2}}dx$
Help me please i'm newbies in Calculus  :blush:

Complete the square. That should indicate the substitution you need- this is in the inverse trig family.

Can you show it please?  :blush:
I have tried for it but...

$ Lx\minus{}x^2\equal{}\left(\frac L2\right)^2\minus{}\left(x\minus{}\frac L2\right)^2$


Please give a lot of solutions to this one.

If $a,b,c>0$ with  $abc=1$  prove that  $(a+b)(b+c)(c+a)\geq (a+1)(b+1)(c+1)$

Thank you

[hide]Expand the inequality to obtain

$a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc \geq abc + ab + bc + ac + a + b + c + 1$

$\Rightarrow (a + b + c)(ab + bc + ac) - abc - (ab + bc + ac) - (a + b + c) - 2abc \geq 0$

$\Rightarrow (a + b + c - 1)(ab + bc + ac - 1) \geq 4abc = 4$

Now apply AM-GM to find that

$a + b + c \geq 3\sqrt[3]{abc} = 3$

and

$ab + bc + ac \geq 3 \sqrt[3]{a^2b^2c^2} = 3$,

and our inequality is proven.[/hide]


Every time I insert a picture in a TeXnic document ([u]I[/u]nsert => Picture...), the final product just shows the place of the file.

For example, I saved "Untitled.png" to my desktop and inserted it as a picture.
The result in the PDF was a page with "C:/Users/(USER)/Desktop/Untitled.png" on it.

Why is this happening?


Given a sequence$\{{x}_{n}\}$ such that ${x}_{1}=2,{x}_{2}=4,{x}_{n+1}={x}_{n}^{2}-{x}_{n-1}({n}\ge{2})$. Is there infinitely many members which is multiple of $2004$ ?

We can define $x_{0}=0$.
$2004=12*167$. It is easy to chek, that $x_{n}(mod 12)$ had period 14.
Therefore sufficiently consider $y_{n}=x_{n}(mod \ 167)$. Because $y_{n+1}=y_{n}^{2}-y_{n-1}$ equavalent to $y_{n-1}=y_{n}^{2}-y_{n+1}$ sequence is periodic. Let T is period and $k=lcm(14,T)$, then $2004|x_{km}\ \forall m\in N$.


If x1, x2, ...,xn are real numbers, then:
x1 + x1(x2 - x1) + ... + xn-1(xn - xn-1) \leq [(n - 1)xn2 + 2xn + n - 1]/2n.
In what case do we have an inequality?

Somenone really likes inequalities. This one is very simple., because all we have to do is to sum up the inequalities x(k+1)(x(k+1)-x(k))<=(kx^2(k+1)+2x(k+1)+k)/2(k+1)-((k-1)x^2(k)+2x(k)+k-1)/2k for k=1,2,...,n-1. We have equality if and only if kx(k+1)-(k+1)x(k)+1=0, from where it is not difficult to find x(k).


Let be m and n positive integers.Show that exist a&b positive integers if:
$(m^{4}-m^{2}+1)(n^{4}-n^{2}+1)=a^{2}+b^{2}$

A little rewording of the problem:
[quote]Let $m$ and $n$ be positive integers. Then there exist positive integers $a$ and $b$ such that \[(m^{4}-m^{2}+1)(n^{4}-n^{2}+1)=a^{2}+b^{2}.\][/quote][hide]\[(m^{4}-m^{2}+1)(n^{4}-n^{2}+1)=\left|(n^{2}-1)(m^{2}-1)+nm\right|^{2}+\left|n(m^{2}-1)-m(n^{2}-1)\right|^{2}.\][/hide]


Hello everyone  :lol:

Sean $ x; y; z$ real numbers satisfying

$ x \equal{} 2y^2\minus{} 1$
$ y \equal{} 2z^2 \minus{} 1$
$ z \equal{} 2x^2\minus{} 1$

Find the amount of securities that can take: $ x \plus{} y \plus{} z$.

Greetings  :D

[quote="TALENTO"]Hello everyone  :lol:

Sean $ x; y; z$ real numbers satisfying

$ x \equal{} 2y^2 \minus{} 1$
$ y \equal{} 2z^2 \minus{} 1$
$ z \equal{} 2x^2 \minus{} 1$

Find the amount of securities that can take: $ x \plus{} y \plus{} z$.

Greetings  :D[/quote]

If the question is :
"Let $ x,y,z\in\mathbb R$ such that :
$ x \equal{} 2y^2 \minus{} 1$
$ y \equal{} 2z^2 \minus{} 1$
$ z \equal{} 2x^2 \minus{} 1$

Find the different values that can take the sum $ x \plus{} y \plus{} z$"

Then here is a rather ugly solution :

Let $ f(x)\equal{}2x^2\minus{}1$ and $ g(x)\equal{}f(f(f(x)))\minus{}x$
The solutions of this system are $ (u,f(u),f(f(u)))$ where $ u$ is any root of $ g(x)\equal{}0$

$ x\plus{}f(x)\plus{}f(f(x))\equal{}x(8x^3\minus{}6x\plus{}1)$

$ g(x)\equal{}(x\minus{}1)(2x\plus{}1)(8x^3\minus{}6x\plus{}1)(8x^3\plus{}4x^2\minus{}4x\minus{}1)$

So the roots :
$ u\equal{}1$ $ \implies$ $ x\equal{}y\equal{}z\equal{}1$ and $ x\plus{}y\plus{}z\equal{}3$
$ u\equal{}\minus{}\frac 12$ $ \implies$ $ x\equal{}y\equal{}z\equal{}\minus{}\frac 12$ and $ x\plus{}y\plus{}z\equal{}\minus{}\frac 32$
$ u$ root of $ 8x^3\minus{}6x\plus{}1$ and so $ x\plus{}y\plus{}z\equal{}0$

If $ u$ is root of $ 8x^3\plus{}4x^2\minus{}4x\minus{}1$, $ f(u)$ cant be root of $ 8x^3\minus{}6x\plus{}1$ (just look at the sum $ u\plus{}f(u)\plus{}f(f(u))$) and so $ 8x^3\plus{}4x^2\minus{}4x\minus{}1$ has 3 roots $ u,f(u)f(f(u))$ and their sum is $ \minus{}\frac 12$

Hence the result : $ x\plus{}y\plus{}z\in\{\minus{}\frac 32,\minus{}\frac 12,0,3\}$


function
logarithms
algebra
functional equation
algebra proposed

Fro each real number $ r$ let $ T_r$ be the transformation of the plane that takes the point $ (x,y)$ into the point $ (2^r x,r 2^r x \plus{}2^r y)$. Let $ F\equal{}\{ T_r | r \in \mathbb{R} \}$. Find all curves $ y\equal{}f(x)$ whose graphs remains unchanged by every transformation in $ F$.

[quote="moldovan"]Fro each real number $ r$ let $ T_r$ be the transformation of the plane that takes the point $ (x,y)$ into the point $ (2^r x,r 2^r x \plus{} 2^r y)$. Let $ F \equal{} \{ T_r | r \in \mathbb{R} \}$. Find all curves $ y \equal{} f(x)$ whose graphs remains unchanged by every transformation in $ F$.[/quote]

We hare looking for functions such that $ f(2^rx) \equal{} r2^rx \plus{} 2^rf(x)$, so for example $ f(2x) \equal{} 2x \plus{} 2f(x)$.

This is a classical functional equation whose general solution may be written as :

$ f(0) \equal{} 0$
$ f(x) \equal{} x(\log_2(x) \plus{} u_1(\{\log_2(x)\})$ $ \forall x > 0$ with $ \{x\} \equal{}$ fractional part of $ x$ and $ u_1(x)$ any function from $ [0,1)\to\mathbb R$
$ f(x) \equal{} x(\log_2( \minus{} x) \plus{} u_2(\{\log_2( \minus{} x)\})$ $ \forall x < 0$ with $ \{x\} \equal{}$ fractional part of $ x$ and $ u_2(x)$ any function from $ [0,1)\to\mathbb R$

And it is easy to check back that these solutions fit the general requirement $ f(2^rx) \equal{} r2^rx \plus{} 2^rf(x)$ $ \forall r\in\mathbb R$

[quote="moldovan"]Fro each real number $ r$ let $ T_r$ be the transformation of the plane that takes the point $ (x,y)$ into the point $ (2^r x,r 2^r x \plus{} 2^r y)$. Let $ F \equal{} \{ T_r | r \in \mathbb{R} \}$. Find all curves $ y \equal{} f(x)$ whose graphs remains unchanged by every transformation in $ F$.[/quote]

I'm sorry but my previous post, as many of you noticed in your replies, was erroneous since it considered that $ r\in\mathbb Z$.

For $ r\in\mathbb R$, we have :
$ f(2^rx) \equal{} r2^rx \plus{} 2^rf(x)$ and so $ f(ax) \equal{} a\log_2(a)x \plus{} af(x)$ $ \forall x$, $ \forall a > 0$ and so :
Taking $ x \equal{} \plus{} 1$ we get $ f(a) \equal{} a\log_2(a) \plus{} af(1)$ $ \forall a > 0$
Taking $ x \equal{} \minus{} 1$ we get $ f( \minus{} a) \equal{} \minus{} a\log_2(a) \plus{} af( \minus{} 1)$ $ \forall a > 0$
Taking $ x \equal{} 0$ we get $ f(0) \equal{} 0$

And so the solution must be in the form :
$ f(0) \equal{} 0$
$ f(x) \equal{} x\log_2(|x|) \plus{} ux$ $ \forall x > 0$
$ f(x) \equal{} x\log_2(|x|) \plus{} vx$ $ \forall x < 0$

And it is easy to check that this form fit the initial requirements :
For $ x \equal{} 0$ , we have $ 0 \equal{} 0 \plus{} 2^r0$
For $ x > 0$, we have also $ 2^rx > 0$ and $ f(2^rx) \equal{} 2^rx\log_2(2^r|x|) \plus{} u2^rx$ $ \equal{} 2^r(x\log_2(|x|) \plus{} ux) \plus{} r2^rx$ $ \equal{} r2^rx \plus{} 2^rf(x)$
For $ x < 0$, we have also $ 2^rx < 0$ and $ f(2^rx) \equal{} 2^rx\log_2(2^r|x|) \plus{} v2^rx$ $ \equal{} 2^r(x\log_2(|x|) \plus{} vx) \plus{} r2^rx$ $ \equal{} r2^rx \plus{} 2^rf(x)$

And sorry, Moldovan, for the wrong previous solution. I hope this did not create problems for you.


Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

[hide] There are 3 choose 2=3 ways to choose which men to put in the office and 2 choose 1=2 ways to choose which woman goes to the office.  When you multiply these, you get 6.  Now, there are 5 choose 3=10 ways to choose 3 people to go to the office.  Therefore, the chance is $\frac{6}{10}=\frac{3}{5}$. [/hide]

[hide]There are $\binom32=3$ ways to choose the men and $\binom22=2$ ways to choose the women for the office.  There are $\binom52=10$ total ways to choose the people for the office, so the probability that there are 2 men and 1 women in the offices is $\frac{2\cdot3}{10}=\frac35$[/hide]


=====
Let $ V$ be a vector space and $ P_1: V\to V$, $ P_2: V\to V$ be operators on $ V$ such that $ P_1^2\equal{}P_1$, $ P_2^2\equal{}P_2$. Show that $ P_1\plus{}P_2\equal{}I$ and $ P_1\circ P_2\equal{}P_2\circ P_1\equal{}0$ if and only if $ V\equal{}\text{im}(P_1)\oplus \text{im}(P_2)$; that is, $ \text{im}(P_1)\cap \text{im}(P_2)\equal{}\{\vec{0}\}$ and $ \forall \vec{v}\in V$ $ \exists \vec{v_1}\in \text{im}(P_1),\vec{v_2}\in \text{im}(P_2)$ such that $ \vec{v}\equal{}\vec{v_1}\plus{}\vec{v_2}$.
=====

I can prove one direction...

