function
algebra
domain
limit
algebra proposed

I have the same !  :) but not on the selection  :(

After seeing the solution, it's seems easy but it's not :)

Can you say $f(x)=y^2$ without worrying about the sign?
At this point, I think that all you can say is:
If there is a value of $x$ with $f(x)=t>0$, then $f(t^{1/2})=t$ and $f(-t^{1/2})=t$
Shouldn't you prove explicitly why these values exist, and why there are no values of $x$ with $f(x)<0$.

I don't remember why but I think I could say that... I will look why

First of all welcome to the forum $LordWings$
Yes I agree with you and such $x$ exists
Take $x=0$ in the original equation then $f(-y^2)=f(f(y))=f(f(-y))=f(y)^2=f(-y)^2 \geq 0$ so for all $x$ then $f(x)\geq 0$

Thank you, AYMANE!
OK, now we know $f(x) \geq 0$. That answers the second part of my question.
But, is that enough? Your solution assumes that for every $t=y^2>0$ there's a value of $x$ with $f(x)=t$.

Thanks AYMANE ^^

[quote="lordWings"]Your solution assumes that for every $t=y^2>0$ there's a value of $x$ with $f(x)=t$.[/quote]
I don't think that assume that... It only assume that there is value such that... I don't think we need to prove that the function is surjectiv on $\mathbb{R^+}$ but we can certainly do it !

That's it, you should prove that the function is surjective on $\mathbb{R^+}$.
You have only proved: If there exists a value of $x$ with $f(x)=y^2$, then $f(y)=y^2$.
You still should prove that it exists. The conclusion can't prove the premise.

Sorry :?, I'm usually wrong about these things.  :blush:
I just feel uncomfortable with the "take $f(x)=y^2$", wich is the key to your reasoning.
If you can take any value of $y$, you're excluding there all the functions that aren't surjective on $\mathbb{R^+}$  (for instance, the bounded ones).
And when you say that, the only things you know about $f(x)$ is that $f(0)=0$ and $f(f(x))=f(x)^2$.
Maybe with $f(f(x))=f(x)^2$, we can prove that the function is surjective on $\mathbb{R^+}$ (or continuous and not bounded), but until we do, I think that the reasoning is not complete.
Am I too complicated? :blush:

What you cannot do with using $f(x)=y^2$ is to "change" the $y$ because you had "fixed" it. I cannot explain more or better :(

hi,

here is my solution :

lets assume that : $f(x) = u$

$y=0 \Rightarrow f(u) = u^2 + f(f(0)) (1)$

$x=y=0 \Rightarrow f(0) = 0 (2)$

$(1),(2) \Rightarrow f(x) = x^2$

 :) .

You clearly have $f(u) = u^2$ with $u=f(x)$.
To say $f(x)=x^2$ you must show that you can obtain all the $x$ with $f(u)$ -> you must prove that $f(u)$ is surjectiv  :D .

P.S Are you the same toetoe how is on the marocan forum "amateur de math" ?

[quote="AYMANE"]$f(x)=y^2$ is a just especial case like $x=y=0$ or $x=y$[/quote]
$f(x)=y^2$ is [b]not[/b] just a special case like $x=y=0$ or $x=y$.

AYMANE has proved that $f(x) \geq 0$ for every $x$. Then, there will always will be a value of $y$ with $f(x)=y^2$ for every $x$. In this case you can use $f(x)=y^2$ as a special case to prove things about $x$ and they will be true for every $x$.

The inverse is not true. We still don't know anything about the range of the function $f$, so there may be some values of $y$ that can't be reached with $f(x)=y^2$ (simply, such $x$ may not exist; for instance, $f(x)$ could be bounded or discontinuous). You can't use $f(x)=y^2$ as a special case to prove things about $y$, thinking that they will be true for every $y$. They will only be true if you prove, [b]beforehand[/b], that $y^2$ belongs to the range of the function $f$ (if you prove it later, it won't mean nothing, it just won't disprove your assumption).

BogG noticed it looking at toetoe's solution. Exactly the same thing happens with toetoe's special case $u=f(x)$: it can only be used to prove things about $u$ if you prove, beforehand,  that $u$ belongs to the range of the function $f$.

So, in general, you can use the special case $s=h(t)$:
-  to prove things about $s$, only if you prove beforehand that $s$ belongs to the domain of the function $h$.
-  to prove things about $t$, only if you prove beforehand that $t$ belongs to the range of the function $h$ (this is more difficult, since we usually know the domain and codomain, but not the range).
You can use special cases like $x=0$ or $x=k$ or $x=y$ to prove things about $x$ or $y$ without worries, because the domain and range of the functions $h(y)=0$, $h(y)=k$ and $h(y)=y$ are both $\mathbb{R}$.
But if you use special cases like $x=f(y)$ or $x=-f(f(y))$ where $f$ is the function that you are trying to find out (remember, we're dealing with an equation), you should be more careful with its domain and range.

Maybe... I'm not yet really sure... but I have another solution !

$x=y=0$ $\rightarrow$ $f(0)=0$
$y=0$ $\rightarrow$ $f(f(x))=f(x)^2$
$f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=f(f(x)-f(y)^2)^2$

$f(z)=z^2$ for all $z \in S = \{f(u)-f(v)^2 | u,v \in \mathbb{R}\}$.

Choose $a$ such that $f(a) \not= 0$ and for exemple with $f(f(x))=f(x)^2$ easy show that $f(a) >0$

Then put $x=a$ $\rightarrow f(f(a)-y^2) - f(y)^2 = f(a)(f(a) - 2y^2)$.

On the right we can have all the value $<0$ then on the left too $\rightarrow$ $\mathbb{R^-} \subset \mathbf{R}$ and then $f(z)= z^2$ for all $z\le 0$

$\rightarrow$ all the value can be reached in $\mathbb{R^+}$ by the non-negativ numbers $\rightarrow$ [b]surjectivity [/b]of $f$ in $\mathbb{R^+}$

And now I think it's more clear  :)

[quote="BogG"]$f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=f(f(x)-f(y)^2)^2$[/quote]
What happened here?
It seems a typo and BogG wants to say
$f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2$
but I think that's not right. I get
$f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2+(f(y)^2-y^4)$.
And then the conclusion wouldn't hold. :(
I hope I'm wrong, or we're back agin.

[quote="toetoe"][quote="BogG"]
P.S Are you the same toetoe how is on the marocan forum "amateur de math" ?
[/quote]
yes,i am  :D .[/quote]

Can you give me the link to this forum :)

Please, discuss this kind of things via PM!

We're not advacing a lot here. Let's get back to the topic.
[quote]Find all the functions $f : \mathbb{R} \to \mathbb{R}$ satisfying for all $x,y \in \mathbb{R}$ $f(f(x)-y^2) = f(x)^2 - 2f(x)y^2 + f(f(y))$.[/quote]
AYMANE has proved so far:
1) If we take $x=y=0$, then $f(0)=0$.
2) If we take $y=0$, then vor every $x$ we have $f(f(x))=f(x)^2$.
3) If we take $x=0$, then for every negative value $t=-y^2$ we have $f(t)\geq 0$.
4) If we take $y^2=f(x)$ ($y^2\in range(f)$), then $f(y)=y^2$.
5) $f(x)=x^2$ is one possible solution of the equation.
More straightforwardly, toetoe proved that:
6) If we take $u=f(x)$ ($u\in range(f)$), then $f(u)=u^2$.
We still don't know anything about the range of $f$ in general. BogG and I have been trying to prove that $range(f)=\mathbb{R^+}$, and that would make $f(x)=x^2$ the only solution.

In fact, nobody noticed the next fact (compatible with facts 1 to 6):
7) $f(x)=0$ is one possible solution of the equation.
To prove a point, in this case, $range(f)=\{0\}$, so we can't assume $range(f)=\mathbb{R^+}$ in general.
Are there any other solutions?
Perhaps we can begin with: Let's suppose there's a $x$ with $f(x)=a\neq 0$...

[quote="lordWings"]What happened here?
It seems a typo and BogG wants to say
$f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2$
but I think that's not right. I get
$f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2+(f(y)^2-y^4)$.
And then the conclusion wouldn't hold. :(
I hope I'm wrong, or we're back agin.[/quote]

Sorry I forgot to say something. Make this change $y$ become $f(y)$

then you have $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ who become $f(f(x) - f(y)^2)=f(x)^2 - 2f(x)f(y)^2+f(f(y))^2$
and with using $f(f(y))=f(y)^2$ you find $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2$  :D

Yes and of cours $f(x) = 0$ is also a solution

I think I finally understand BogG's reasoning, but it's really convoluted!
(I'll change the name of some variables to avoid confusions.)

With the last post, he's proved (taking $x=u$, $y=f(v)$ inside the equation, and using fact 2) that:
8) If there are $u,v\in\mathbb{R}$ with $s=f(u)-f(v)^2$, then $f(s)=s^2$.

Then we go back to his previous post:
We know from fact 7 that $f(x)=0$ is one possible solution.
Any other solution will have at least one $x_0$ with $f(x_0)=z\neq 0$.
If we take $x=z$ inside the equation (along with facts 2 and 6), we get $f(z^2-y^2)=z^4-2z^2y^2+f(y)^2$.
This can be rewritten in this way: $f(z^2-y^2)-f(y)^2=z^2(z^2-2y^2)$.
We must remember that $z$ is fixed, but $y$ can be any real number.
So, there are $u=z^2-y^2$ and $v=y$ with $s=f(u)-f(v)^2=(u+v^2)(u-v^2)=u^2-v^4\le z^4$.
Using fact 8 with this last result, we get:
9) If there is one $x_0$ with $f(x_0)=z\neq 0$, then for every $s\le z^4$, $f(s)=s^2$.
This fact includes that for every $s\in\mathbb{R^-}$, $f(s)=s^2$. In this case, $f$ would be surjective on $\mathbb{R^+}$, because for every $k\in\mathbb{R^+}$, there is $s=-k^{1/2}\in\mathbb{R^-}$ with $f(s)=k$, and then, with case 4, $f(y)=y^2$ for every $y$.