[hide="Partial Solution"]
=====
$ \boxed{P_1\plus{}P_2\equal{}I, P_1\circ P_2\equal{}P_2\circ P_1\equal{}0 \implies V\equal{}\text{im}(P_1)\oplus \text{im}(P_2)}$

It is evident that $ \forall \vec{v}\in V$, $ \vec{v}\equal{}P_1(\vec{v})\plus{}P_2(\vec{v})$, where $ P_1(\vec{v})\in \text{im}(P_1)$ and $ P_2(\vec{v})\in \text{im}(P_2)$, so that $ V\equal{}\text{im}(P_1)\plus{}\text{im}(P_2)$. Thus it suffices to show that $ \text{im}(P_1)$ and $ \text{im}(P_2)$ are disjoint.

Suppose $ \vec{v}\in \text{im}(P_1)\cap \text{im}(P_2)$, so that $ \exists \vec{v_1},\vec{v_2}$ such that $ P_1(\vec{v_1})\equal{}\vec{v}$ and $ P_2(\vec{v_2})\equal{}\vec{v}$. We wish to show that $ \vec{v}\equal{}0$. Since $ P_1\plus{}P_2\equal{}I$, we have:

\[ \begin{cases}
P_1(\vec{v}_1)\plus{}P_2(\vec{v}_1)&\equal{}\vec{v}_1\\
P_1(\vec{v}_2)\plus{}P_2(\vec{v}_2)&\equal{}\vec{v}_2
\end{cases}\]

\[ \implies \begin{cases}
\vec{v}\plus{}P_2(\vec{v}_1)&\equal{}\vec{v}_1\\
P_1(\vec{v}_2)\plus{}\vec{v}&\equal{}\vec{v}_2
\end{cases}\]

Applying $ P_1$ to the first line and $ P_2$ to the second, we get:

\[ \begin{cases}
P_1(\vec{v})\plus{}P_1(P_2(\vec{v}_1))&\equal{}P_1(\vec{v}_1)\\
P_2(P_1(\vec{v}_2))\plus{}P_2(\vec{v})&\equal{}P_2(\vec{v}_2)
\end{cases}\]

\[ \implies \begin{cases}
P_1(\vec{v})&\equal{}\vec{v}\\
P_2(\vec{v})&\equal{}\vec{v}
\end{cases}\]

Applying $ P_2$ to the first line yields $ 0\equal{}P_2(\vec{v})\equal{}\vec{v}$, so that $ \vec{v}\equal{}0$, as desired. $ \blacksquare$
=====
[/hide]

...but don't know where to go for the other direction. I'm also concerned because my partial solution doesn't appear to use the fact that $ P_1,P_2$ are projection operators.  :(

It is good that you cannot prove the inverse statement, because it is false   :wink: . Let us give a counterexample in $ V \equal{}\mathbb R^2$. Define for every $ x \equal{} (x_1, x_2) \in \mathbb R^2$ the following two progections:
\[ P_1 x \equal{} (x_1, 0), \,\, P_2 x \equal{}(0, x_1 \plus{} x_2).\]
Their images are just the coordinate axes, so $ V\equal{}\text{im}(P_1)\oplus \text{im}(P_2)$, but $ P_1\plus{}P_2 \neq I$.

Thank you very much for your response. The problem was removed from my problem set by the professor shortly after your post.  :lol:


We can't really do anything about it unless you specify whether it is cyclic or symmetric.

[quote="rachitgoel"]Let $a\geq b\geq c$ then,

$a^2b\geq c^2a\geq b^2c$

$\frac{1}{c}\geq\frac{1}{b}\geq \frac{1}{a}$

As these two sequences are similarly sorted,

$ \frac{a^{2}b}{c}+\frac{b^{2}c}{a}+\frac{c^{2}a}{b}\geq \frac{a^2b}{b}+\frac{b^2c}{c}+\frac{c^2a}{a}$.[/quote]
$c^2a\geq b^2c\iff ca\geq b^2$
?

@sqing If $b^2\geq ca$ the inequality is very trivial.

[quote="arqady"][quote="Cezar Lupu"]Let $a,b,c>0$.Prove the following inequality:

    $\frac{a^{2}b}{c}+\frac{b^{2}c}{a}+\frac{c^{2}a}{b}\geq\sum a^2 $ ;)[/quote]
It's wrong, of course![/quote]
It's wrong?
$\frac{a^{2}b}{c}+\frac{b^{2}c}{a}+\frac{c^{2}a}{b}\geq a^2 +b^2+c^2$
$\iff a^3b^2+b^3c^2+c^3a^2\ge a^3bc+b^3ca+c^3ab$
$\iff a^3(b-c)^2+b^3(c-a)^2+c^3(a-b)^2\ge 0.$

@arqady I am not seeing why is this wrong??
Please tell.

[quote="Cezar Lupu"]Let $a,b,c>0$.Prove the following inequality:

$\frac{a^{2}b}{c}+\frac{b^{2}c}{a}+\frac{c^{2}a}{b}\geq\sum a^2 $ ;)[/quote]
it's true when $(a-b)(a-c)(b-c) \ge 0$
proof
we have : $(a-b)(a-c)(b-c) \ge 0$
then : \[a^{2}b+b^{2}c+c^{2}a \ge a b^2+a^2 c+b c^2\\\\\Rightarrow \frac{a}{c}+\frac{c}{b}+\frac{b}{a}\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\\\\\Rightarrow \left (ab+ac+bc  \right )\left (  \frac{a}{c}+\frac{c}{b}+\frac{b}{a} \right ) \ge \left (ab+ac+bc  \right )\left ( \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right )\\\\\Rightarrow \frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b}\geq \frac{b^2a}{c}+\frac{c^2b}{a}+\frac{a^2c}{b}\\\\\Rightarrow \frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b}\geq \frac{1}{2}\left ( \sum a^2\left ( \frac{b}{c}+\frac{c}{b} \right ) \right )\geq \sum a^2\]
done.

@arqady If the problem was wrong,
then how is my solution above wrong??

[quote="arqady"][quote="aditya21"]
but by tchebycheff inequality
we have
              $a^3b^2 + b^3c^2 + c^3a^2 >= (a^2 + b^2 + c^2) (ab^2 + bc^2 + ca^2) / 3$
[/quote]
It's wrong! Try $c=0.001$, $a=1$ and $b=1000$.[/quote]
but @ arqady, its correct by tchebycheff inequality.
do you see any flaw in my solution other than what you are saying?

[quote="aditya21"]but @ arqady, its correct by tchebycheff inequality.[/quote]
Id est, you have got a contradiction on mathematics. My congratulations!
[quote="aditya21"]
do you see any flaw in my solution other than what you are saying?[/quote]
What about my counterexample?  :wink:

@arqady can you please tell the flaw in my solution??
Thanks.

@aditya21 see this:
Chebyshev states that
If
$a_1\geq a_2\geq.........\geq a_n$ and
$b_1\geq b_2\geq b_3.........\geq b_n$
then,
$\frac{(\sum a_1b_1)}{n}\geq \frac{\sum a_1}{n}\cdot \frac{\sum b_1}{n}$
Your statement is:
$a^3b^2 + b^3c^2 + c^3a^2 \geq (a^2 + b^2 + c^2)\cdot \frac{(ab^2 + bc^2 + ca^2)}{3}$
Let $a\geq b\geq c$
For your statement to be true

$ab^2\geq bc^2\geq ca^2$
$\implies bc^2\geq ca^2$
$\implies bc\geq a^2$
So,
clearly your statement is wrong.

@anybody Can you please tell why my sol is wrong??

[quote="rachitgoel"]

@anybody Can you please tell why my sol is wrong??[/quote]
Your solution is the following:
[quote="rachitgoel"]Let $a\geq b\geq c$ then,

$a^2b\geq c^2a\geq b^2c$

$\frac{1}{c}\geq\frac{1}{b}\geq \frac{1}{a}$

As these two sequences are similarly sorted,

$ \frac{a^{2}b}{c}+\frac{b^{2}c}{a}+\frac{c^{2}a}{b}\geq \frac{a^2b}{b}+\frac{b^2c}{c}+\frac{c^2a}{a}$.[/quote]
You say that if $a\geq b\geq c$ so $c^2a\geq b^2c$, which is wrong.

@arqady I made 2 cases
1. $b^2\geq ca$.
For which the inequality is obvious.

2. $b^2\leq ca$.
For which I wrote the solution.
Now please tell why my solution is wrong.
Thanks.

rachitgoel, write your full solution, please.

[quote="aditya21"]but there no flaw in my solution, i think you wrongly calculated it @ arqady[/quote]
please take a calculator, and try to calculate it yourself   :lol:

[quote="rachitgoel"]Let $a\geq b\geq c$ then,

$a^2b\geq c^2a\geq b^2c$
[/quote]
[b]1.[/b]

 ok, your solution is true when $a^2b\geq c^2a\geq b^2c$

                  what is your proof when  $a^2b\geq b^2c\geq c^2a$  :D

[b]2.[/b]

Since it is not symmetric inequality , $a\ge b\ge c$ is not enough  :D

I still don't see why my solution is wrong!!
Can somebody please help me??

[quote="rachitgoel"]@Mirus Here is my solution for $ a^2b\geq b^2c\geq c^2a $
This implies $b^2\geq ca$
Replace $b^2$ for $ca$ in the original inequality ans see what you get.[/quote]
show us, how did you replace it and how did you finish it.
what about the case, $a\ge c \ge b$ ?

@Mirus see this
$\frac{a^2b}{c}\geq a^2$ (As $b\geq c$)

$\frac{b^2c}{a}\geq \frac{ca\cdot c}{a}=c^2$

$\frac{c^2a}{b}\geq c^2$ (As $a\geq b$).

Add the three inequalities.

Also I think we can take $a\geq b\geq c$!! Can't we??

[quote="rachitgoel"]@Mirus see this
$\frac{a^2b}{c}\geq a^2$ (As $b\geq c$)

$\frac{b^2c}{a}\geq \frac{ca\cdot c}{a}=c^2$

$\frac{c^2a}{b}\geq c^2$ (As $a\geq b$).

Add the three inequalities.[/quote]

After adding three inequalities, in the right side we get $a^2+2c^2$

[quote="rachitgoel"]

Also I think we can take $a\geq b\geq c$!! Can't we??[/quote]
it is not enough, because the inequality is not symmetric.

@Mirus Thanks!! I think I know now why my solution is wrong.


Let $a_{n}$ and $b_{n}$ be sequences and there exist $\lim_{n\to\infty}a_{n}=a,\lim_{n\to\infty}b_{n}=b$. Prove that:
$\lim_{n\to\infty}\frac{a_{1}b_{n}+a_{2}b_{n-1}+...+a_{n}b_{1}}{n}=ab$

$\textrm{Limits}\subset \textrm{Calculus}$.  Therefore, problems involving limits should be placed in the calculus forum.


This is quite complex.
Suppose I have two 3x3 matrices represented as squares on a grid. The squares have nine circles in them, at positions 1,1 ; 1,2 ; 1,3 ; 2,1 ; 2,2 ; 2,3 ; 3,1 ; 3,2 ; 3,3. Square A is located in the upper left corner, placed in such a way that its circle at position 3,3 intersects with the circle at 1,1 in Square B (located in the center of the grid.)
OK. Now, Square A will move towards the right until it reaches the stage at which its circle at 3,1 intersects with the circle of Square B at 1,3. Next, Square A will be shifted down so that points 2,1 and 3,1 on Square A intersect with points 1,3 and 2,3 on Square B, respectively. Now Square A will move all the way to the left until it reaches the stage at which points 2,3 and 3,3 on Square A intersect with points 1,1 and 2,1 on Square B, respectively. Once again, Square A will be moved down and it will shift towards the right, move down, etc. The process ends when Square A is located at the bottom right corner, its circle at point 1,1 intersecting with the circle at point 3,3 of Square B. In sum, Square A will follow a snake trail/zig-zag (from the left all the way to the right, down one unit, then all the way to the left, down one unit, then all the way to the right, etc.)

OK. In total, there are 81 intersections (3^4). 9 of those occur at Stage 1 (while it is moving to the right), 18 occur at Stage 2 (moving to the left), 27 occur at stage 3 (moving to the right once again), 18 at Stage 4, and finally 9 more at Stage 5.

I've begun to make a table that illustrates the concept. It maps the intersections of the points.

[b]Square B       /       Square A[/b]

[b]#.....Row...Col...Row...Col[/b]
----------------------------
[b]01[/b].....	1.....		1.....		3.....		3
[b]02[/b].....	1.....		1.....		3.....		2
[b]03[/b].....	1.....		2.....		3.....		3
[b]04[/b].....	1.....		1.....		3.....		1
[b]05[/b].....	1.....		2.....		3.....		2
[b]06[/b].....	1.....		3.....		3.....		3
[b]07[/b].....	1.....		2.....		3.....		1
[b]08[/b].....	1.....		3.....		3.....		2
[b]09[/b].....	1.....		3.....		3.....		1
[b]10[/b].....	1.....		3.....		2.....		1
[b]11[/b].....	2.....		3.....		3.....		1
[b]12[/b].....	2.....		2.....		1.....		3
[b]13[/b].....	2.....		1.....		1.....		2
[b]14[/b].....	2.....		2.....		3.....		2
[b]15[/b].....	2.....		2.....		3.....		1

(etc.)

Now, I would like to model this into a function... Actually, four functions. For example, I would like to supply the # and function 1 returns the SquareA.Row value; function 2 returns SquareA.Column value; function 3 returns SquareB.Row value; and function 4 returns SquareB.Column value.
NOTE: The table is just an example. I want the function to be modeled mathematically, and NOT based on the table. Basically I need to find the pattern...!

Can somebody help me?

I know that the way I explained all this makes it more complicated than it really is. Just try picturing in your mind how Square A moves about Square B.

In order to simplify things a bit, consider the matrix to be a 1-dimensional imitation of a 2d matrix, in such a way that element 1,1 will be 1; 1,2 will be 2; 1,3 will be 3; 2,1 will be 4; and so on.
In this case, the table would consist of two variables, one for Square A and one for Square B

[b]B.....A[/b] oops --> Please take note that A is the mobile square!!!
-------
1.....9
1.....8
2.....9
1.....7
2.....8
3.....9
2.....7
3.....8
3.....7
(etc.)

Alright, a much much simpler way to go about solving (and understanding!) this is to take a look at the attached excel spreadsheet.
I've arranged the numbers in such a way as to make the pattern visible.
Now, I need a function that can give me those numbers!


Let a pentagon $ABCDE$ satisfies $DE=DC;AE \perp ED;BC \perp CD$;and  $F \in AB$ such that $\frac{AF}{BF}=\frac{AE}{BC}$
prove that: $\widehat{ADE}=\widehat{FCE}$

This problem is solved on the forum. There were two solutions (one of which is mine :) ). So try to search!

(hint: http://www.mathlinks.ro/Forum/viewtopic.php?t=22138 )


symmetry
inequalities
number theory
relatively prime
AMC

$ a\minus{}7b\plus{}8c\equal{}4$
$ 8a\plus{}4b\minus{}c\equal{}7$
what's $ a^2\minus{}b^2\plus{}c^2$?

i've seen this before...
where did you find this???

i know i solved it somehow, but i forget....

ugh....
i need better memory.,....

i'm still struggling....
i'll edit it later..

MMPC problem.
[hide="hint"]Since there are 3 variables, but 2 equations, you can't solve for the variables independently.[/hide]

[hide="Approach 1"] Noting the symmetry in the coefficients, we rewrite the two equations as

$ a + 8c = 4 + 7b \implies a^2 + 16ac + 64c^2 = 16 + 56b + 49b^2$
$ 8a - c = 7 - 4b \implies 64a^2 - 16ac + c^2 = 49 - 56b + 16b^2$

Adding the equations,

$ 65a^2 + 65c^2 = 65 + 65b^2 \implies$
$ a^2 - b^2 + c^2 = \boxed{1}$. [/hide]
[hide="Approach 2"] A general approach is to parameterize the solution set of the first two equations (the intersection of two planes, which is a line).  So:

$ 64a + 32b - 8c = 56$
$ 65a + 25b = 60 \implies 13a + 5b = 12$

To keep coefficients integer, let $ a = 4 + 5t, b = -8 - 13t$, where $ t$ is our parameter.  Then

$ c = 8a + 4b - 7 = 32 + 40t - 32 - 52t - 7 = -7 - 12t$

and

$ \begin{eqnarray*}
a^2 - b^2 + c^2 &=& (4 + 5t)^2 - (8 + 13t)^2 + (7 + 12t)^2 \\
 &=& (-4 - 8t)(12 + 18t) + (7 + 12t)^2 \\
 &=& 49 + 168t + 144t^2 - 48 - 168t - 144t^2 \\
 &=& 1$

Of course, we could parameterize in any number of ways.  The simplest way is to let, say, $ c = t$, but we'll get rational coefficients if we parameterize naively. [/hide]

[quote="cdymdcool"]I'm sorry, I don't understand this "To keep coefficients integer, let , where  is our parameter".[/quote]
In general, the solution to $ ax\plus{}by\equal{}c$, where $ a$ and $ b$ are relatively prime, is $ a\equal{}a_0\plus{}bt, b\equal{}b_0\minus{}at$, where $ (a_0, b_0)$ is one solution to the equation and $ t$ is a parameter.

A bit about the motivation of the problem:

Notice that the intersection of the first and second equations is a line.

Furthermore, the third equation [when set equal to a positive constant] is a hyperboloid of one sheet. But we know that is a ruled surface [it contains an entire line].

Hence the problem writers chose the line determined by the first two equations to be one of the lines that lie along the third equation equal to a constant.

Nice, Al.  I didn't consider the interpretation as a real ruled surface, but in hindsight it was clear.

[hide="Approach 3 - the fastest one yet!"] The problem implies that $ a^2 \minus{} b^2 \plus{} c^2$ is a constant independent of particular values of $ a, b, c$; all you have to do, then, is find a particular set of $ a, b, c$ that work and plug in!

You can do this in any number of ways depending how much work you feel like doing.  A straightforward method is just to set, say, $ b \equal{} 0$, and solve the resulting system.  We have

$ a \plus{} 8c \equal{} 4 \implies 8a \plus{} 64c \equal{} 32$
$ 8a \minus{} c \equal{} 7 \implies 65c \equal{} 25 \implies c \equal{} \frac{5}{13}$
$ a \equal{} 4 \minus{} 8c \equal{} \frac{12}{13}$

Then $ a^2 \minus{} b^2 \plus{} c^2 \equal{} \frac{5^2}{13^2} \plus{} 0 \plus{} \frac{12^2}{13^2} \equal{} 1$ as expected. [/hide]

Hmm...I am getting a strange result there...note that the expression cannot be equal to a negative constant because that would be hyperbolid of two sheets, which is not a ruled surface. Hence we get the following strange inequality:

Let $ x_1\equal{}a_1 t\plus{} b_1$.

If $ (a_1 t\plus{}b_1)^2\plus{}(a_1 t\plus{}b_2)^2\minus{}(a_3 t\plus{}b_3)^2\equal{}\beta$, for all $ t$, then $ \beta \ge 0$!?

$ (a_1)^2\plus{}(a_2)^2\minus{}(a_3)^2\equal{}0$
$ a_1b_1\plus{}a_2b_2\minus{}a_3b_3\equal{}0$.
Then $ (b_1)^2\plus{}(b_2)^2\minus{}(b_3)^2\ge 0$!?


Harvard college Gauss Princeton articles

Just wondering, because this hit my mind the other day, how much does one's school affect one in applying to the more elite colleges? I'd also like to know how much it might compare to things like APs, GPA, etc.

thanks!

Not at all, in my opinion. You go to Choate, I think--check out the acceptance rates there for, say, Harvard, and see if the percentage of people accepted is any different from Harvard's average. Then take into account the fact that someone who applies to Harvard from Choate first had to go through Choate's admission process, is likely to be pretty well-off financially and so has "more opportunities" or whatnot, is supposed to have received a better than normal education, is supposed to be prepared for college (boarding schools are called prep schools for a reason), has a college counselor who doesn't serve 500+ people, etc.
You'd think, with all that, that Choate's acceptance rate would be higher than the national average (around 15%, I think, not sure), but I'm guessing it's probably not much different, if at all.

Admissions officers aren't going to accept you just because you went to a supposedly better school. In fact, if two applicants are pretty much "equal" everywhere (same interests, same SATs, same extracurriculars, etc.) except in what kind of school they went to (one went to a top high school and the other went to a pretty poor public one), I think college admission officers would probably rather take the one who went to the poorer school, because he "overcame his surroundings" or whatever (sorry, I think that's just nonsense, but probably what they really do say).

Indeed, going to a top school they might expect more from you (actually, I don't believe this, but it's possible). Or, whereas you might have shined at an average public school and gotten extraordinary teacher recommendations, you might be only average at an elite.

Do schools like Andover and Exeter have their acceptance rates simply because they're well-known schools? Probably not. The top people who go there are just a lot better than normal.

My point: don't choose what school you go to because of college. Go to a private school because you want the experience, because you can find more motivated kids there, because they have the classes your local school doesn't offer. Or go to a public school if you're more independent, if you're motivated enough to learn on your own, if you can take classes at a nearby college, if whatever.

For example, I went to a (pretty good) public school my first two years of high school and then transferred to a (top) private one because I thought I'd learn more there. Nope. I have to say I wish I were still at the public (I'm more of an independent type of person, so I actually learned more at my old school, and the "nerdy" types of people I usually hung around with just aren't there at my new one (replaced by preppy kids...=))). But that's just me.

Thanks for the response.

Heh, yeah, sometimes I wish I stayed at my old school...

care to expand on that idea?
besides you keep talking about "expensive private schools" and "plain public schools" as if no others exist. some people actually do work for their opportunities (ie- public schools that are magnet-school-esque but harder to get into).

It's good to see Gauss202 posting again. I'd like to hear from him what the difference is.

I homeschool my children because that allows them to develop a lot of intrinsic motivation. I also arrange a certain number of group classes organized by parents for each of my two oldest children. And my oldest son is in a (public) university-based accelerated math program for middle-and-high-school students. The one thing I would NEVER do is send any of my children to a high school like the one I attended.

The quality and rigor of a school's academic program is taken into account in most admission decisions. Not that that means you should necessarily uproot yourself and move to a better school than what you're attending now - as long as your school is decent and offers classes and activities that legitimately prepare you for college.

But if you go to a school with an obviously weak program or where no one has gotten into the kind of school that you want to attend in say, the last five years, it could be a factor.

If you notice, a large number of the students who attend "elite" Universities come from the same high schools each year. Part of that may be that guidance counselors are putting in a lot of work on behalf of their students, or have a good relationship with admissions officers at certain schools. And part of it is that, well, good students tend to come from good schools.

Students from schools with reputations of having a strong or rigorous academic program will get an advantage over students from schools with weak programs because Universities will at least have a sense of the kind of preparation that they have had.

As an example, there are quite a few students who attend Princeton from New York City, but in my time there I don't think I [i]ever[/i] met more than one or two who graduated from a New York City public school (aside from Stuy or Bronx Sci).

Yet for elite universities, proximity helps (somewhat??)

6 students from Princeton High were accepted to Princeton University.................on early decision!!!!!!!!!

[quote="gauss202"]If you notice, a large number of the students who attend "elite" Universities come from the same high schools each year. Part of that may be that guidance counselors are putting in a lot of work on behalf of their students, or have a good relationship with admissions officers at certain schools. And part of it is that, well, good students tend to come from good schools.