That's it: BogG has proved that, if the solution is not the trivial $f(y)=0$, then it must be $f(y)=y^2$ for every $y$.

There's one question left.
At the Swiss IMO where this problem was proposed, were they supposed to solve it like BogG? Is it the most simple way to solve it?
I think I would have never solved it that way in a limited time. Step 8 seems difficult and pointless until it's been used (deus ex machina) at step 9. Kudos to BogG.
Of course, without these steps, we have already proved that $f(y)=0$ and $f(y)=y^2$ are solutions (maybe worthy of some points), but we haven't proved that these are the only solutions.

The last solution is not mine.. I only have copy it from the Swiss solution...

[quote="BogG"]Find all the functions $f : \mathbb{R} \to \mathbb{R}$ satisfying for all $x,y \in \mathbb{R}$ $f(f(x)-y^2) = f(x)^2 - 2f(x)y^2 + f(f(y))$.[/quote]
It's easy to show that, $f(f(x)) =  f(x)^2$
Let, $S=\{x\ |\ f(x)=x^2\ \}$, then, $f(x)\in S$ for all $x$.
Now, since $f(f(f(y)))=f(f(y))^2=f(y)^4$, $y\longrightarrow f(y)$ in given equation implies,
$f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2$
So, $f(x)-f(y)^2 \in S$ for all $x,\ y$.
Now, from the given equation and $f(f(y))=f(y)^2$, we get, $f(f(x)-y^2)-f(y)^2=f(x)^2-2y^2f(x)$
So, $\{f(x)^2-2y^2f(x)\ |\ x,\ y\ \in\ \mathbb{R}\ \} = (-\infty, f(x)^2] \subseteq S$ for all $x$
So, $f(u)=u^2$ for all $u\leq f(x)^2$.
Or, $\lim_{u\longrightarrow -\infty} f(u) = \infty$
Hence, $\lim_{u\longrightarrow -\infty} (-\infty, f(u)^2] = \mathbb{R} \subseteq S$

So, $S=\mathbb{R}$, Done! :)

Let $P(x,y)$ denote the given assertion. Since $f\equiv 0$ works, assume $f(u)\neq 0$ for some $u.$ Clearly $f(0)=0$ and $f(f(x))=f(x)^2.$

Note that $f(u)^4-2f(u)^2x^2$ can be any $z \leq 0$, so $P(f(u),x)$ implies $z=f(v)-f(x)^2$, for $v=f(u)^2-x^2.$ Then $P(v,f(x))$ implies $f(z)=z^2$ for all $z\leq 0.$ Then $P(z,y)$ shows $f(x)\equiv x^2;$ which works.


invariant
combinatorics unsolved
combinatorics

Numbers $ 1, 2, . . . , 2009^2$ are written in the cells of a $ 2009\times 2009$ table. It is allowed to apply the following transformations: exchange places of any two rows or any two columns, or reverse a row or column. (When a row or column is reversed, the first and last entry exchange their positions, so do the second and second last, etc.) Is it possible that, after finitely many such transformations, arbitrary two numbers exchange their positions and no other number changes its position?

Answer to the first post: no. Suppose the numbers are in the trivial order ($ 1, 2, ..., 2008$ and $ 2009$ are in the first line, $ 2010, 2011, ..., 4017$ and $ 4018$ are in the second line, ... and $ 2009^2$ is in the last line). If two numbers are in the same column in the start, then they are always in the same column (invariant!). Suppose $ 1$ and $ 2009^2$ have changed and all the other numbers have not. Then $ 2009$ is fixed. But now $ 1$ is not in the same row that $ 2009$. Contradiction!

Answer to second post: yes.

[quote="feliz"] If two numbers are in the same column in the start, then they are always in the same column (invariant!). [/quote]

I don't think so...


In several variants of chess, a camel jumps in the opposite corner of a $ 2$ by $ 4$ rectangle. Prove that on an arbitrarily large chessboard a camel can return to its original position only after an even number of jumps.

[hide]Color the square the camel is on black, and let that square be 0,0. Let every square with coordinates (x,y) such that x,y=0 mod 2 be black. The camel's jumps always take it to a different color, since each move shifts its position by 3 and 1 in either direction, which are 1 mod 2. Thus the coordinate of the camel does not change mod 2 after an even number of jumps. An odd number of jumps would land the camel on an uncolored square, and an even number would land it on a colored square. The starting square is colored, so we are done.[/hide]


Can you write any any conditional as an expression using only absolute values?
For example
a>b would be sometihng like
(abs(a-b)/(a-b)+1)/2

Your a>b example doesn't quite work, because it will error if a=b.

What exactly do you mean by a "conditional"? What about isprime(p)? That would probably not be expressable.


In a right triangle with integer sides the radius of the inscribed circle is 12. Compute the longest possible hypotenuse.

[hide]
The solution I tried first worked very straight-forwardly, but it raises even more questions:

The shorter leg has to be longer than 24 (but should be minimized to maximize the hypotenuse); a length of 25 gives us an integer length for both the other leg and for the hypotenuse (maximized at 313, the solution), but interestingly enough, as long as the length of the shorter leg leads to a length of the longer leg that is also an integer, with $ a$ (length of shorter leg) ranging from 25->28, the hypotenuse has integer length. I figured out that the longer leg has length $ 24*\frac {a \minus{} 12}{a \minus{} 24}$, which implies that the shortest valid ($ > 24$) integer length for the shorter leg that gives a non-integral length for the longer leg is 29, but why the hypotenuse should be an integer for all $ a$ from 25 to 28 eludes me for the moment, although I did find the expression for the length of the hypotenuse in terms of the shorter leg. $ (\sqrt {\frac {a^4 \minus{} 48a^3 \plus{} 1152a^2 \minus{} 13824a \plus{} 82944}{a^2 \minus{} 48a \plus{} 576}})$

Any thoughts?
[/hide]

Oh, you guys are doing it differently than I did, my method:

[hide]
$ A \equal{} A\\
rs \equal{} bh/2\\
rp \equal{} bh\\
12(a\plus{}b\plus{}c) \equal{} ab\\
a \plus{} b \plus{} \sqrt{a^2\plus{}b^2} \equal{} \frac{ab}{12}$
Finding $ b$ for a constant $ a$:
$ (\frac{a}{12} \minus{} 1)b \minus{} a \equal{} \sqrt{a^2\plus{}b^2}\\
(\frac{a\minus{}12}{12})b \minus{} a \equal{} \sqrt{a^2\plus{}b^2}$
Squaring:
$ (\frac{a^2\minus{}24a\plus{}144}{144})b^2 \minus{} (\frac{a\minus{}12}{6})ab \plus{} a^2 \equal{} a^2 \plus{} b^2\\
(\frac{a^2\minus{}24a}{144})b^2 \equal{} (\frac{a\minus{}12}{6})ab$
And since $ b>0$, we can divide by $ b$ and eliminate the extraneous zero root.

$ (\frac{a^2\minus{}24a}{144})b \equal{} (\frac{a\minus{}12}{6})a\\
b \equal{} (\frac{a\minus{}12}{6})a * (\frac{144}{a^2\minus{}24a})\\
b \equal{} 24*(\frac{a\minus{}12}{a\minus{}24})$

[/hide]

The rest of my thoughts are in the first post.

[hide]Since the radius of a circle inscribed in a right triangle is (a+b+sqrt(a^2+b^2))/2, we have that the sides sum to 24.  So, a+b-24 = sqrt(a^2+b^2).  Square both sides to get a^2+b^2-48a-48b+576=a^2+b^2, so 48a+48b=576+2ab, or ab-24a-24b+288=0.  Add 288 to both sides to get ab-24a-24b+576=288.  Then factor to (a-24)(b-24)=288.  To maximize a^2+b^2, we want to minimize one of the variables.  So, let a=25, since a-24 must be an integral factor of 288 and a is positive.  Then, b-24=288, so b=312.  Then the hypotenuse is sqrt(a^2+b^2)=sqrt(25^2+312^2)=313.  *this last part is easily verified since it's just the primitive pythag triplet of the form n (n^2-1)/2, (n^2+1)/2 for n=25*[/hide]

edit: didn't read the hidden on the first post so i didn't see what you were looking for, but there's a solution anyway.


I need some out of the box physics questions which have really simple answers in the end. Basically, I want questions which involve a high degree of lateral thinking and really are a test of true intelligence! They may involve higher math and concepts - it doesn't matter. [size=150]Just [b]OUT OF THE BOX!!![/b][/size]

Any suggestions!! Please post if you know any good sites/papers etc!!!

... I don't really know what does "out of the box" mean but we have some physics competitions for high school people here in Slovakia and Czech Republic and the Czech one even has English translations so feel free to check out

http://fykos.mff.cuni.cz/en/

they're really nice... well, at least some of them :) .

Is out of the box the same as non-routine?

Sort of. Those requirong considerable lateral thinking is what I would call them. Lateral thinking means taking a no. of things in mind and logically answering!!!