Students from schools with reputations of having a strong or rigorous academic program will get an advantage over students from schools with weak programs because Universities will at least have a sense of the kind of preparation that they have had.
[/quote]

That said, schools must also consider the background of their entering class and thus can only accept a few students from each school every year. So this means that for some students, it will be harder for them to get into a competitive university if they attend an elite high school instead of the average neighborhood high. I can't seem to find the specific articles right now, where students and admissions officers discuss how Student X increased his/her chances by attending Neighborhood High instead of Elite Academy because it set him/her apart from many other students of similar abilities, but if someone else comes across these articles, perhaps they can post them here.

But the flip side of the coin is, can Neighborhood High provide an adequate number of challenges? Can your teachers and peers at Neighboorhood High motivate and challenge you to be the best you can be? Are the opportunities to live life abundant? For many people here, the answer to the first two questions is no. And if that's your answer, then you might want to ask yourself what you want out of high school. Is it merely another motion you go through on your way to adulthood? Do you want to be challenged academically, socially, athletically, and/or artistically?

Personally, I would never choose to sacrifice four years of my life so that I can attend Dream University X because by the time I enter college, my needs and desires may be different. But I'm not you, and if it's important for you to attend Dream University X, then by all means pursue that dream with everything you've got.

But no matter where you go to school, remember, colleges and universities are looking for exceptional people, people who can excel in their chosen fields. (Ok, they also say that they like to see well-rounded applicants.)

[quote="codeblue87"]I'd like to think it doesn't matter much.[/quote]

Too bad it does,  :(

Hey mcalderbank, your father is a professor at Princeton isn't he?

yes, but I would not like to go to Princeton. It is too close.


Prove that there is no solution for the equation $x^2+5=y^3$ in integers $x,y$

This problem can be easily proved by using Legendre symbol.

$\Leftrightarrow x^2 +4 = (y-1)(y^2 + y +1)$

For arbirtary odd prime $p$ such that $p|(x^2+4)$, $p$ must have $4k+1$ form.

However, For the cases $y=4k, 4k+1, 4k+2, 4k+3$, right side contains $4k+3$ form prime.

Contradiction!

$Q.E.D.$

There is some mistake in your solution, because for $y=4k+3$ there is not necessary such a prime divisor oo the right hand side. But yes, this is the main idea.

I missed that! :)

But cuz $x^2+4= 0 or 1 (mod 4)$

$y$ can't be $4k+3$

$Q.E.D.$

I have searched for it, but it's as good as impossible to find such things...   :?

I have a very similiar solution (or maybe the same?)
$x^2+4=(y-1)(y^2+y+1)$
simple checking yields that x is even and $y=1(mod 4)$.
Then $y^2+y+1=3(mod 4)$, so it must have a prime factor p such that $p=3(mod 4)$. This means that $x^2+4=0(mod p)$. However, this is impossible, since
$(\frac{-x^2}{p}) = (\frac{-1}{p}) = (-1)^{\frac{p-1}{2}} = -1$.

Yes, this was also my solution  :)
I haven't thought about it till now, but this equation is claimed to have a solution modulo every integer. (it's given as an example that Hasse's global-local principle is wrong for equations of degree greater than two)

Hi,
May be he wants to say [tex]x^n+y^n=z^n[/tex]  :D

I've checked the statement about solutions modulo every positive integer, so feel free to prove it.


Hello!

I am asking this question again, quoting on everyone who also has this problem

I want to take the Amc 10 B, but my school doesn't offer it, all the colleges are too far away, and i dont no what to do.

i need help and quick!!

It stinks, but you're pretty much stuck at this point.

I recommend just doing some networking in the area, see if any of the nearby high schools offer the test.  If not, as above poster said, your pretty much stuck...

If somebody takes the AMC at a school that is not theirs, will the student be affiliated with the test center or with their regular high school?

Ask your math teacher to proctor the test for you. You should be able to get the school to pay the registration fee / exam bundle; but if that is an issue at all simply offer to pay the cost involved. All things considered, it is fairly minimal.

I tried to take 12 B but that was on a finals day and I wanted to take after the finals
but my math teacher wouldn't let me

They are right. I'm pretty sure you have to take it on the specified day, the 25th of February.


Peter Wiggins stands on an escalator that brings him up one floor in 20 seconds.  When he gets to the top he starts to walk down at a constant rate.  It takes Peter 20 seconds to walk back down the up escalator.  Peter then walks back up at the same pace that he walked down.  How long does it take Peter to walk up the up escalator?

[hide]let x=rate of escalator, y be rate of peter, and s=number of steps in escalator. we're trying to find t, which is the time he took walking back up.

20x=s           (1)

20(y-x)=s      (2)

t(x+y)=s        (3)

20x=20y-20x   (1=2)

2x=y

so t(x+y) becomes t(3x), and because s=20x, t(3x)=20x, so t=20/3[/hide]

i think i screwed up.

Here's the way I did it.

[hide]Since the escalator moves up one floor in 20 seconds and Peter went down it in the same amount of time, he moves twice as fast as the escalator.

When Peter is going up the up elevator Peter is moving 3 times the speed of the elevator (the speed of the elevator+Peter's speed of walking (2*the speed of the elevator)).

Thus the escalator moves up one floor in 20 seconds so Peter moves up one floor in 20/3 seconds.[/hide]

I tried to use speed but it became too confusing.

I just realized that this isn't a proof at all... :(

[color=cyan]I don't see why that isn't a proof . . . I mean, you didn't use algebraic reasoning, but you used a perfectly solid series of logical steps to get from the givens to an answer.  Why isn't that a proof?[/color]

Tare, I agree with JBL.  Your reasoning is good enough to be considered a proof.  Very often in a mathematical paper simple algebraic reasoning suffices for a line of symbols.  It's the reasoning that is important.

So mine was a proof after all.

Um...so did I (and s0mp) get it right?

[quote="MCrawford"]Yes, but I'm not always going to tell you if you got it right.  I would rather let the students discuss some of these problems without my commentary at all.  Amongst the Mathcounts students here, you are as a group generally capable of critiquing nearly every solution.[/quote]
I didn't know that.
I wish there were a list of intentions for this site somewhere...

call his moving rate y, and the escalator rate x.

20x = D [D the length of escalator].
(y-x)*20=D=20x, or x=y-x or y=2x.

Thus at the rate 3x,   3x*t=D=20x, t=20/3


Hey,

What is the most viewed and replied thread in the forum? I would say that the three champions would be, considering mainly the replies

(1) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15270]AOPS Trivia Challenge[/url]: 401 Replies, 10052 Views
(2) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28104]AOPS Trivia Invitational[/url]: 187 Replies, 3053 Views
(3) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19092]The probability of GOD[/url]: 171 Replies, 4588 Views

Can any thread beat those? (some can beat the first two in Views, but what about Replies?)

Yeah but those posts are kind of....short.  :D

i remember in the tare days that there was a thread about 28 or 9 pages long called "how to turn girls on" (got a lot of views too) or something like that
forgot what happened to it though, but pretty certain it wasnt deleted

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=12588]Kerry or Bush[/url]  234 Replies 3904 views

[quote="paladin8"]You obviously haven't seen the [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=25915]Word Association Game v2[/url][/quote]
Well, I didn't find the Word Association Game v1, and I still don't know what are the rules they mention!!

Look at a couple of posts + add the name of the thread and you obtain the simple rule of the "game" :)

[quote="djimenez"]Well, I didn't find the Word Association Game v1, and I still don't know what are the rules they mention!![/quote]

"v2" is used to denote the continuation of this tradition by the (primarily) Texas ARML team from our original CodyBowl forum. When that crashed, we got a new name (NOML) and got a forum here. Since that was v1, this new thread is v2. v1 is now dead.

The rules are pretty much common sense. Just read a few posts in succession.

[quote="xxreddevilzxx"][quote="paladin8"]You obviously haven't seen the [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=25915]Word Association Game v2[/url][/quote]

i think hes talking about just this forum... like the other topics forum or the Games forum itself.[/quote]
No, I was talking about the whole site in general.

[quote="paladin8"]You obviously haven't seen the [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=25915]Word Association Game v2[/url][/quote]
I think  it has an anti-spam protection or something like that. I mean, if you see, as well as in the test forum, in that thread, if you post, it does not increase your post count!  :D

[quote="Valentin Vornicu"][quote="djimenez"]I think  it has an anti-spam protection or something like that. I mean, if you see, as well as in the test forum, in that thread, if you post, it does not increase your post count!  :D[/quote]Quite ingenious, isn't it? :P :P[/quote]Well, it seems to be a good mesure, then you don't have people "overspamming" the thread, that somehow is for spamming. If it continues that way, by the end of the year that thread is going to have more posts than Grobber  :D  :P

[quote="djimenez"]Well, it seems to be a good mesure, then you don't have people "overspamming" the thread, that somehow is for spamming. If it continues that way, by the end of the year that thread is going to have more posts than Grobber  :D  :P[/quote]

The only reason it isn't already pwning Grobber's post count is because of the lack of participation. We used to average a post like every 5 minutes during daylight hours. The high activity in the afternoons would offset the lower activity during school hours.

Anyways...threads with lots of posts...

Okay there was this one...like "What type of girls do math nerds like" or something like that that was just massive when AoPS was starting out. Man those were the days. There were like 20 of us (jojobird being one of them) that would spam *cough* I mean...er...actively contribute *cough* to the Etc. board. Then there were also threads like "FieryTycoon and Rippledance" and um..."What do you guys do at parties" and..."LoTR vs. HP" and stuff like that. Sadly, most of those are gone...and our posts from those threads were removed from our post count. (I give sole blame for me not being above 2500 posts to the removal of that portion of my post count. :D )

Hehe..I remember the days when I could log on to aops in the afternoon and see like 10 new threads in Etc. :-)

[quote="djimenez"]
[quote="paladin8"]You obviously haven't seen the [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=25915]Word Association Game v2[/url][/quote]
I think  it has an anti-spam protection or something like that. I mean, if you see, as well as in the test forum, in that thread, if you post, it does not increase your post count!  :D[/quote]

Whoa. I didn't even notice. No wonder, or I would have like 50 more posts ;)

[quote="GoBraves"][quote="djimenez"]

(2) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28104]AOPS Trivia International[/url]: 187 Replies, 3053 Views
[/quote]

Its AOPS Trivia [b]Invitational[/b] not international :D[/quote]Upss!! Corrected! :blush:  :blush:

Nothing much but beginning to grow with actual "math" contented thread.

[url=http://www.artofproblemsolving.com/Forum/topic-33505.html]Problem Solving Marathon 2[/url]

First posted - April 13

Current date: April 27

Total Number of Replies: 168

Views: 1873

Major point: This is a math thread.. Not like a thread in Fun & Games Thread and it's only been two weeks.. So, I consider this thread to be quite active..

[quote="Silverfalcon"]Major point: This is a math thread.. Not like a thread in Fun & Games Thread and it's only been two weeks.. So, I consider this thread to be quite active..[/quote]Looks good, let's see how does it go in a few weeks! and the thread itself seem very nice.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5423]USAMO 2004[/url] >> 10,000 views (few replies though).
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5462]Romanian IMO TSTs 2004[/url] >> 14,000 views

These are math threads :D

Also the young [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34685]Getting Nervous?[/url] in the AMC forum :) with 200 replies in just a couple of days, and >>7,000 views and growing.


function
combinatorics unsolved
combinatorics

how many permutations of $ \{1,...,n\}$ satisfy the following condition:
there exist $ 1 < m < n$, such that first $ m$ numbers in the permutation are a permutation of $ \{1,...,m\}$

The complementary object is "indecomposable" or "connected" permutations, sequence [url=http://www.research.att.com/~njas/sequences/A003319]A003319[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  It's not hard to give a generating function or recursion for these numbers, but you're not going to get a closed formula.

By PIE, there are $ \sum_{j\equal{}1}^{n\minus{}1} \sum_{k\equal{}j}^{n\minus{}1} (\minus{}1)^{j\plus{}1}k!(n\minus{}k)!\binom{k\minus{}1}{j\minus{}1}$


Find the intersection of the lines $ 9x\minus{}4y\equal{}30$ and $ 7x\plus{}y\equal{}11$. Express your answer as an ordered pair $ (x,y)$.


The AMC 8 2009 statistics are up [url=http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2009-8a/2009-amc8stats.shtml]here[/url].

I noticed that in all the states' statistics, way at the bottom, there is a problem difficulty chart. On number 12, there is a star (marking the right answer) by both D and E, while the correct answer is just D.

Discuss anything else here.

Now we definitely know that this year's test was harder.
There were only 163 perfect scores this year as opposed to 334 last year.

That's exactly what I was going to say.

It could be that they misjudged the importance of calculators in 2008, the first year without calculators.

I agree that this is evidence that this year's was harder than last year's. Here are some interesting facts I noticed:
 - Less than half as many perfects than last year
 - About 0.1% of the participants got perfects
 - Both 5th grade perfects were from California
 - On number 14, over 70% of the people put C, and only 12.91% got it right.
 - About 23% of the perfects were from California

Is there a website where it gives what score you must have in order to achieve the top 1%, 5%, etc?

No, but I think it can be calculated. Just go to Global Statistics and look at the number of people who got each score. The top 1% mark falls on a score of 21 and the top 5% mark falls exactly between a score of 17 and 18. So I think the cutoffs would be:

Distinguished Honor Roll - 21
Honor Roll - 18

Hmm yeah, just go here to calculate any percentiles:
http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2009-8a/2009statestats/GLOBALSTATS.HTML

Hmm, this year's test was hard. (I'm in 9th grade) Darn, our state is again without a perfect. GJ to all those 25er's, and also everyone else!

Gah.. I am so anxious.
It says on AL, that 1 male 7th grader got a 25, which could be me, because the next high 7th grade male score is a 21...I hope.

It's cubeX! Well couldn't you probably tell in a 4 problem discrepancy?

Darn... So many people got higher then me in my state.... Even though I sucked on the AMC 8 this comes as a threat in the mathcounts area.  :wallbash_red: blehhhhhhhhh  :bomb:

my score of 22 (FAIL) puts me at like top 15 in NC. 4th and 5th grader in my state got 23!!! :o

[quote="Maybach"]Wow.  My score of 21 puts me in the top 30 in Indiana.  Maybach is very surprised.[/quote]

My score also puts me at first in my school, and the first time someone from my school has gotten Distinguished Honor Roll since 2003.

Wow. You guys did so much better than me. I only got a 20. GJ to all those who got perfects!  :D

Well, I guess 20 is an OK score when all the other 6th graders in my school got an average of 11.2.  :wink:

>.> I registered at my old school for the AMC 8 but I changed schools about 3 days before the test. Didn't get to take it. I skipped seventh grade as well and my schools didn't offer AMC 8 to 6th graders.

[quote="k00lperson"]thanks powerofpi, but It didn't have how many people got honor roll distinction...?
Well, they showed the statistics, but I already knew the statistics :P[/quote]
If you want approximations, take .05 (5%) times # of participants (for honor roll). If you want exact numbers, then I don't know.

i am one of the fifth graders that got a perfect score
it wasn't easy
congratulations to everyone who took it :)

cricket 229, im in NC too and got 25 ^.^ (eat it)

When will the nationwide honor roll be posted?

huh? We already got ours.like the top 10%,top 5%, and top 1% in nation..Our teacher posted them up :o

according to the official statistics, number 14 was the hardest problem; only 12% of the people got it.

Yes (link [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1693409#1693409]here[/url]). I think many people put 50 because they disregarded that each speed was traveled at over different time periods.

[quote="The Hobbit"]No one in my state got a perfect so that made me feel a bit better but there where some 24s and 23s...  :|[/quote]

Same here.
I got a 24, the same score as what I got in 6th grade.  :mad:

I made a stupid mistake on one of the problems.  :oops:

My state (California) had 33 perfects!  And I wasn't one of them.  :mad:

As I posted in another thread:

The Distinguished Honor Roll (top 1%) level is 20 (compared to 22 last year).
The Honor Roll (top 5%) level is 17 (compared to 19 last year).
The Achievement Honor Roll (6th Grade and below) level is 15.

Hmm that should be made a sticky.

When will they post the national individual honor roll?

I won Ohio, but its so wierd how California's so smart. How do you get a 25 as a 5th grader? I got a 22 as a 6th grader last year.


When is the deadline to send in the Entry and Permission Forms? Also, can we send in the forms with our Round 1 solutions?

[quote="calc rulz"]Yes, you can send them in with your Round 1 solutions.[/quote]

That's correct; and for those who already sent theirs in, that's ok, too.  We haven't filed the permission forms we have received yet, so don't panic if the USAMTS page says we haven't received yours yet.

[quote="1/(ln x)"]I don't have a fax machine, can I just scan my entry form as a pdf and email it with my solutions?[/quote]

Sure.

[quote="FrznTears"]Can you send in the entry forms with Round 2 problems and do you have to qualify Round 1 to go to Round 2?[/quote] Yes you can, and no you don't have to qualify. :D


algebra
difference of squares
special factorizations

Problem 1
If the repeating decimal $0.696969...$ is written in its simplest fractional form,
what would the sum of the numerator and denominator be ?

Problem2
Find the value of $1000^2$ - $999^2$ + $998^2 -997^2$ ...... +$2^2$ - $1^2$

:-)

[hide="1"]
The first I think is just $\frac{69}{99}$ which simplifies to $\frac{23}{33}$. Therefore, the sum of the numerator and denominator is $23+33 = 56$.
[/hide]

[hide="2"]
For the second one, take the difference of squares for each pair of terms, and you will get:
$1999 + 1995 + 1991 +...+ 3$, which sums to $500500$
[/hide]

[quote="xxxyyyy"]Problem 1
If the repeating decimal $0.696969...$ is written in its simplest fractional form,
what would the sum of the numerator and denominator be ?[/quote]

[hide]$0,69696969...=\frac{23}{33}$. Then, the answer is $23+33=\boxed {56}$[/hide]

[hide="Note"]
For #1, one of the ways to verify $0.\overline{69}=\frac{69}{99}=\frac{23}{33}$ is as follows:

Let $x=0.696969...$.
Then $100x=69.696969...$ (Move decimal place)
So $99x=69\Longleftrightarrow x=\frac{69}{99}=\frac{23}{33}$

You can also use $a+ar+ar^2+...=\frac{a}{1-r}$ ($|r|<1$).
[/hide]

I think these questions should be in the Middle School forums.  :roll:  :D


sequence $ \{a_{n}\}$ , $ a_{1}=1$ , $ \mathbf{S}_{n}=\sum_{i=1}^{n}a_{i}$ , and $ a_{n}^{2}=n\cdot \mathbf{S}_{n}$

so , what's the general formula of $ a_{n}$

Shouldn't it be $ \mathbf{S}_{n}=\sum_{i=1}^{n-1}a_{i}$? How can you determine the second term in the sequence without having to use it's non-existent self?

If so, then the sequence would be $ 1,1,2,4,8,16\ldots$, and the formula would be $ 2^{n-2}$ for $ n>1$.


function
calculus
integration
logarithms
complex analysis
complex analysis unsolved

Hi,
I've tried real hard to solve this 1 but... i guess i am no good enough
so... here it is

Let f be an Entire Function so that for each z exists (Re(z) != 0) ==>> (f(z) < |Re(z)|^-0.5
prove that f is constant.

I guess it is somehow related to Liouville's theorem but... a couldn't find any way to bound f near the imaginary axis.
(I used Cauchy's formula to represent f's values as integrals - but... it was not much help for me)

I could use some help

thanks

A similar problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174622]has been solved[/url]. I don't know a solution that does not use subharmonicity.

I think there is another way to solve this (we did not learn about subharmonicity but still we are requested to solve this without it)/
i think we should somehow evaluate the integral representing f's values on the imaginary axis.

What's wrong with the good old Cauchy formula for the Taylor coefficients at $ 0$? :? Just write the integrals and estimate them using the given majorant! That the integral of the majorant over the circle is improper, shouldn't bother you: it converges and that's all that really matters. Also, you can use the same idea to estimate $ f$ on the imaginary axis and then use Liouville.

I should have paid more attention to the difference between this and the older thread (where the majorant was not integrable).  :oops:

Even in the old thread (with $ 1/|\Re z|$), you could get away without using subharmonicity. Just consider $ f(z)(z\minus{}ia)(z\minus{}ab)$ and use the Cauchy integral formula for the circle with diameter $ [ia,ib]$ to estimate $ f$ near the imaginary axis. Of course, the truth is that the conclusion holds whenever $ |f(z)|\le H(|\Re z|)$ where $ H: (0,\plus{}\infty)\to(0,\plus{}\infty)$ is a positive decreasing function such that $ \int_0\log\log H<\plus{}\infty$ (Levinson? :?) but proving that would, indeed, require much more advanced tools.


[quote="Interval"]An air rescue plane averages 300 mph in still air.  It carries enough fuel for 5 hours of flying time.  If, upon takeoff, it encounters a wind of 30 miles per hour, how far can it fly and return safely?

Assume that the wind remains constant.[/quote]
[hide="Hint"] I'm assuming the 30 mph wind in parallel to the flight path. If the plane travels $x$ hours going there, it will take $5-x$ hours coming back. How far will it travel forward in terms of $x$? How far will it travel back in terms of $x$? You must get these 2 distances to be equal. [/hide]

Are you saying for me to equate the distance formula to itself?


probability
calculus
probability and stats

Let Bt be Brownian motion and fix t0>0
Prove that Bt=B(to+t)-B(t0) is Brownian motion.
[/quote]

This follows immediately due to stationary increments.

It seems simple, but I do not have proper literature and I have difficulties to solve the problem.
So could you recommend me literature? Or give me some useful link
Thank you

Well it depends on your level in probability I would recommand Karatzas and Shreve's book on brownian motion and stochastic calculus
or if you are less advanced the book of Bernt Oksendal "Stochastic Differential Equations: An Introduction with Applications" is very good introduction.

regards

thank you,
it helped me a lot. I already had the book of Oksendal, but what I need is more  solved examples, because this book has no sufficiently solved examples
Could you recommend me some literature with solved examples?


The two polynomials $(x) =x^4+ax^3+bx^2+cx+d$ and $Q(x) = x^2+px+q$ take negative values on an interval $I$ of length greater than $2$, and nonnegative values outside of $I$.  Prove that there exists $x_0 \in \mathbb R$ such that $P(x_0) < Q(x_0)$.

This is incomplete:

We also have the condition P(x)<0 iff Q(x)<0 (I saw this in a file taken from Kalva and which contains the problems used in the Russian olympiad in 2001).

I don't think it said the polynomials were monic in the file I saw, but I'll solve this problem.

Q(x)=(x-x_1)(x-x_2), with x_1 < x_2 reals s.t. x_2-x_1>2. P(x)=(x-x_1)(x-x_2)(x-z)(x-z'), where z is a number from C\R and z' is its conjugate. This follows from the fact that P(x)<0 iff Q(x)<0, so you see, this is an important condition.

Let's assume P(x)>=Q(x) for all real x. This means that
|x-z|>=1 for all x>x_2 and x<x_1
|x-z|<=1 for x in (x_1,x_2)
Since |x-z| is a continuous function on R, it means that |x_1-z|=|x_2-z|=1. Now x_2-x_1=|x_2-x_1|=|(x_2-z)-(x_1-z)|<=|x_2-z|+|x_1-z|=2, but x_2-x_1>2, so we've got a contradiction, our assumption is false, so there must exist a real x_0 s.t. P(x_0)<Q(x_0).