No Suggestions??? Please HELPPPPPPP!!!!!!!!!!! :(


Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} abc$. Prove that
  $ (\sqrt[3]{(a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)} \minus{} 1)^3 \geq (a^2 \plus{} 2bc \minus{} 1)(b^2 \plus{} 2ca \minus{} 1)(c^2 \plus{} 2ca \minus{} 1) \geq (ab \plus{} bc \plus{} ca \minus{} 1)^3$

[quote="2424"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} abc$. Prove that
  $ (\sqrt [3]{(a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)} \minus{} 1)^3 \geq (a^2 \plus{} 2bc \minus{} 1)(b^2 \plus{} 2ca \minus{} 1)(c^2 \plus{} 2ca \minus{} 1) \geq (ab \plus{} bc \plus{} ca \minus{} 1)^3$[/quote]
It's not easy and interesting! :)
No one like it? :lol:

[quote="quykhtn-qa1"][quote="2424"]Let $ a,b,c > 0$ such that $ a + b + c = abc$. Prove that
  $ (\sqrt [3]{(a^2 + 2bc)(b^2 + 2ca)(c^2 + 2ab)} - 1)^3 \geq (a^2 + 2bc - 1)(b^2 + 2ca - 1)(c^2 + 2ca - 1) \geq (ab + bc + ca - 1)^3$[/quote]
First,we prove: $ \sqrt {((a^2 + bc)(b^2 + ca)(c^2 + ab))^3} - 1)^3 \ge (a^2 + 2bc - 1)(b^2 + 2ca - 1)(c^2 + 2ca - 1)$
Put $ a^2 + 2bc = x^3;b^2 + 2ca = y^3;c^2 + 2ab = z^3$,we only need to prove:
$ (xyz - 1)^3 \ge (x^3 - 1)(y^3 - 1)(z^3 - 1)$
${ < = > \sum \ (y - z)^2(x^2(xy + yz + zx) - x - y - z)} \ge 0$
Which 's obvious true becase : $ x > 1;xy + yz + zx \ge \sqrt {3xyz(x + y + z)}$; $ x + y + z \le \sqrt [3]{9a^2b^2c^2} < 3\sqrt [3]{a^2b^2c^2} \le xyz$
then,we have Q.E.D
Second,we prove: $ (a^2 + 2bc - 1)(b^2 + 2ca - 1)(c^2 + 2ab - 1) \ge (ab + bc + ca - 1)^3$
Assume $ a \ge b \ge c$,we have $ ab \ge ac \ge bc > 1$
and $ c^2 + 2ab - 1 \ge ab + bc + ca - 1$
so,we only need to prove that:
$ (a^2 + 2bc - 1)(b^2 + 2ca - 1) \ge (ab + bc + ca - 1)^2$
$ < = > (a - b)^2(2(a + b)c - c^2 - 1) \ge 0$
Which 's obvious true[/quote]
your proof is very nice,quykhtn-qa1! :)
And we also have:
$ 8(a^2+5bc-3)(b^2+5ca-3)(c^2+5ab-3)\geq27(ab+bc+ca-1)^3$ with $ a,b,c>0$ such that $ a+b+c=abc$. :)


$ a_{i}$ ha adady tabiei hastand betorike baraye $ i \equal{} 1,2,...,n$ darim $ 1 \leq a_{i} \leq i$ va hamchenin darim $ a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \equal{} 2k$ .sabet konid mitavan $ a_{i}$ ha ra be do daste tory tagsim konim ke majmoe do daste ba yekdigar barabar bashad ?

hokm ra ba esteghra rooye $ n$ sabet mikonim
Baraye $ n\equal{}2$ tanha halate momken $ a_{1}\equal{}a_{2}\equal{}1$ mibashad banabarin badihist ke baraye $ n\equal{}2$ hokm sadegh ast.
Farz konid baraye $ n\equal{}k$ hokm sadegh bashad hal baraye $ n\equal{}k\plus{}1$ sabet mikonim.
2halat darim
$ 1)$ agar $ a_{2}\equal{}1$ bashad :$ a_{1}\equal{}1$va$ a_{2}$ ra dar 2 goroohe mokhtalef gharar midahim hal banabar farze esteghra mitavan $ k\minus{}1$ adade baghimande ke majmooeshan zoj ast ra be $ 2$ gorooh taghsim kard.banabarin hokm sadegh ast.
$ 2)$ agar$ a_{2}\equal{}2$ bashad farz konid majmooe $ a_{i}$ha $ 2t$ bashad darin soorat bedune darnazar gereftane $ a_{2}$ tebghe farz $ k$ adade baghimande ra be $ 2$ daste ba majmooe $ t\minus{}1$ taghsim mikonim sepas dar dastei ke dar an $ a_{1}\equal{}1$ gharar darad $ a_{2}$ ra gharar dade va $ a_{1}$ ra dar dasteye digar gharar midahim,dar insoorat niz hokm sadegh ast.

That was a very nice problem . I enjoyed it !!!!!!!!

ممنون از سؤال بسیار زیباتون


function
algebra
polynomial
inequalities
logarithms
domain
real analysis

Is it known whether there exists a real number $P$ such that for any positive reals $a, b$, the function $f_p(a,b) = \left(\frac{a^p+b^p}{2}\right)^{\frac{1}{p}}$ is such that $\frac{\partial^2}{\partial p^2}\left[f_p(a,b)\right] \le 0$ (*) for all $p \ge P$?

Normalizing, it suffices to consider only the case $0 < a \le 1 = b$, although even then a straight-forward computation yields a terrible forest of polynomials, logs, and exponentials.

I am hoping that $P = 1$ is satisfactory.

Edit- Unfortunately, after carefully recomputing everything, I think I probably made an error in my previous method for showing the case $p=1$. :(

Sigh. (See first post.)

The core idea was to use, for $0 < \gamma \le 1$, the bounds on the natural log: $\frac{\gamma - 1}{\sqrt{\gamma}} \le \ln(\gamma) \le \frac{2(\gamma - 1)}{\gamma + 1}$ (derived from the usual AM-logarithmic-GM inequality $\frac{x+y}{2} \ge \frac{x-y}{\ln(x) - \ln(y)} \ge \sqrt{xy}$.) fedja, I will try to verify your equivalent statement.  (Can you explain how that is equivalent? It's much cleaner than what I obtained.)

[quote="fedja"]If you've done the case $p=1$, you've done all the work! The reason is that $f_{\theta p}(a,b)=f_{\theta}(a^p,b^p)^{1/p}$, so
\[ \begin{aligned} p^2\frac{\partial^2}{\partial q^2}f_q(a,b)\bigr|_{q=p}&= \frac{\partial^2}{\partial \theta^2} [f_\theta(a^p,b^p)^{1/p}]\bigr|_{\theta=1} \\ &=\frac 1p \left(\frac 1p-1\right) f_\theta(a^p,b^p)^{\frac 1p-2}\left[\frac{\partial}{\partial \theta} [f_\theta(a^p,b^p)\right]^2\Bigr|_{\theta=1}+\frac 1p f_\theta(a^p,b^p)^{\frac 1p-1}\frac{\partial^2}{\partial \theta^2} [f_\theta(a^p,b^p)]\bigr|_{\theta=1}\,. \end{aligned}  \]
Now it remains to note that each term is non-positive if $p\ge 1$.
By the way, I still wonder how you've done the case $p=1$. It is equivalent to the inequality
\[ \ln\frac{1+x}{2}+\frac 12\left[\ln\frac{1+x}{2}\right]^2+x \ln\frac{1+x}{2x}+\frac x2\left[\ln\frac{1+x}{2x}\right]^2\le 0  \]
for all $x\ge 0$. I managed to get a computer-assisted proof of it showing that $x=1$ is a root of multiplicity $4$ (this can be done by hand) and that there are no other roots in some domain on the complex plane containing $[0,1]$ (this I did using the argument principle letting the computer to evaluate the corresponding contour integral). Do you have a better approach?[/quote]

I have found a solution using this result!

It's pretty clear $x=0$ is not a root of the LHS since $2 < e^2$. In particular, the LHS is strictly negative in that case. Now consider $P(x) = \ln\frac{1+x}{2}+\frac 12\left[\ln\frac{1+x}{2}\right]^2+x \ln\frac{1+x}{2x}+\frac x2\left[\ln\frac{1+x}{2x}\right]^2$ over the positive reals. We know that $P(x)$ has $x=1$ a root of multiplicity 4. Suppose $x_0 > 0$ and $x_0 \ne 1$ is also a root of $P(x)$. Then by Rolle's theorem there exists a positive $x_1$ between $x_0$ and 1 for which $P'(x) = 0$. Also note that $x=1$ is a root of multiplicity 3 of $P'(x)$. Then applying Rolle's theorem a second time shows that there exists a positive real $x_2$ between $x_1$ and 1 for which $P''(x)$ is zero. But $P''(x)$ has a $x=1$ as a root of multiplicity 2. In fact, we compute

$P''(x) = -\frac{x\ln\left(\frac{1+x}{2}\right)+\ln\left(\frac{1+x}{2x}\right)}{x(x+1)^2}$

We have that $Q(x) : = x(x+1)^2 P''(x)$ has $x_2$ as a root, and $1$ as a root of multiplicity 2. Applying the same strategy once more, $R(x) = Q'(x)$ has a root $x_3$ between $x_2$ and 1, and 1 as a root of multiplicity 1. We compute $R(x) = -1 + \frac{1}{x} + \ln(2) - \ln(1+x)$. But clearly $R(x)$ is strictly monotonic and decreasing. Thus, 1 is the unique root of $R(x)$, and it follows that $x_3$ is not a root of $R(x)$. Contradiction. Unwinding, our assumption that another root of $P$ exists must be false; thus, $x=1$ is the unique root of $P(x)$. Because it is easily checked that $P(0)<0$ and that $x=1$ has multiplicity 4, we have $P(x) \le 0$ for all $x \ge 0$, as desired. $\Box$

I am very much interested in seeing the work that shows this is equivalent to the case $p=1$!

[quote="Mildorf"]I am very much interested in seeing the work that shows this is equivalent to the case $p=1$![/quote] Just differentiate $\left(\frac{1+x^p}2\right)^{1/p}$ twice with respect to $p$ and plug in $p=1$. After some algebraic manipulations you'll get the expression I wrote (I hope I haven't made a silly mistake somewhere...).