Let $I=[s,t]$ with $t-s \ge 2$. Now observe that $Q(x)>0$ if and only if $x \not \in I$ so $Q$ vanishes at $s,t$. Similarly $P$ also vanishes at $s,t$. Suppose $P(x) \ge Q(x)$ for all $x \in \mathbb{R}$. Now let $f \in \mathbb{R}[X]$ with $P(x)=f(x) \cdot q(x)$.

[b]Claim.[/b] $f$ has a real root other than $s,t$.

(Proof) Since $Q(x) \cdot \left(f(x)-1\right) \ge 0$ we see $f(x)>1$ if and only if $x \not \in I$. Consequently, $f(x)-1$ vanishes at $s,t$ so $$f(x)=(x-s)(x-t)+1=x^2-x(s+t)+st+1$$ for all $x \in \mathbb{R}$. But the discrimant of this quadratic is $(s+t)^2-4st-4=(s-t)^2-4 \ge 0$ proving the claim. $\blacksquare$

Finally, we see $P$ must only have two real roots $s$ and $t$ but it now has a third root, contradiction!


algebra
polynomial
calculus
calculus computations

Let f(x) be a polynomial with complex coefficients such that all its roots lie in the upper half plane. Prove that all the roots of f'(x) also lie in the upper half plane.

It's a special case of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=152998]this theorem[/url], which is strong enough to prove the general case. Follow the links for a proof.


A set of magnetic strips and balls are connected in an alternating fashion, as shown.
The length of the first four items in the chain is 14 cm, and the length of the first
seven items is 25 cm. What is the length of the first 25 items in the chain?

[asy]import graph;
draw((5,1)--(9,1));
draw((5,-1)--(9,-1));
draw(Circle((4,0),1.7));
draw((-1,1)--(3,1));
draw((-1,-1)--(3,-1));
draw(Circle((10,0),1.7));
draw((11,1)--(15,1));
draw((11,-1)--(15,-1));
dot((16,0));
dot((18,0));
dot((20,0));
dot((-2,0));
dot((-4,0));
dot((-6,0));[/asy]

First four items takes 14 cm, next three (from 4 to 7) take 11 cm.  So each three "items" take up 11 cm.  25-7=18, meaning there's 6 more 11 inch segments.  25+6*11=91.

However, the game says the answer is 88. Hmm....

Edit: Misinterpreted the word "items."

If you let strips be $ s$ and balls be $ b$, then looking at the picture we can see that:
$ 2s \plus{} 2b \equal{} 14$
and
$ 4s \plus{} 3b \equal{} 25$
Solving that gives $ b \equal{} 3$ and $ s \equal{} 4$ so $ 13 \times s \plus{} 12 \times b \equal{} 13 \times 4 \plus{} 12 \times 3 \equal{} \boxed{88}$

[quote=futuregenius123]YOUR FORMULA IS WRONG![/quote]

Who's?


Prove that for every $ n\in \mathbb{N} \ n\ge 1$ and $ a,b,c \in \mathbb{R}$ such that
$ a,b \ge 1$ and $ c\ge 0$ the following inequality holds:
$ \ (ab \plus{} c)^{n} \minus{} c \le ((b \plus{} c) ^{n} \minus{} c) a^{n}$  :wink:

The inequality is equivalent to
$ c(a^n \minus{} 1) \leq (ab\plus{}ac)^n \minus{} (ab\plus{}c)^n$
or
$ c(a\minus{}1)(a^{n\minus{}1}\plus{}...\plus{}1) \leq c(a\minus{}1)(\sum_{k\equal{}0}^{n\minus{}1} (ab\plus{}ac)^k(ab\plus{}c)^{n\minus{}1\minus{}k})$.
And the inequality follows, since it's obvious that $ (ab\plus{}ac)^k(ab\plus{}c)^{n\minus{}1\minus{}k} \geq a^k$ ($ a, b \geq 1$ and $ c \geq 0$).


logarithms
inequalities
inequalities proposed

[quote="Marius Mainea"]Let $ x,y,z > 0$ such that $ xyz \equal{} 1$. Prove that
\[ \frac {xy}{1 \plus{} z^2} \plus{} \frac {yz}{1 \plus{} x^2} \plus{} \frac {zx}{1 \plus{} y^2}\geq\frac {3}{2}.
\]
[/quote]
The following reasoning works here:
$ \sum_{cyc}\frac {yz}{1 \plus{} x^2}\geq\frac {3}{2}\Leftrightarrow\sum_{cyc}\left(\frac {1}{x \plus{} x^3} \minus{} \frac {1}{2} \plus{} \ln x\right)\geq0.$ :wink:
By the way for all $ k\geq0$ and for all positive $ x,$ $ y$ and $ z$ such that $ xyz\equal{}1$ holds:
\[ \sum_{cyc}\frac {yz}{k \plus{} x^2}\geq\frac {3}{k\plus{}1}\]

[quote="arqady"]By the way for all $ k\geq{0}$ and for all positive $ x,y$ and $ z$ such that $ xyz\equal{}1$ holds:
\[ \sum_{cyc}\frac {yz}{k \plus{} x^2}\geq\frac {3}{k\plus{}1}\][/quote]
Can you prove it? :wink:

Sunkern_sunflora, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1202304#1202304
The same idea. :wink:

Oh No! Links again! :(
Can't you post your proof without links arqady? I can't view such links.
What happened to the two problems I sent you as my unsolved problems? Should I resend them?

[quote="arqady"] http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1202304#1202304 [/quote]
Hello, this link redirects [b]this[/b] very page :P  :rotfl:

[quote="Sunkern_sunflora"][quote="arqady"]By the way for all $ k\geq{0}$ and for all positive $ x,y$ and $ z$ such that $ xyz \equal{} 1$ holds:
\[ \sum_{cyc}\frac {yz}{k \plus{} x^2}\geq\frac {3}{k \plus{} 1}
\]
[/quote]
Can you prove it? :wink:[/quote]
$ \sum_{cyc}\frac {yz}{k \plus{} x^2} \equal{} \sum_{cyc}\frac {(yz)^2}{kyz \plus{} x^2yz}\ge \frac {(xy\plus{}yz\plus{}zx)^2}{k(xy \plus{} yz \plus{} zx) \plus{} xyz(xy \plus{} yz \plus{} zx)}\ge$
$ \ge \frac {xy \plus{} yz \plus{} zx}{k \plus{} \frac {xy \plus{} yz \plus{} zx}{3}}$
we need prove that $ \frac {xy \plus{} yz \plus{} zx}{k \plus{} \frac {xy \plus{} yz \plus{} zx}{3}}\ge \frac {3}{k \plus{} 1}$
$ \Leftrightarrow ab \plus{} bc \plus{} ca \ge 3$ which is true by am-gm

[quote="bunhiacovski"][quote="Sunkern_sunflora"][quote="arqady"]By the way for all $ k\geq{0}$ and for all positive $ x,y$ and $ z$ such that $ xyz \equal{} 1$ holds:
\[ \sum_{cyc}\frac {yz}{k \plus{} x^2}\geq\frac {3}{k \plus{} 1}
\]
[/quote]
Can you prove it? :wink:[/quote]
$ \sum_{cyc}\frac {(yz)^2}{kyz \plus{} x^2yz}\ge \frac {(xy \plus{} yz \plus{} zx)^2}{k(xy \plus{} yz \plus{} zx) \plus{} xyz(xy \plus{} yz \plus{} zx)}$[/quote]

 I dont think so  :blush:

[quote="Marius Mainea"][/quote][quote="bunhiacovski"][quote="Sunkern_sunflora"][quote="arqady"]By the way for all $ k\geq{0}$ and for all positive $ x,y$ and $ z$ such that $ xyz = 1$ holds:
\[ \sum_{cyc}\frac {yz}{k + x^2}\geq\frac {3}{k + 1}
\]
[/quote]
Can you prove it? :wink:[/quote]
$ \sum_{cyc}\frac {(yz)^2}{kyz + x^2yz}\ge \frac {(xy + yz + zx)^2}{k(xy + yz + zx) + xyz(xy + yz + zx)}$[/quote][quote="Marius Mainea"]

 I dont think so  :blush:[/quote]
apply Cauchy-Schwarz inequality  :wink:

bunhiacovski, it can be
$ \sum_{cyc}\frac {(yz)^2}{kyz \plus{} x^2yz}\ge \frac {(xy \plus{} yz \plus{} zx)^2}{k(xy \plus{} yz \plus{} zx) \plus{} xyz(x\plus{}y\plus{}z)}$ :wink:


In triangle $ABC$ we have $\angle ABC = 2 \angle ACB$. Prove that

a) $AC^2 = AB^2 + AB \cdot BC$;

b) $AB+BC < 2 \cdot AC$.

The first part is: http://www.mathlinks.ro/Forum/viewtopic.php?p=288310#p288310

For the second part: $AC^2=AB \cdot (AB+BC) \Longrightarrow \frac{AC}{AB} \cdot AC = AB+BC$.
Thus the inequality becomes: $\frac{AC}{AB} \cdot AC < 2AC \Longleftrightarrow AC<2AB$.
Now $AC < 2AB \Longleftrightarrow \sin{2C} < 2\sin{C} \Longleftrightarrow 2\sin{C}\cos{C} < 2\sin{C} \Longleftrightarrow \cos{C}<1$. Because $\angle{C}< 90^o$, so equality cannot hold and the inequality is strict. Thus the problem is solved.


geometry
geometric transformation
reflection
similar triangles
geometry unsolved

the locus of points whose distances to two given lines $ m, n$ are proportional to two given segments is the union of two lines meeting at the intersection point of $ m, n$ (if $ m, n$ are parallel, which is not the case in your picture, the locus will be the union of two parallel lines). so each pair of lines produces a locus of points, and the desired locus is the union of the (finitely many) intersection points of these pairwise loci...

(if all three lines are parallel, note that the locus may be empty...)

"the locus of points whose distances to two given lines are proportional to two given segments is the union of two lines meeting at the intersection point"

i don´t see how can construct these lines   :(

let's say you're given two lines $ l_1, l_2$ that intersect at $ O$, and two segments of lengths $ d_1, d_2$, respectively, and that you want the set of points whose distances to $ l_1, l_2$ are proportional to $ d_1, d_2$, respectively. draw a line $ a$ parallel to $ l_1$ at distance $ d_1$ from $ l_1$ (there are two ways to draw the line $ a$ -- one on either side) and a line $ b$ parallel to $ l_2$ at distance $ d_2$ from $ l_2$ (again, two ways to draw $ b$). $ a$ and $ b$ will intersect at a point $ P$. since we have two choices for $ a$, and two for $ b$, we'll actually get $ 2\cdot 2 \equal{} 4$ different points $ P$, one in each of the "quadrants" formed by the lines $ l_1$ and $ l_2$. but it's not hard to see (via similar triangles) that points in opposite quadrants will be reflections of each other across $ O$. connecting $ O$ to your points $ P$ by straight lines then produces two lines through $ O$ which form your locus (again, use similar triangles to show that this is in fact your locus... )


Prove that numbers 49, 4489, 444889, ..., obtained placing number 48 in the middle of the previous number, they are the squares of whole numbers.


function
functional equation
algebra
IMO Shortlist

I have a solution to this one (not mine) and I can post it if you want (it's not short but not very long).
If you are still interested in trying it yourself here are 2 hints:
1)Prove that $g$ is surjective (that's not very easy).
2)Therefore, it takes value $0$, using it solve the rest of a problem (that's the easier part).

Hello !

My way to prove that g(x) may be not surjective is very short :

$(f(x)=0, g(x)=0)$ is a solution in which g is not surjective.

--
Patrick

I think that you're totally misunderstanding e.lopes and TomciO! When $f(0) = 0$ (that's the first thing I tried) it's easy to prove that $g$ takes the value $0$. In the other case, I think it may be that $g$ is surjective.

Exactly, in fact, I wasn't clear in my previous post. We need only that $g$ takes value $0$, if $f(0)=0$ then we can do it by hand, in other case we prove that $g$ is surjective.