Find the all integer solution of $ 2n^3\minus{}m^3\equal{}mn^2\plus{}11$

[quote="puujee"]Find the all integer solution of $ 2n^3 \minus{} m^3 \equal{} mn^2 \plus{} 11$[/quote]
Rewrite it as
$ (n\minus{}m)(2n^2\plus{}nm\plus{}m^2)\equal{}11$
But $ 2n^2\plus{}nm\plus{}m^2\geq0$ so
$ n\minus{}m\equal{}1$ and $ 2n^2\plus{}nm\plus{}m^2\equal{}11$ which gives $ n\equal{}2$ and $ m\equal{}1$
or $ n\minus{}m\equal{}11$ and $ 2n^2\plus{}nm\plus{}m^2\equal{}1$ which has no solutions...
So the only solutions are $ (n,m)\equal{}(2,1)$


$p>5$ is a prime, prove or disprove; for any integer $a$ there exists integers $b, c$ such that $a+b^{2}=c^{2}(mod p)$
edit: $b, c$ is not divisible by $p$

Generalisation: If $n$ is odd and $a,b,c$ are integers with $\gcd(n,a)=\gcd(n,b)=1$, then there are integers $x,y$ with $ax^{2}+by^{2}\equiv c \mod n$.

It is true for all p. If $p>2$ works $c=\frac{(a+1)(p+1)}{2},b=\frac{(a-1)(p+1)}{2}$. If p=2 it is obviosly.

zetax your generalisation was very useful thanks if you know the proof i will be pleased to know that
also i forgot to say that b and c is not divisible by p that is why i said $p>5$

we take N odd integre

if $a=2s+1\ (odd)$
$a\equiv 2s+1[N]\equiv ((s+1)-s)((s+1)+s)[N]\equiv (s+1)^{2}-s^{2}[N]$
so $a+s^{2}\equiv (s+1)^{2}[p]$
if $a=2s\ (even)$
$a\equiv a+N[N]$ and we have $a+N=2k+1\ (odd)$
so $a+k^{2}\equiv (k+1)^{2}[N]$

then $\forall N\ odd\ integre$;$\forall a$,there exists integers $b$ and $c$, such that $a+b^{2}\equiv c^{2}[N]$


inequalities
geometry
trigonometry
angle bisector
geometry solved

In a triangle $ABC$ let $D$ be the intersection of the angle bisector of $\gamma$, angle at $C$, with the side $AB.$ And let $F$ be the area of the triangle $ABC.$ Prove the following inequality:

\[2 \cdot \ F \cdot \left( \frac{1}{AD} -\frac{1}{BD} \right) \leq AB.\]

If the sides are $a, b, c$ as usual, then noting that $AD=\frac{bc}{a+b}$ and $BD=\frac{ac}{a+b}$, then we need $ab \sin  \gamma \left(\frac{a+b}{bc}-\frac{a+b}{ac}\right) \leq c$, or $(a+b)(a-b)\sin \gamma \leq c^2 = a^2 + b^2 - 2ab \cos \gamma$.  If $\cos \gamma \leq 0$ then this is obvious from $a^2 \geq a^2 \sin \gamma$. Otherwise, it follows from $b^2(1+\sin \gamma)+\frac{a^2 \cos ^2 \gamma}{1+\sin \gamma} \geq 2ab \cos \gamma$ and $\sin \gamma + \frac{\cos ^2 \gamma}{1+\sin \gamma} = 1$.

[color=red][Moderator edit: Typo corrected (- mistyped for +).][/color]

Nice solution, Fiachra. Here is a simpler way to prove that $\left(a+b\right)\left(a-b\right)\sin C\leq c^2$:

By the Mollweide formulas, we have $\frac{a+b}{c}=\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}$ and $\frac{a-b}{c}=\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}}$, and thus

$\frac{\left(a+b\right)\left(a-b\right)}{c^2}=\frac{a+b}{c}\cdot\frac{a-b}{c}=\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}\cdot\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}}=\frac{\sin\frac{A-B}{2}\cos\frac{A-B}{2}}{\sin\frac{C}{2}\cos\frac{C}{2}}=\frac{2\sin\frac{A-B}{2}\cos\frac{A-B}{2}}{2\sin\frac{C}{2}\cos\frac{C}{2}}=\frac{\sin\left(A-B\right)}{\sin C}$,

where we have applied the formula sin (2x) = 2 sin x cos x. Now, trivially, $\sin\left(A-B\right)\leq 1$ and sin C > 0; thus,

$\frac{\left(a+b\right)\left(a-b\right)}{c^2}=\frac{\sin\left(A-B\right)}{\sin C}\leq\frac{1}{\sin C}$.

Multiplication with $c^2$ yields $\left(a+b\right)\left(a-b\right)\leq\frac{c^2}{\sin C}$, and thus $\left(a+b\right)\left(a-b\right)\sin C\leq c^2$, completing the proof.

  darij


Find the greatest and smallest value of $ f(x,y)\equal{}y\minus{}2x$, if $ x,y$ are distinct nonnegative real numbers with $ \frac{x^2\plus{}y^2}{x\plus{}y} \le 4.$

It is equavalent to $ (x\minus{}2)^2\plus{}(y\minus{}2)^2\le 8,x,y\ge 0$.
Greatest mean, when $ x\equal{}0,y\equal{}4, f\equal{}4$, smallest mean, when, $ x\equal{}2\minus{}2\sqrt 2sint, y\equal{}2\plus{}2\sqrt 2 cost, f\equal{}\minus{}2\plus{}2\sqrt 2(2sint\plus{}coat)\equal{}2\sqrt{10}\minus{}2.$


Find the $ x$-intercept point of the line $ 3x\plus{}5y\equal{}20$. Provide your answer as an ordered pair.  Express the x and y coordinates as common fractions as needed.

The $ x$-intercept is simply where $ y\equal{}0$.  So we get $ 3x\equal{}20\Rightarrow x\equal{}\frac{20}{3}$.  So the ordered pair we desire is $ \left(\frac{20}{3},0\right)$.


quadratics
analytic geometry
graphing lines
slope
conics
parabola
parameterization

For all your why asking needs:

Why are expressions with a degree of two known as quadratics?

Why is m used to designate slope?

http://mathforum.org/dr.math/faq/faq.terms.html for the slope question.

Can you explain what exactly the [b]parameter[/b] is?

The root of "quadratic" is the greek word for four; in this case meaning "square". Thus a quadratic equation is one in which a variable is squared (and nothing more complicated happens). The name has nothing to do with the geometry of a parabola.

The name from before analytic geometry is "conic"; calling such an object quadratic is a part of the language of analytic geometry.

[quote="Farazy"][quote="Magnara"]I would guess that a quadratic is named such because it parabola is determined by 4 points.  Remember that they were naming these things before analytic geometry.  And it really is 4 points, not 3.  It's only 3 if you're given that it opens either "up" or "down."   Consider a parabola that opens diagonally.[/quote]

Why would you need to use 4 points when it opens diagonally? and what is the 4th point?[/quote]

What I meant was when you don't know which way it opens.  If I gave you three points, infinitely many parabolas would pass through those points, each opening in a different direction.

But, I was wrong anyway.


number theory
greatest common divisor
modular arithmetic

[hide="hint which is nearly solution"]Consider any prime which divides exactly one of b or c[/hide]

Using digger's hint:

[hide]
Obviously no prime may divide both $b$ and $c$.  Say that a prime $p$ divides $b$ but not $c$.  It clearly also divides $b^{2}$ and $bc$.  So $c^{2}-b^{2}\equiv c^{2}\pmod p$.  But $p$ does not divide $c$, so $p$ does not divide $c^{2}$, thus $p$ does not divide $c^{2}-b^{2}$.

Say that a prime $q$ divides $c$ but not $b$.  Clearly $q$ does not divide $b^{2}$.  However, $q$ does divide $c^{2}$ and $bc$.  So $c^{2}-b^{2}\equiv-b^{2}\pmod q$.  But $q$ does not divide $b^{2}$, so $q$ does not divide $c^{2}-b^{2}$.
[/hide]


Show that $(p_{1}-1)(p_{2}-1)...(p_{n}-1)+(p_{1}+1)(p_{2}+1)...(p_{n}+1)>2p_{1}p_{2}...p_{n}$, where the $p_{i}$ are primes and $n \geq 2$.

I'm interested in what methods there are...I did it using convexity  :roll:

What's the pattern in the second term?  Is it $(p_{i}-1)$ or $(p_{i}-i)$?

[hide="Directly"] It's actually fairly obvious just expanding.

Of course we have the two terms $p_{1}p_{2}... p_{n}$.  Every other term takes the form of some product of primes.  It will be positive in the second expansion, and in the first expansion it will alternate signs depending on which primes are used.

Any terms that are negative in the first expansion are cancelled out in the second, and any terms that are positive are added to that corresponding term in the second, so the sum of the other terms is positive.

I suspect $p_{i}$ can be replaced by any set of natural numbers. [/hide]


function
Euler
number theory unsolved
number theory

Given that Moebius function $ \mu(n)$ is defined in the regular way, prove that $ \sum_{d^2|n}\mu(d)\equal{}\mu^2(n)$ and furthermore,

$ \sum_{d^k|n}\mu(d)$ is $ 0$ if $ m^k|n$ for some $ m>1$, and is $ 1$ otherwise.

Let $ S$ denote the set of primes $ p$ dividing $ n$ such that $ p^k|n$.
The condition $ m^k|n$ fore some $ m > 1$ is equivalent to $ S \neq \emptyset$.
The sum is equivalent, by the definition of the Moebious function (we sum only over products of distinct primes, such that their k'th power divides $ n$, i.e., over all the $ d$'s that are products of elements of $ S$):
$ \sum_{d^k|n}\mu(d) \equal{} \sum_{A \subseteq S}( \minus{} 1)^{|A|} \equal{} \sum_{i \equal{} 0}^{|S|} ( \minus{} 1)^{i} \binom{|S|}{i}$.
If indeed $ S \neq \emptyset$, the last sum equals $ (1 \minus{} 1)^{|A|} \equal{} 0$ (by the binomial theorem).
Otherwise, the only term in the sum is $ ( \minus{} 1)^{0} \equal{} 1$ (the term that corresponds to the empty set).
The case $ k \equal{} 2$ follows from the example, because $ \mu(n)^2 \equal{} 1$ if $ m^2|n$ for some $ m > 1$.

I suspect that there's an easier proof...

EDIT: another proof:
For $ n \equal{} \prod p_i^{a_i}$, define $ b_i \equal{} a_i$ if $ a_i \ge  k$, $ b_i \equal{} 0$ otherwise, and $ f_k(n) \equal{} \prod p_i^{b_i}$.
We get:
$ \sum_{d^k|n}\mu(d) \equal{} \sum_{d|f_k(n)}\mu(d)$

It is well-known that $ \sum_{d|n}\mu(d) \equal{} 0$ for all $ n > 1$, and equals 1 for $ n \equal{} 1$ (actually, this is equivalent to the first solution).  The result follows.

The arithmetic function that returns $ 0$ if $ m^k | n$ and $ 1$ otherwise has Euler product $ \prod_{p \in \mathbb{P}} \left( 1 \plus{} p^{ \minus{} s} \plus{} ... \plus{} p^{ \minus{} (k \minus{} 1)s} \right) \equal{} \prod \frac {1 \minus{} p^{ \minus{} ks}}{1 \minus{} p^{ \minus{} s}} \equal{} \frac {\zeta(s)}{\zeta(ks)}$, and the arithmetic function associated to $ \frac{1}{\zeta(ks)}$ is precisely $ \mu(\sqrt[k]{n})$ if $ n$ is a $ k^{th}$ power and $ 0$ otherwise.  The result follows.


calculus
integration
linear algebra
matrix
vector
linear algebra unsolved

$A \in M_n(\mathbb{Z})$, a symetric matrix with entries equal to $0$ or $1$.
Suppose that:
1/$Tr(A)=0$
2/$J$, the matrix with $1$ everywhere, belongs to the algebra generated by $(I,A,A^2,..,A^k,...)$

Prove that $max_{i} | \lambda_{1 \leq i \leq n} |$ is an integer. (the $\lambda_i$ 's are the eigenvalues)

So $A$ is a graph, and it's connected. The examples I know are strongly regular graphs; we should probably have $AJ=JA=kJ$ for some $k$. No proof yet, but this feels interesting.

Since $J$ is in the algebra generated by $A,$ we have that $A$ and $J$ commute. $AJ$ has constant columns, and $JA$ has constant rows. Since $AJ=JA$ and has both constant rows and constant columns, then it has constant elements, and is thus a multiple of $J$. Since each element of $AJ$ is the number of $1$'s in some row of $A,$ that multiple is an integer.

Thus $AJ=JA=mJ$ for some $m\in\mathbb{N}.$

If $v$ is the vector of all $1$'s, then $Av=mv,$ so $m$ is an eigenvalue.

But a simple norm calculation, using, say, the sup norm, shows that

$||Ax||_{\infty}\le m||x||_{\infty},$ where $||x||_{\infty}=\max_j|x_j|.$

The spectral radius must be less than or equal to any operator norm. This is a standard argument. In this case, suppose $x$ is an eigenvector of $A$ with $||x||_{\infty}=1.$

Then $|\lambda|=||\lambda x||_{\infty}=||Ax||_{\infty}\le m||x||_{\infty}= m.$

So $m$ is the spectral radius.

This argument never used the zero trace (the fact that $A$ is a graph.)

yes, you're right ...
this is because my friend built this little exercice in order to use some basic results on graph.
That is: if a graph is connex and $k$-regular, the spectral radius of the adjency matrix is $k$.
And a graph is $k$ regular connex iff $J$ is in the algebra generated by the adjency matrix.

but Kent ,where you use the fact that Tr(A)=0 ? and alekk what about this notion of Graph and regularity ... are you studing in ENS ?

As I said above, I never used the hypothesis of zero trace, and in fact that hypothesis is not needed for the result.

(It does make sense if you are thinking about graphs. It's just that you don't need it to prove this.)


help me with this INEQUALITY!$p,q,r>0 ,p+q+r=1$ prove:
$2+9pqr\ge7(pq+qr+rp)$

- http://www.mathlinks.ro/Forum/viewtopic.php?p=37204#37
- http://www.mathlinks.ro/Forum/viewtopic.php?p=529863#529863
- http://www.mathlinks.ro/Forum/viewtopic.php?p=463809#463809
- http://www.mathlinks.ro/Forum/viewtopic.php?p=176821#176821
- ...


Hi, what is wrong with the following "proof":

1 = squareroot((-1)(-1)) = squareroot(-1)*squareroot(-1) = i*i = -1

hence:

1 = -1

*cough*latex*cough*

The problem is that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is only true when $a$ and $b$ are not both negative. This is due to the fact that every number besides 0 has two square roots: a positive and a negative one (if imaginary numbers can be called positive or negative). The general equation which is true for all $a$ and $b$ is $\sqrt{ab}=\pm\sqrt{a}\sqrt{b}$.

This problem has already been answered in the Getting Started forum. There is no need to post a solution here.

Unless you want to increase your post count.  ;)

Thanks.


Let be $ a,b,c > 0$ such that $ abc \equal{} 1$ . Prove that :

$ \frac {a}{c(a^3 \plus{} 1)} \plus{} \frac {b}{a(b^3 \plus{} 1)} \plus{} \frac {c}{b(c^3 \plus{} 1)}\ge \frac {a}{c(a^3 \plus{} b^3)} \plus{} \frac {b}{a(b^3 \plus{} c^3)} \plus{} \frac {c}{b(c^3 \plus{} a^3)}$

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ abc \equal{} 1$ . Prove that :

$ \frac {a}{c(a^3 \plus{} 1)} \plus{} \frac {b}{a(b^3 \plus{} 1)} \plus{} \frac {c}{b(c^3 \plus{} 1)}\ge \frac {ab}{1 \plus{} c^2} \plus{} \frac {bc}{1 \plus{} a^2} \plus{} \frac {ca}{1 \plus{} b^2}$[/quote]
Thank you for your gift but I think that your inequality is wrong.
Try $ a\equal{}4$ and $ b\equal{}c\equal{}\frac{1}{2}.$ :wink:

Sorry , correct it is :

$ \frac{a}{c(a^3\plus{}1)}\plus{}\frac{b}{a(b^3\plus{}1)}\plus{}\frac{c}{b(c^3\plus{}1)}\ge \frac{a}{c(a^3\plus{}b^3)}\plus{}\frac{b}{a(b^3\plus{}c^3)}\plus{}\frac{c}{b(c^3\plus{}a^3)}$ , with $ abc\equal{}1$ .

We can rewrite it to:
$ \sum\ \frac{1}{a(b\plus{}c)} \geq\ \sum\ \frac{1}{a^2\plus{}bc}$
(obviously trues by Voni Schur)
@ alex2008: I think you shouldn't transform this original ineq to a uglier with $ abc\equal{}1$. It's only take more time for the reader  :wink:


Try to solve this "task":

Let $a_i >0 , i=1,2,3,...,5n$ where $n$ is an integer greater or equal to 1. If $a_{5k-3} + a_{5k-2} + a_{5k-1} + a_{5k} = 1$ for
any integer $k \geq 1$, then show that ${a_1}^2+{a_2}^2+...{a_{5n}}^2 \geq a_1 + a_6 + ... + a_{5n-4}$.

 :ninja:


Careful counting yields $ \boxed{67}$, does anyone get something else?

Smallest size of squares: $ 5\cdot4\plus{}4\equal{}24$
Double the side length of the side length of the smallest squares: $ 4^2\plus{}5\plus{}4\equal{}25$
Triple the side length of the side length of the smallest squares: $ 4$
Two times the side length of the second largest type of squares: $ 9$
Three times the side length of the second largest type of squares: $ 4$
Largest square: $ 1$
So adding up, we get $ \boxed{67}$.


Find all functions f from the reals to the reals such that f(x+y)=F(x)+f(y) for all x,y in the reals satisfying the constraint of continuity everywhere.

$f(x)=f(1)\cdot x$

We can prove that $f(x)=f(1)\cdot x$ for $x\in\mathbb{Q}$; by density and continuity, this proves it for the reals and we're done.

Ha! However, prove that by density and continuity that such a function on the rationals necessarily hold on the reals, if it really isa true.

Bomb

This is well known.  Am I doing a homework problem for you?

Anyways, for every $x\in\mathbb{R}\setminus\mathbb{Q}$, take a sequence $\{x_n\}$ with $x_i\in\mathbb{Q}$ such that $x_n\to x$.  By continuity, we have $f(x)=\lim_{n\to\infty} f(x_n)=\lim_{n\to\infty} f(1)\cdot x_n\to f(1)x$

Thanks. Has anyone heard off MBST.

Bomb


induction
continued fraction
algebra proposed
algebra

Find all $ f: \mathbb{R} \to \mathbb{R}$ such that:
(i) $ [f(x)]\equal{}[x]$ for all $ x \in \mathbb{R}$
(ii) $ f(x\plus{}1)\equal{}f(x)\plus{}1$ for all $ x \in \mathbb{R}$
(iii) $ f(x)f(\frac{1}{x})\equal{}1$ for all $ x \in \mathbb{R} \minus{}\{0\}$

[quote="jedaihan"]Find all $ f: \mathbb{R} \to \mathbb{R}$ such that:
(i) $ [f(x)] \equal{} [x]$ for all $ x \in \mathbb{R}$
(ii) $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$ for all $ x \in \mathbb{R}$
(iii) $ f(x)f(\frac {1}{x}) \equal{} 1$ for all $ x \in \mathbb{R} \minus{} \{0\}$[/quote]

Since $ f(1)^2 \equal{} 1$ and $ [f(1)] \equal{} [1]$, we get $ f(1) \equal{} 1$ and, using $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$, $ f(n) \equal{} n$

Consider $ u > 0$ and $ x \equal{} [a_0;a_1;a_2; ...]$ a normalized continued fraction representation of $ x$ (not ending with 1, except for $ x \equal{} 1$)
Consider then $ v \equal{} f(u) \equal{} [b_0;b_1;b_2; ...]$ a normalized continued fraction representation of $ f(x)$

$ [f(u)] \equal{} [u]$ $ \implies$ $ a_0 \equal{} b_0$

$ f(u \minus{} a_0) \equal{} f(u) \minus{} a_0$ $ \implies$ $ f([0;a_1;a_2; ...] \equal{} [0;b_1;b_2; ...]$

Then, since $ f(\frac {1}{x}) \equal{} \frac {1}{f(x)}$, $ f([a_1;a_2;...]) \equal{} [b_1;b_2;...]$

And it's easy to show with induction that $ a_i \equal{} b_i$ $ \forall i$ (using, for rational numbers, whose represention is limited, the fact that $ f(n) \equal{} n$)

So $ f(x) \equal{} x$ $ \forall x > 0$
So $ f(x) \equal{} x$ $ \forall x$ (use $ f(1 \plus{} 0) \equal{} 1 \plus{} f(0)$ for $ f(0)$, for example)


geometry
calculus
integration
trigonometry
LaTeX
geometric series

A triangle is inscribed in a circle. The verticies of the triangle divide the circle into three arcs of lengths 3,4 and 5. What is the area of the triangle?