I found one more nice and short way to solve our equation, my friends!

We will prove first that $g(i) = 0$, for any i.

If f(0) = 0, we find that $f(x+g(0)) = g(x)$. Take $x =-g(0). g(-g(0)) = f(0) = 0$.

If f(0) isn't, is more difficult. See:
Take x = 0 in the original equation. We see that f is surjective and g is injective!, because $f(g(y)) = g(0)-y.f(0)$

In original equation, put $x = g(x)$. We find $f(g(x)+g(y)) = g(x)f(y)-yf(g(x)+g(g(x))=g(x)f(y)-y(g(0)-xf(0))+g(g(x))$

This is symetric, so, $g(x)f(y)-y(g(0)-xf(0))+g(g(x)) = g(y)f(x)-x(g(0)-yf(0))+g(g(y))$ (*).

Cause f is surjetive, we can take $y = m$, where $f(m) = 0$.

Do this in (*)!

We find $-f(0)m+g( g(x) ) = g(m) f(x)-f(0)x+g(g(m))$. Putting $g(m) = p$ and $g(g(m))+f(0)m = q$,

this becomes $g(g(x)) = pf(x)-f(0)x+q$.

Substituting back into (*) we find  $g(x) f(y)+pf(x) = g(y)f(x)+pf(y)$ (**)

Let y = 0 in (**). We find g(x) = $\frac{(f(0)-p)f(x)}{g(0)}+k$. So, g is surjective! and we can take g(i) = 0! for one real i.

_

Now, we will prove that f and g are linears!
So, in the original equation, let y = i!

We find $g(x) = (i+1)f(x)-f(i)x$. Substituing back into original equality, we find:

$f(x+g(y)) = (i+1-y)f(x)+x(f(y)-f(i))$ (***)

Let $y = i+1$ in (***).

We find that $f(x+r) = x(f(i+1)-f(i))$, where $f = g(i+1)$! So, $f$ is linear!  :)

In the equality g(x) = (i+1)f(x)-f(i)x, we see the that g is also linear!

_

Now, is easy, let f(x) = ax+b and g(x) = cx+d, and good luck with the computations!

The solutions are f(x) = g(x) = 0 and $f(x) = \frac{k(x-k)}{k+1}$ , $g(x) ={k(x-k)}$! for a real k!

suppose that $P(x,y)$ be the following assertion:$f(x+g(y))=xf(y)-yf(x)+g(x)$
$P(x,0):f(x+g(0))=xf(0)+g(x)$. (put $g(0)=c$)      $(1)$
$P(x+t+c,y):f(x+c+t+g(y))=(x+t+c)f(y)-yf(x+c+t)+g(x+c+t)$
$P(x+c,y):f(x+c+g(y))=(x+c)f(y)-yf(x+c)+g(x+c)$.now subtracting two equations and use of $(1)$ we have:$tf(0)+g(x+t+g(y))-g(x+g(y))=tf(y)-y(tf(0)+g(x+c+t)-g(x+c))+g(x+t)-g(x)$.
so if the previous assertion be $Q(x,y)$,then
$Q(x,0):g(x+t+c)-g(x+c)=g(x+t)-g(x)$ $(2)$
then with use of $(2)$ we have $Q(x,1):g(x+t+g(1))-g(x+g(1))=t(f(1)-f(0))$ so for all $x$ we have:$g(x+t)-g(x)=ta$ (put $a=f(1)-f(0)$) put $x=0$ we have:$g(t)=at+c$ so:($put b=f(0)$)
$f(x+c)=bx+ax+c$ so $f(x)=(a+b)x+c(1-a-b)$ now by putting it into main equation we conclude that:$a(a+b)=-c(1-a-b)=-b$ and $f(x)=(a+b)x+b,g(x)=ax+c$.or for some real $a\neq -1$.$f(x)=\frac{a(x-a)}{a+1},g(x)=a(x-a)$.

[quote="perfect_radio"]Find all pairs of functions $ f, g : \mathbb R \to \mathbb R$ such that $ f \left( x \plus{} g(y) \right) \equal{} x \cdot f(y) \minus{} y \cdot f(x) \plus{} g(x)\quad (1)$ for all real $ x, y$.[/quote]
Let $f(0)=a$ and $g(0)=b.$ Taking $x=0$ in $(1),$ we get \[f\big(g(y)\big)=-ay+b,\quad \forall y \in \mathbb R. \quad (2)\] There are two cases to consider:

[b]Case 1: $\mathbf{a=0.}$[/b] Taking $y=0$ in $(1),$ we get \[f(x+b)=g(x),\quad \forall x \in \mathbb R. \quad (3)\] From this, we deduce that $g({-b})=0.$ Now, replacing $y=-b$ in $(1),$ we have \[f(x)=x\cdot f({-b})+b\cdot f(x)+g(x),\quad \forall x \in \mathbb R. \quad (4)\] Taking $x=-b$ in $(4),$ we get $f({-b})=g({-b})=0$ and hence, the equality $(4)$ can be rewritten as \[(1-b)\cdot f(x)=g(x),\quad \forall x \in \mathbb R. \quad (5)\] Taking $x=0$ in $(5),$ we get $b=g(0)=0.$ Thus, the identities $(2)$ and $(5)$ can be written as \[f\big(g(x)\big)=0,\quad \forall x \in \mathbb R\quad (6)\] and \[f(x)=g(x),\quad \forall x \in \mathbb R. \quad (7)\] From $(6)$ and $(7),$ we get $g\big(g(x)\big)=0,\, \forall x \in \mathbb R.$ Now, replacing $y$ by $g(x)$ in $(1),$ we can easily deduce that $f(x)=g(x)=0,\, \forall x \in \mathbb R.$ These functions satisfy our condition.

[b]Case 2: $\mathbf{a \ne 0.}$[/b] From $(2),$ we can easily see that $f$ is surjective and $g$ is injective. Replacing $x$ by $g(x)$ in $(1)$ and using $(2),$ we get \[f\big(g(x)+g(y)\big)=g(x)\cdot f(y)+axy-by+g\big(g(x)\big),\quad \forall x ,\,y \in \mathbb R. \quad (8)\] Changing the position of $x$ and $y$ in $(8),$ we get \[g(x)\cdot f(y)-by+g\big(g(x)\big)=g(y)\cdot f(x)-bx+g\big(g(y)\big),\quad \forall x,\,y \in \mathbb R. \quad (9)\] Replacing $y=0$ in $(1),$ we have \[f(x+b)=ax+g(x),\quad \forall x \in \mathbb R. \quad (10)\] Replacing $x$ by $g(x)$ in $(10),$ we get \[g\big(g(x)\big)+a\cdot g(x)=f\big(b+g(x)\big)=b\cdot f(x)-x\cdot f(b)+g(b),\] or \[g\big(g(x)\big)=b\cdot f(x)-a\cdot g(x)-x\cdot f(b)+g(b),\quad \forall x \in \mathbb R. \quad (11)\] Plugging this result into $(9),$ we get \[\begin{aligned} g(x)\cdot f(y)-by+{}&b\cdot f(x)-a\cdot g(x)-x\cdot f(b)=\\ &=g(y)\cdot f(x)-bx+b\cdot f(y)-a\cdot g(y)-y\cdot f(b).\quad (12)\end{aligned}\] Since $f(b)=b$ (we can easily obtain this by setting $y=0$ in $(2)$), the above identity can be written as \[g(x)\cdot \big[ f(y)-a\big] +b\cdot f(x)=g(y)\cdot \big[ f(x)-a\big] +b\cdot f(y),\] or \[\big[ g(x)-b\big] \big[ f(y)-a\big]=\big[ g(y)-b\big] \big[ f(x)-a\big],\quad \forall x,\,y \in \mathbb R. \quad (13)\] Since $f$ is surjective, there exists $y_0$ such that $f(y_0) \ne a.$ Taking $y=y_0$ in $(13),$ we get \[g(x)-b=\frac{g(y_0)-b}{f(y_0)-a}\big[f(x)-a\big],\quad \forall x \in \mathbb R.  \quad (14)\] If $g(y_0)=b,$ then we have $g(x)=b,\,\forall x \in \mathbb R$ which is a contradiction since $g$ is injective. So we must have $g(y_0)\ne b.$   From this and the previous results, we can easily prove that $f$ and $g$ are bijectives. Thus, the exists a unique number $c$ such that $g(c)=0.$ Taking $x=y=c$ in $(1),$ we have $f(c)=g(c)=0.$ Continuously, replacing $y=c$ in $(1),$ we get \[(1+c)\cdot f(x)=g(x),\quad \forall x \in \mathbb R. \quad (15)\] Since $g$ is bijective, it is clearly that $c \ne -1,$ from which it follows that \[f(x)=\frac{1}{1+c}\cdot g(x),\quad \forall x \in \mathbb R. \quad (16)\] Now, replacing $y=1+c$ in $(1)$ and using $(15),$ we get \[f\big(x+g(1+c)\big)=x\cdot f(1+c)=\frac{x\cdot g(1+c)}{1+c},\quad \forall x \in \mathbb R. \quad (17)\] Replacing $x=c-g(1+c)$ in $(17),$ we get \[g(1+c)\cdot \big[c-g(1+c)\big]=0.\] Since $g$ is bijective, we have $g(1+c)\ne 0$ and hence, it follows that $g(1+c)=c.$ From this, we have \[f(x+c)=\frac{cx}{1+c},\quad \forall x \in \mathbb R.\quad (18)\] Replacing $x$ by $x-c$ in $(18),$ we get \[f(x)=\frac{c}{1+c}(x-c),\quad \forall x \in \mathbb R.\] It follows that \[g(x)=c(x-c),\quad \forall x \in \mathbb R.\] These functions satisfy our condition.

Notice f(g(x))=g(0)-f(0)x. (1)
If f(0)=0 we get g(-g(0))=0.
Otherwise f is surjective and g is injective. taking x=g(x) and considering y as a consant, we get an equation. if we change the value of y in this equation, we find that g(x)=0 for some x. So g(a)=0 for some a always.
Then in our original equation f(x)=(f(x+g(a))=xf(a)-af(x)+g(x) so f(x)=(g(x)+xf(a))/(a+1), unless a=-1 but in this case we finish easily. And we get f(a)=af(a)/(a+1). so if a=0 we finish easily, otherwise f(a)=0 and we get g=cf for a constant c. also f is bijective and from here we finish easily.

answer: linear f and g=cf for a constant c. additional equations must be satisfied by the coefficients of f, we obtain these relations by just plugging in.

I proved the $f(0) = 0$ the same way as CanVQ. I have a slightly different approach in proving $f(0) \neq 0$, specifically in proving bijectivity of $f$ and $g$

[b][size=150]f(0) is not equal to 0[/size][/b]

Then $f(g(y)) = -y*f(0) + g(0)$ (1).

[b]Lemma: f and g are both bijective[/b]

Proof:

Clearly $f \circ g$ is a bijective function as it's linear. Thus we can immediately conclude from it's injectivity that $g$ is injective and from it's surjectivity that $f$ is surjective.

For any $c \in \mathbb{R}$ note that if $f(c) = a$ we have $f(c) = a = \frac{a-g(0)}{f(0)} * f(0) + g(0) = f(g(\frac{g(0)-a}{f(0)}))$ and since $f(0)$ is non-zero we have that there necessarily exists some real number a satisfying $g(a) = c$ for all real c proving surjectivity

Now, we had $a,b \in \mathbb{R}$ such that $f(a) = f(b) = c$. Then, we know there exist real m and n such that $g(m) = a$ and $g(n) = b$

Then we would have $f(g(m)) = f(a) = -m*f(0) + g(0)$ and $f(g(n)) = f(b) = -n*f(0)+g(0)$ Thus we have $m=n$ and $a=b$ so f is injective. This completes the lemma.