Let $R$ be the radius of circumscribed circle, the area is $\frac{3+\sqrt{3}}{4}R^2$

First draw a diagram. O is the center of the circle, arc AB=5, arc BC=3, and arc AC=4.

Well, since none of the arcs are more than half of the triangle, the triangle is acute. Because the circumference is 12, the radius of the circle is $\frac{6}{\pi}$.

$\angle{BOC} = 3/12 = 90^\circ$ $[AOC]= \frac{1}{2}(OA)(OC)sinBOC=\frac{(\frac{6}{\pi})^2}{2}=\frac{18}{\pi^2}$

The other two parts can be found similarly and the area is $\frac{27+9\sqrt{3}}{\pi^2}$

Penguin Integral, nice problem!  :D

[quote="4everwise"]Actually, the arc lengths are 3,4,5. That doesn't necessarily form a 3,4,5 right triangle does it?  :huh:[/quote]

no it desnt..oops

4everwise,can u explain the part from [AOC]........ :D

[quote="PenguinIntegral"]I have another one, very easy, but I think very cool.

Draw an equalateral triangle.  Then draw an equalateral triangle by connecting the midpoints of that triangle, then draw an equalateral triangle by connecting the midpoints of [b]that[/b] triangle, and so on.

Assuming the starting triangle has side length 1, what is the length on all the triangles combined.

[hide="Hint"]Geometric series[/hide][/quote]
$3+3\left(\frac{1}{2}\right)+3\left(\frac{1}{2}\right)^2+...$

$=\frac{3}{1-\frac{1}{2}}$

$=6$

[quote="shadysaysurspammed wrote"]4everwise,can u explain the part from [AOC]........ [/quote]

Gladly.  :D

O is the center of the circle and is inside $\triangle ABC$ I connected O to each of the vertices to form three seperate triangles. Then I found the area of each of them and added them together.

To find the area of $\triangle AOC$, I used the formula $A=\frac{ab\sin c}{2}$, where the a and b were the radii and c was $90^\circ$. And we all know that $\sin 90^\circ=1$ ;)

hey frt can you tell me how you got to the second line of your latex? did you apply Gp ? in that case please tell how ya got the 2nd step cos i dunn know much GP :D

Well, I used the formula of infinite geometric series.

$a+ar+ar^2+ar^3+...=\frac{a}{1-r}$ ($|r|<1$)

In this case, $a=3$ and $r=\frac{1}{2}$, so $\frac{3}{1-\frac{1}{2}}=6$

[quote="anirudh"]the three sides have to be 3, 4, and 5. Use the half base*height formula to get the area 6. The area of the triangle is 6. :)[/quote]

are you talking about the first one because if you are, your wrong

First one? I thought that there was only one :(

This is the second one :) (See the post #9)

[quote="PenguinIntegral"]Draw an equalateral triangle.  Then draw an equalateral triangle by connecting the midpoints of that triangle, then draw an equalateral triangle by connecting the midpoints of [b]that[/b] triangle, and so on.

Assuming the starting triangle has side length 1, what is the length on all the triangles combined.[/quote]

Yes I agree the $2$nd one is $6$
[hide="Solution"]
Sum to infinity of a sequnce having values between $-1<r>1$
So,
The triangle perimetres starting from $1$ are,
$3,1.5,0.75.....$
$3(1+0.5+0.25....)$
$3(2)$
$6$
Answer=$\boxed6$
[/hide]

[quote="eminem"]but hey anirudh , the arcs are 3,4,5 how the heck did the sides of the triangle become 3,4,5??[/quote]

he misread the problem. He thought it was sides, not arc.


calculus
integration
calculus computations

When you evaluate an improper integral with say an infinite discontinuity between your limits of integration, and if one side of the expression for the improper integral is undefined or divergent, then does the improper integral automatically not converge?

Because what if you combined the 2 halves of the limit expression used to evaluate the improper integral and things canceled out, leaving a convergent expression?

My high school text simply says that even if one side is divergent, the entire improper integral is divergent.

Your textbook is right (meaning that it follows the conventional definition of improper integrals).
[quote]what if you combined the 2 halves of the limit expression used to evaluate the improper integral and things canceled out[/quote]
Then you would dealing with the [url=http://en.wikipedia.org/wiki/Cauchy_principal_value]principal value[/url] of the improper integral. They are useful, but usually don't appear in introductory calculus texts.


LaTeX
inequalities
inequalities unsolved

Let $a,b,c$ three positive reals such that $abc=1$. Prove that :

$\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}}+\frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}}+\frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}}>=\sqrt{2}$

ha ha well known problems
let $a=\frac{x}{y}$, $b=\frac{y}{z}$,$c=\frac{z}{x}$
the ineq turn to:
$\sum\sqrt{\frac{xz}{2xy+2yz+zx}}\geq1$

let a=xz.....
 so we get:
$\sum\sqrt{\frac{a}{2b+2c+a}}\geq1$

our ineq is equivalent to
$\sum\sqrt{\frac{1}{2\frac{b+c}{a}+1}}\geq1$

let $x=\frac{b+c}{a}$....

so$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$

let x+1=1/u, y+1=1/t,z+1=1/k

we gain:
$\sum\sqrt{\frac{u}{2-u}}\geq1$
with u+t+k=1

easily we have :
by Cauchy
$\sqrt{\frac{u}{2-u}}\geq u$
sum them up we get the answer :D  :rotfl:
cool but famous :ninja:

[quote="Labiliau"]$... >= \sqrt{2}$[/quote]

Use the \geq (or \leq) command in LaTeX rather than typing >= (or <=).
You will get $\geq$ (or $\leq$) instead of $>=$ (or $<=$).

As I know, this is for Serbia and Montenegro mo 2004  :P

This is one of the original "Serbian" solutions, I believe:
Notice that $\sqrt{1 + 2x} < 1 + x$ for all $x > 0$.
Hence \[\sum{\frac{1}{\sqrt{1 + 2a + \frac{2}{b}}}} > \sum{\frac{1}{1 + a + \frac{1}{b}}} = 1,\]
since $abc = 1$. Hence we are done. Equality cannot occur.

Why did my reply vanish?

I just said I think he meant strict inequality $>$, as I cannot see any equality case.

[quote="Peter VDD"]Why did my reply vanish?[/quote]

Peter, I think we agreed that no useless posts should be made anymore in this forum...


I'm looking for a graph theory book that is elementary but has challenging problems
suggestions? :)

Try [u]Introduction to Graph Theory[/u] by Trudeau


least common multiple
number theory unsolved
number theory

Prove or disprove the following statement:
If $ x_1 < ... < x_k < n$ are positive integers such that the least common multiple of any two of the numbers $ x_1, ... , x_k$ is greater than $ n$, then $ 1/x_1 \plus{} ... \plus{} 1/x_n \leq 1$.

[quote="udyam.bnghs"]your problem should be x(1)+x(2)+...+x(k)<=1, otherwise what is x(n)???

the statement is false! just take n=6, k=3.and x(1)=2,x(2)=3,x(3)=5, you'll get a counterexample!!!!!!!!!!    right????[/quote]
$ lcm(2,3) \equal{} 6$ which is not greater than $ 6$..

[quote="udyam.bnghs"]your problem should be x(1)+x(2)+...+x(k)<=1, otherwise what is x(n)???

right????[/quote]

Of course it should be $ \frac{1}{x_1} \plus{} \ldots \plus{} \frac{1}{x_k}\leq 1$


Starting the first week of January I will be teaching a course focusing on problems from the AHSME (the old AMC-12) and the AIME.  I have picked out well over 100 problems from past contests.  The focus for the AHSME problems will be on those that are the 10 hardest on any single competition.  A few easier problems will be posted on the boards.  AIME problems will cover all difficulty levels.

I have picked out several problems of many different varieties, so hopefully we will maximize coverage of problem solving techniques and topic areas.


function
calculus
derivative
calculus computations

I tried to solve a problem of analytical mechanics and got this system of differential equations where i'm stuck. Can anyone help me solve it?
$ \left\{ {\begin{array}{*{20}c}
   {\ddot \varphi _1  \plus{} \frac{g}{l}\varphi _1  \plus{} \frac{{ca^2 }}{{ml^2 }}(\varphi _1  \plus{} \varphi _2 ) \equal{} 0}  \\
   {\ddot \varphi _2  \plus{} \frac{g}{l}\varphi _2  \plus{} \frac{{ca^2 }}{{ml^2 }}(\varphi _1  \plus{} \varphi _2 ) \equal{} 0}  \\
\end{array}} \right.$
$ \varphi _1$ and $ \varphi _2$ are just functions of time and this is what I mean when I say solve this system. Find $ \varphi _1 \equal{} \varphi _1 (t)$ and $ \varphi _2 \equal{} \varphi _2 (t)$
In physical notation $ {\ddot \varphi _1 }$ is simply the second derivative of $ \varphi _1$ with respect to time $ t$. The same thing for $ {\ddot \varphi _2}$. I must also note that as far as the person who mathematically solves the system should be concerned, $ \frac{g}{l}$ and $ \frac{{ca^2 }}{{ml^2 }}$ are all constant numbers.
Any help would be appreciated.

Subtract and add the two equations and solve the resulting equations for $ \varphi_1 \minus{} \varphi_2$ and $ \varphi_1 \plus{} \varphi_2$, respectively.  Also, you may not have noticed, but there's a post at the top of this forum titled "PLEASE do not post calculus problems in this forum!"   :wink:


For any positive integer $ n$, let $ \tau(n)$ be the number of divisors of $ n$ greater than $ 2008$. For instance, $ \tau(10000)\equal{}3$ because $ \frac{10000}{1}\equal{}10000$, $ \frac{10000}{2}\equal{}5000$ and $ \frac{10000}{4}\equal{}2500$ are greater than $ 2008$. (In particular if $ n<2008$ then $ \tau(n)\equal{}0$)
Now consider the sequence $ (a_i)$ where $ a_i \equal{} 0$ if $ \tau(i)$ is even and $ a_i \equal{} 1$ if $ \tau(i)$ is odd.
Decide whether this sequence ever becomes periodic or not.

$ f(n)$  is the number of divisor of n great than M and $ \tau(n)$ is the number of divisor
Call A is a subset of natural number such that all divisor different from 1  of them great than M.
The first we will prove following property :
Let $ m\in A$ and $ n$  is a natural number such that $ \gcd(m,n) = 1$ then we have  :
\[ f(mn) = \tau(m).\tau(n) + f(n)
\]
Proof :
Call d>1 is a divisor of $ mn$
If $ d|m$ then $ d > M$ . The number of d in this case is $ %Error. "taugh" is a bad command.
(m).%Error. "taugh" is a bad command.
(n)$
If $ \gcd(d,m) = 1$ then $ d|n$ ,the number of d in this case is  $ f(n)$
Therefore
\[ f(mn) = \tau(m).\tau(n) + f(n)
\]
Now consider your problem :
Suppose this sequence is periodic T then :
\[ f(m + T)\equiv f(m) (\mod 2)
\]
Because $ T|T^2$ so $ T^2$ is also satisfy :
\[ f(mT^2)\equiv f(T^2) (\mod 2)
\]
Chose $ m^2$ is a element of A and $ \gcd(m,T) = 1$ then
\[ f(m^{2}T^{2}) = \tau(T^2).\tau(m^2) + f(T) (\mod 2)
\]
We know that $ \tau(x^2)\equiv 1(\mod 2)$
Then
\[ f(m^{2}T^2)\equiv f(T^2) + 1(\mod 2)
\]
This is contradiction
So this sequence can not be periodic .
Problem claim .


Solve the system  of the equations:
$\{ \begin{array}{l}uv^{3}+vy^{3}=14 \\ ux^{2}+vy^{2}=5\\ ux+vy=2 \\u+v=1\end{array}$

I'm going to assume that that first line should be $ux^{3}+vy^{3}= 14$.  In which case:

Let $a_{k}= ux^{k}+vy^{k}$.  Observe that

$a_{k+1}= (x+y ) a_{k}-xy a_{k-1}$

Let $A = x+y, B = xy$.  This gives

$5 = 2A-B$
$14 = 5A-2B \implies$
$A = 4, B = 3$

So $x, y$ are roots of the polynomial

$t^{2}-4t+3 = 0 \implies \boxed{ x = 1, y = 3 }$

Then we have

$u+v = 1$
$u+3v = 2 \implies$
$\boxed{ v = \frac{1}{2}, u = \frac{1}{2}}$

Thanks vey much.I have a similar problem:
Solve the system  of the equations:
$\{ \begin{array}{l}uv^{3}+vy^{3}=10 \\ ux^{2}+vy^{2}=6\\ ux+vy=4 \\u+v=3\end{array}$

It's similar.  We have

$6 = 4A-3B$
$10 = 6A-4B \implies$
$A = 3, B = 2$

So $x, y$ are roots of

$t^{2}-3t+2 = 0 \implies \boxed{ x = 1, y = 2 }$

Which gives $\boxed{ u = 2, v = 1 }$.


cos x  cos2x   cos4x   cos8x------------  cos((2^n)x)

you should know that $\sin 2x=2\sin x \cos x$
so that :
$\sin 2^{n+1}x=2\sin 2^{n}x \cos 2^{n}x=4 \sin 2^{n-1}x \cos 2^{n-1}x \cos 2^{n}x=\cdots = 2^{n+1}\sin x \cos x \cos 2x \cos 4x \cdots \cos 2^{n}x$
so the required sum is $\frac{\sin 2^{n+1}x}{2^{n+1}\sin x}$


calculus
integration
logarithms
algebra
partial fractions
calculus computations

How to integrate this.

$ \int x^2 \frac{(a\minus{}x)}{(a\plus{}x)} dx$.

[hide="First Hint"]Break down the $ \frac {a \minus{} x}{a \plus{} x}$ part with partial fractions,
[hide="Which breaks down to"]$ \frac {a \minus{} x}{a \plus{} x} \equal{} \frac {2a}{a \plus{} x} \minus{} 1$[/hide][/hide]

[quote="kan"]How to integrate this.

$ \int x^2 \frac {(a \minus{} x)}{(a \plus{} x)} dx$.[/quote]

subsitute x = x-a and break the fraction in the integral into two parts.  Expand-divide and then integrate.

[hide="Second hint"] after breaking down and multiplying with $ x^2$, use the difference for squares on the fractional part.[/hide]

hello, we have $ \frac{x^2(a\minus{}x)}{a\plus{}x}\equal{}\minus{}2a^2\plus{}2xa\minus{}x^2\plus{}\frac{2a^3}{a\plus{}x}$.
Sonnhard.

[quote="kan"]How to integrate this.

$ \int x^2 \frac {(a \minus{} x)}{(a \plus{} x)} dx$.[/quote]
Is the answer:

$ ax^2 \minus{} 2a^2 x \minus{} \frac{x^3}{3} \plus{} 2a^3 \ln (a \plus{} x)$ $ ???$

hello, i have $ \int x^2\frac {a \minus{} x}{a \plus{} x}\,dx \equal{} \minus{} 2a^2x \plus{} ax^2 \minus{} \frac {x^3}{3} \plus{} 2a^3\ln|a \minus{} x| \plus{} C$.
Sonnhard.

I'm sure you've made a typo, Sonnhard: it's $ 2a^3\ln |a \plus{} x|$, isn't it?

[hide="solution"]$ \int x^2\frac {a \minus{} x}{a \plus{} x}dx \equal{} \int x^2\frac { \minus{} a \minus{} x \plus{} 2a}{a \plus{} x}dx \equal{} \int x^2\left( \minus{} 1 \plus{} \frac {2a}{a \plus{} x}\right) \equal{} \\ \minus{} \int x^2 dx \plus{} 2a\int \frac {x^2}{a \plus{} x}dx \equal{} \minus{} \int x^2 dx \plus{} 2a\int \frac {x^2 \minus{} a^2 \plus{} a^2}{a \plus{} x}dx \equal{} \\ \minus{} \int x^2dx \plus{} 2a^3\int \frac {1}{x \plus{} a} \plus{} 2a\int(x \minus{} a)dx \equal{} \frac { \minus{} x^3}{3} \plus{} 2a^3\ln (x \plus{} a) \plus{} ax^2 \minus{} 2a^2x$[/hide]

hello, yes it is, you have right. I will correct it, thank you.
Sonnhard.


When I try to compile my solutions in TeXnicCenter nothing happens.

I have noticed that the box that used to say LaTeX=>PDF is now blank and I don't seem to have an output profile. I think this might have something to do with the problem.

I've been using TeXnic center for about a year and I've never had this problem before.

Try rebuilding your output profiles: select "Define Output Profiles..." in the Build menu, and click on the "Wizard..." button in the lower-left.  Then follow the instructions shown.

If this doesn't work, you might try reinstalling TeXnic center (and if you haven't reinstalled it in the past year, then this is probably worth doing anyway because the new version has some cool new features).

I reinstalled TeXnic center and it works now.

Thanks.


calculus articles MIT college Stanford AMC AIME

We're offering

(intro) physical science

honors chemistry

honors biology

[to be determined]

to the kids in our homeschooling group, over four years. Some of the kids may take the AP chemistry test at the end of the biology school year, with some additional chemistry practice. (That would be at the end of the eighth-grade year for my son.)

The order I took when I was a kid was

9th grade "general science"

10th grade biology

11th grade chemistry

12th grade physics.

I should have gone to a tougher high school--that would have been good for all the decades that have come since.

The sequence I took was:

9: Honors Biology (I lived in the states at that point in time)
10: Chemistry, Computer Science
11: IB Chemistry, Physics
12: IB Physics, IB Computer Science

in my state:
for gifted students

Bio
Chem
Physics/AP Bio/AP chem
Physics/AP Bio/AP chem
Physics/AP Bio/AP chem/environmental sci/
(bio phy and chem are required)
for me:
8: bio
9: Chem
10: ap chem
11:ap phy
12:don't know

At my school, the standard is

9:  Biology

10:  Chemistry (they won't let you take AP Chem this year)

11:  Physics (some people take AP this year)

12:  No science Requirement

For me in particular:

9:  AP Physics as well as the Biology requirement

10:  AP Chemistry (they let me skip basic level)

11 (now):  AP Biology

12:  I dunno yet, maybe astronomy?

Here's how I took sciences:

9th: Physical Science (waste of time  :( )
10th: Chemistry and Biology
11th: AP Physics B
12th(what I'm signed up for): AP Chem

8. honors bio
9. honors physics/honors chem
10. ap chem
11. ap bio (the BEST class EVER)
12. ap physics

for me:
9th: physical science honors
10th: biology honors, chemistry honors
11th: ap biology
12th: signed up for ap chemistry and physics (not enough room in my schedule for ap  :(  )

I battled my school for my schedule... :)
9th grade: Science IV (what everybody takes... One trimester each of Bio, Chem and Physics.)
10th grade: AP Chemistry (had to study Chem over the summer to take it), Physics (plus independent study for AP B)
11th grade: AP Bio (they'll probably let me skip bio, too..) and possibly Physics C at EPGY.
12th grade: I will hopefully be at college then.....