$\square$

Now this means there exists a unique value $a$ such that $g(a) = 0$. Plug in $x=y=a$ into the original equation to get $f(a) = g(a) = 0$. (2)

Now, plug in just $y=a$ and use the result from (2) to obtain $f(x) = -a*f(x) + g(x)$ or $(a+1)f(x) = g(x)$ (3)

Plugging (3) into (1) reveals $f((a+1)f(x)) = -x*f(0)+(a+1)f(0)$. Now notice that when $x=a+1$ we have that $f((a+1)f(a+1)) = 0$. Thus $f(a+1) = \frac{a}{a+1}$ and thus also $g(a+1) = a$ (4)

Now, plug (3) back into the original equation and we get

$f(x+(a+1)f(y)) = xf(y) - yf(x)+(a+1)f(x)$. Plug in $y=a+1$ and use (4) and we get:

$f(x+a) = \frac{xa}{a+1}$ or $f(x) = \frac{(x-a)a}{a+1}$ and thus from (3) $g(x) = (x-a)a$ for some real constant $a$ which completes the problem.

$\square$

(note that since $(a+1)*g(0) = f(0)$ and we have that $g(0) \neq 0$ and $f(0) \neq 0$ this means that $a + 1 \neq 0$ and $a \neq -1$.)

[hide=solution][quote=ISL 2000 A3]Find all pairs of functions $ f : \mathbb R \to \mathbb R$, $g : \mathbb R \to \mathbb R$ such that \[f \left( x + g(y) \right) = xf(y) - y  f(x) + g(x) \quad\text{for all } x, y\in\mathbb{R}.\][/quote]
Say P be the assertion,
$P(0,y) \implies f(g(y))=-yf(0)+g(0)$
Putting $-f( 0)=p,g(0)=q$ gives $f(g(y))=py+q$.
$P(g(x),y) \implies f(g(x)+g(y))=g(x)f(y)-yf(g(x))+g(g(x))-------(2)$.
Exchanging $x,y$ in $(2)$ gives,
\begin{align*} g(y)f(x)-xf(g(y))+g(g(y))=g(x)f(y)-yf(g(x))+g(g(x)) \\
g(y)f(x)-x(py+q)+g(g(y))=g(x)f(y)-y(px+q)+g(g(x))\\
g(y)f(x)-xq+g(g(y))=g(x)f(y)-yq+g(g(x))------(3)\\
\end{align*}
Now ,putting $y=0$ in $(3)$ gives
$g(g(x))=pg(x)+q(f(x)-x)+g(g(0))$
Putting the value of $g(g(x))$ in $(3)$ gives,
\begin{align*} g(y)f(x)+pg(y)+qf(y)-qy-qx=g(x)f(y)+pg(x)+qf(x)-qx-qy \\
g(y)f(x)+pg(y)+qf(y)=g(x)f(y)+pg(x)+qf(x) \\
g(y)f(x)+pg(y)-qf(x)-pq=g(x)f(y)+pg(x)-qf(y)-pq \\
(g(y)-q)(f(x)+p)=(g(x)-q)(f(y)+p)  \\
\end{align*}
This concludes $f,g$ are linear $f(x)=ax+b,g(x)=cx+d$ for some real constants $a,b,c,d$.
If $f,g$ are constant then $f,g \equiv 0$
and hence we are done       $\blacksquare$[/hide]
[hide=hmmmm..]This was impossible without BJV sir hints was impossible for me ig :blush:.It wasn't enough though,you have to somehow force your brain to think :rotfl:. Guy

This took way too long T^T

Let $x=0$ to get $f(g(y))=-yf(0)+g(0)$ which is linear, so $f(x)$ is surjective and $g(x)$ is injective.

Let $y=0$ then $f(x+g(0))=xf(0)+g(x).$ Suppose $x=-g(0)$ then $f(0)=-f(0)g(0)+g(-g(0))$ which implies that $g(-g(0))=f(0)(1+g(0)).$ Thus, if $f(0)=0$ then $g(-g(0))=0.$

Let $x=g(a)$ while $y=b.$ Then \[f(g(a)+g(b))=g(a)f(b)-bf(g(a))+g(g(a))=g(b)f(a)-af(g(b))+g(g(b)).\] Note that we can let $f(a)=0$ to give us $g(g(b))=pf(b)+\text{linear}.$ Thus, $pf(a)-g(a)f(b)=pf(b)-g(b)f(a)$ which can be arranged to form $g(a)=xf(a)+y$ for some constants $x,y$ with $x\neq 0.$ Thus $g$ is surjective unless $g$ is $0$.

Now we let $g(y)=0$ so $f(x)(1+y)=xf(y)+g(x).$ Now, let $y=y+1$ to get $f(x+g(y+1))=xf(y+1)-(y+1)f(x)+g(x)=xf(y+1)-xf(y)$ so $f(x)$ is clearly linear, and so $g(x)$ is also linear. Thus, we finish by blindly plugging in coefficients into the original equation to find that $f(x)=\frac{c}{1+c}x-\frac{c^2}{1+c}$ and $g(x)=cx-c^2$ for all $c\neq -1$

Solved with Jeffrey Chen and Max Lu.

The answer is \(f(x)\equiv\frac c{c+1}(x-c)\) and \(g(x)\equiv c(x-c)\) for \(c\ne-1\), which work. Now we show these are the only solutions.

Let \(P(x,y)\) denote the assertion. Note by \(P(0,x)\) that \[f(g(x))=-f(0)\cdot x+g(0)\quad\forall x.\tag{\(\star\)}\]

[color=red][b]Claim:[/b][/color]      We have      \[\big[g(y)-g(0)\big]\cdot\big[f(x)-f(0)\big]     =\big[g(x)-g(0)\big]\cdot\big[f(y)-f(0)\big].\]

[i]Proof.[/i]      By \(P(g(x),y)\) and \((\star)\), we have     \begin{align*}         f(g(x)+g(y))         &=g(x)f(y)=f(0)xy-g(0)y+g(g(x)).     \end{align*}     Combining with \(P(g(y),x)\) gives      \[g(y)f(x)+g(g(y))-g(0)x     =g(x)+f(y)+g(g(x))-g(0)y.\tag{\(\dag\)}\]

Note \((\star)\) gives \(f(g(0))=0\),     so \(P(g(x),0)\) and \(P(g(0),x)\) give     \[g(g(x))+g(x)f(0)=f(g(x)+g(0))=g(0)f(x)-g(0)x+g(g(0)).\]     Substituting \(g(g(x))\) into \((\dag)\),     we have     \[g(y)f(x)+g(0)f(y)-f(0)g(y)=g(x)f(y)+g(0)f(x)-f(0)g(x),\]     which arranges into the desired. \(\blacksquare\)

[b]First case:[/b] \(f(0)=0\). Then \((\star)\) gives \(f(g(x))=g(0)\) for all \(x\), but \(P(x,0)\) gives \(f(x+g(0))=g(x)\). Hence \[f(f(x))=f(g(x-g(0)))=g(0)\] for all \(x\). In particular \(x=0\) gives \(g(0)=0\), so \(f(x)\equiv g(x)\) and \(f(f(x))\equiv0\). Now \(P(x,f(x))\) gives \(f(x)\equiv0\).

[b]Second case:[/b] \(f(0)\ne0\). By \((\star)\), \(f\circ g\) is a nonconstant linear function, so \(g\) is injective and \(f\) is surjective. But the claim gives that \(g\) and \(f\) are linear in each other, so \(f\) and \(g\) are both bijective.

In particular, \(g(c)=0\) for some \(c\). By \(P(c,c)\), we also have \(f(c)=0\). By \(P(x,c)\), we have \((c+1)f(x)\equiv g(x)\), and by \(P(c,y)\), we have \[f(c+(c+1)f(y))=cf(y).\] Since \(f\) is bijective, we are done.

[quote=mathverse06]Does anyone know the official solution for this? I was like wow...takes forever[/quote]

Here's the kalva solution:
[url]https://prase.cz/kalva/short/soln/sh00a3.html[/url].

Let's try to solve it myself, too.

[b][u]Claim 1:[/u][/b] If $g(x)=0$ for some $x$ then $f$ and $g$ are linear.
[i]Proof.[/i] Let $g(z)=0,$ for some real $z.$
$P(x,z)\implies g(x)=(z+1)f(x)-xf(z).$
$P(x,z+1)\implies f\text{ and }g \text{ are linear.}$
Now it's easy to get that, $f(x)=\frac{kx-k^2}{1+k}$ and $g(x)=kx-k^2$
are indeed solutions for all $x\in \mathbb{R}$ and $k\neq -1.$
---------------

[b][u]Claim 2:[/u][/b] $\exists z\in \mathbb{R}: g(z)=0.$
[i]Proof.[/i] We do casework on $f(0).$

[color=#f00]Case 1:[/color] $f(0)=0.$
Then $P(x,0)\implies f(x+g(0))=g(x).$ Take $u=-g(0).$
So $g(u)=f(0)=0.$ Q.E.D.

[color=#f00]Case 2:[/color] $f(0)\neq 0.$
Then it is easy to find that $f$ is surjective and $g$ is injective.
Comparing $P(g(y),x)$ and $P(g(x),y)$ yields
$g(x)f(y)-y(g(0)-xf(0))+g(g(x))=g(y)f(x)-x(g(0)-yf(0))+g(g(y)). ~~~~~(*)$
$\exists v\in \mathbb{R}:f(v)=0.$
$y\to v$ in $(*)$ yields,
$g(g(x))=f(x)g(v)-xg(0)+g(g(v))+vg(0).$
$(*)\implies g(x)f(y)+g(v)f(x)=g(y)f(x)+g(v)f(y).$
$y\to 0 \implies g(v)+\frac{g(0)-g(v)}{f(0)}f(x)=g(x).$
We have $g(0)\neq g(v)$ since $g$ is injective and $v\neq 0,$ thus $g$ is surjective. Q.E.D.


If $ f(x) \equal{} 3\sqrt{x}$, find all x for which $ f(x) \equal{} f^{\minus{}1}(x)$

so obviously I've found $ f^{\minus{}1}(x) \equal{} \frac{x^2}{9}$

setting the two equations equal to each other,

$ 3\sqrt{x} \equal{} \frac{x^2}{9}$

$ 3^3\sqrt{x} \equal{} x^2$

$ 3^6x \equal{} x^4$

$ 3^6 \equal{} x^3$

$ x \equal{} 3^2 \equal{} \boxed{9}$

but $ x \equal{} 0$ is obviously another solution, but how would I get this algebraically?

good thing the AIME doesn't rely too much on just algebra and stuff.. XD

ahh nvm. the future helped me in the Indiana forum:

$ 3^6x \equal{} x^4$

$ 3^6x \minus{} x^4 \equal{} 0$

$ x(3^6 \minus{} x^3) \equal{} 0$

so $ x \equal{} 0$ or $ x \equal{} 3^2 \equal{} 9$.

[quote]
$ 3^6x \equal{} x^4$

$ 3^6 \equal{} x^3$ [/quote]

in this step you assume $ x<0$ or $ x>0$ because you can't divide by zero ....

'How you get this algebraically" is move all terms to one side and solve as a polynomial instead of dividing by $ x$.

Thanks guys, I'll keep your tips in mind :lol:

Wait.
$ f^{\minus{}1}(x)\equal{}(f(x))^{\minus{}1}$ right?

No the negative exponent on the $ f$ isn't actually an exponent - it represents the inverse function, so that:

$ f^{\minus{}1}(f(x)) \equal{} 1$

$ f(x)$ reflected over the line $ y \equal{} x$ is $ f^{\minus{}1}(x)$

..etc.

[quote="undefined117"]
$ f^{ \minus{} 1}(f(x)) \equal{} 1$
[/quote]

You mean

$ f^{\minus{}1}(f(x)) \equal{} x$

[quote="SimonM"][quote="undefined117"]
$ f^{ \minus{} 1}(f(x)) \equal{} 1$
[/quote]

You mean

$ f^{ \minus{} 1}(f(x)) \equal{} x$[/quote]
XD heh yeah sorry