Everyone else:
9th: Science IV
10th: Biology (a few people take AP Chem, another few take Anatomy&Physiology with Bio)
11-12th: Anatomy&Physiology, Microbiology, Physics for the majority. Some people will take IB or AP Sciences.

What kind of schools do you people go to?  Do any of you go to magnet schools?  I go to a tiny private school in rural MD, so I don't really have a perspective on other schools.

[quote="sky9073"]
12th grade: I will hopefully be at college then.....
[/quote]

Interesting, I actually considered graduating a year early as well, but I decided not to. . .I'm a junior this year, and I've opted to take a couple classes in my highschool next year, while spending the rest of my time working part-time at the Johns Hopkins Applied Physics Lab.  I've talked to people with experience in this kind of thing, and alot of them say it's a better idea to stay in highschool for senior year than to graduate early.

I go to an American "international" school with 160 people per class.

Of course, I'd love to go on that senior trip to Cancun with everyone else, but.... One of the reasons I'm thinking of applying to college early is that my family will probably be moving somewhere after my junior year. If we're moving anyway, why not college? well, that what I thought, at least. And living in Mexico w/o speaking fluent Spanish really limits my opportunities for anything outside school. Plus, the academic level of my school isn't anywhere near impressive..

I'm a senior. I went to a bad middle school, where everyone takes the same things (none of them advanced; used to be a selective 9th grade math, but the teacher moved on and no one took his place). My high school requires 1 year each of biology, chemistry, and physics, and a 4th year of a lab science.

I did
honors biology
honors physics
AP chemistry
AP physics C (small jump from honors, imo)

A regular person might go like this
biology
chemistry
physics
astronomy/electronics (most popular)

also offered (I don't think they can replace the core 3 years): organic chemistry, physics in sports, nutritional sciences, horticulture, AP psychology, earth sciences, environmental sciences

The following are technically under the math department, but whatever:
AP Java, visual basic, computer science, research literacy (required in freshman year unless you take an honors science, waste of time), TV production, C++

For our school (I think this is pretty common in Texas) your first science class depends on what math you are in.  If you are taking Algebra I as a freshman:

IPC (integrated physics and chemistry)
(pre-ap)Biology
(pre-ap)Chemistry
(pre-ap)Physics (optional, only 3 science credits are required for graduation in Texas)
the pre-aps are optional

If you have an algebra I credit entering freshman year:
(pre-ap) biology
(pre-ap) chemistry
(pre-ap) physics
pre-aps are optional

The kids at my school who are more interested in science or are looking to do something science related at college often take something like this:
Pre-AP Bio
Pre-AP Chemistry
Pre-AP Physics AND AP Chemistry
AP Physics (our school offers both levels) AND AP Biology.

ok, how come there are all of those "pre-AP" courses? how come there is "pre-AP" bio but no AP bio? doesn't that kind of defeat the purpose, if there is "pre-AP" and not AP?

i go to a "special" school. its the most specialest school ever. AP classes are very much the norm here. i think i will graduate with about 8 AP courses. maybe 9, if i take AP Programming, I haven't decided yet. considering programming REALLY wont help me in college. at all.

This is what I'm probably going to do

9 - Enriched life science (Biology)
10 - Enriched Physical science (Chemistry)
11 - AP Biology
12 - AP Chem and either Physics at OSU or if they have the AP at my school, at my school

9th grade i took hon. bio
In 10th grade i'll take honors chem and honors physice
for the rest of HS, maybe ap enviromental and ap physics (maybe more?)

I had to battle my school to let me out of a pre-calc class this year (my junior year) so that I could take an independent study on multivariable calculus/preparing for calculus BC Test. Everything I know after freshmen geometry is self taught  ;) .

That being said, I wish I took AP Physics C this year  :|

9th : Honors Bio
10th: Honors Chem
11th: Honors Physics (huge waste of time)
12th: AP Physics C, AP Chemistry, AP Computer Science AB

I'm really kicking myself for not taking intiative earlier... coulda been doing a cool independent study next year

General Science
Biology
Chemistry H/R
Physics R/B AP
Chemistry AP
Biology AP

In 9th grade, one can take General Science or Biology.  Sometimes, a student will take General Science in 9th grade and take Biology in 10th grade.

After Biology, comes some form of Chemistry (Honors or Regular).

After Chemistry, it's really up to the student on what he or she wishes to do.  Many students take elective science courses, while others will take one of the "real" science courses.  Now, here's where it gets tricky.  Because of a faculty shortage, AP Chemistry and AP Physics B are now offered in alternating years (the teacher is the same).  However, AP Biology is offered every year because they make sure it doesn't get cancelled--it's our most popular course (but the one where people do the worst on the AP).

With that said, there have been rare occassions where a student has jumped around in the order, though, in recent memory, that only includes myself and a friend one year my senior.

My friend:
9   AP Biology
10 H. Chem + AP Physics B
11 AP Chem
12 2nd semester took 1st semester Java at a local college (equivalent to first semester of AP Comp Sci AB)

Me:
9   H. Chem
10 AP Chem IS + AP Bio
11 AP Physics B
12 nothing planned

I suppose I should describe the classes a little before one gets the wrong idea.  Basically, the courses are crap.  AP Biology has been a longstanding joke, often being called Bio 1.5.  The teacher of many years retired after my freshman year, leaving us with one of the other bio teachers, who is one of the most incompetant people I have ever (she does mean well, though).  In fact, she won't be teaching AP Bio next year, because the administration has sort of pushed her to leave the station and return to her prior classes.

The Chem teacher does a decent job, but a strange one.  She is very good at reenforcing concepts with adequete problems, but I now realize that there was a fundamental problem with her teaching style.  She was great with those who "already got it," but couldn't really help those who didn't understand.  Anyway, she really knows what she's talking about though, unlike the soon leaving AP Bio teacher.

The Chem teacher is also the AP Physics B teacher.  I sat in on her AP Physics B class a few times as a freshman, and it wasn't organized at all.  You see, the teacher was sort of given the class because no one else could handle it.  It's too bad she really didn't have a handle on what she should be doing either (no one in that class took the AP they only covered around 2/3s of the material at the end of the year).  However, this year, she was much more organized, but I found at the end of the year a problem--she had tried to teach us too much.  We never did nuclear, thermo, and we did fluids in around 1.5 weeks.  On the other hand, we had gone indepth into rotational mechanics, moment of intertia, and a few other things that are Physics C only topics.

We used to have an AP Comp Sci class, but that was thrown out my freshman year because not enough people signed up that year (I was signed up  :mad: ).

BTW, I'm a big believer in the benefits of excellent education early in life.

My school has an extremely screwed up AP system. It's basically required that inorder to take an AP science class, you have to take the regulars or honors class first. This made my schedule look like:

9: Physical Science Honors
10: Biology Honors
11: Chemistry Honors
12: will be taking Chem AP (everyone goes in this order)

As you can see, using the school's system, the maximum number of science APs you can take in high school is 1. Others don't even have 1 AP, since they choose to take regular Physics instead of Chem AP. This retarded system made me independent study Physics B in order to compensate.

On another note, my school says that, since the UCs don't consider freshmen year in GPA, you shouldn't bother to take any lab science freshmen year, since it won't count. That's why I have Physical Science in freshmen year.

ah. our school is the same way (have to take high school classes before AP) but then we take chem and physics in the same year, and biology in eighth grade. so we can take three APs.

My school hates me. In our school, before any AP science, you need bio, chem, and physics honors; it must be in that order , and each is a prerequiste for the other. You're also strongly discouraged from doubling up in the sciences, and our coursebook warns us that AP science is reserved for the senior year only. Now take a look at my schedule:
Freshman year: Honors Bio (Still thought school's word was sacrosanct)
Sophomore year: Honors Chem and Physics (I start fighting back)
Junior year: Physics C (The beauracrats scream) and Honors Astronomy (best class ever)
Senior year (Guess when I started AoPS): AP Chem, Honors Science Projects (Research class), EPGY Physics: Light and Heat (and maybe modern physics), and astronomy at either my local community college or Northwestern.

The typical college-prep sequence at my school is Biology, Chemistry, Physics which either begins in 9th or 10th grade, depending on the student. Unlike most schools, there is no physical/general science class required in order to take biology. The only pre/co-requisite for Biology is to be at least concurrently enrolled in Algebra 1 or Algebra 1B.

The track most of the gifted students take is:
9) Honors Biology
10) Honors Chemistry
11) AP Chemistry and/or AP Biology
12) Physics or AP Physics B

i'm sort of an average math/science guy in smart-people classes :(...this is what i took:

9) Biology/ Algebra 1
10) Chemistry/ Geometry
11) Physics/ Algebra 2
12) AP Physics B/ Trigonometry w/Math Analysis AND AP Statistics (this is my current year)

so yeah, me = noob.

Can u take ap physics C exam and ap physics b exam in the same year? My school is a queer IB school that only lets people take ap when they go through bio, chem and physics in that order. (i gotta kill them)

im a sophmore and im in honors physics and chem but next semester going to independent study some real physics  :D. if plan to take both physics c and b at the same year is that possible?

also, does light and heat count for high school at epgy?

[quote]Can u take ap physics C exam and ap physics b exam in the same year?[/quote]

No. The College Board schedules them at the same time.

[quote="nigma"]Does anyone know if classes have to adhere to a standard curriculum to use the titles Accelerated or Honors? Are courses labeled as accelerated generally considered to be the same as those labeled honors?[/quote]

Nothing of that kind is standardized in the United States. One school's "honors" course might have the same curriculum as another school's "regular" course--or even an easier curriculum. The same applies to the label "advanced." That's why I always ask for very specific information about the textbook, syllabus, and number and kind of lab experiments when comparing courses across schools.

[quote="aznphatso"]You can take them in the same year if you talk to your AP coordinator who can sign you up for one during the regular testing sc hedule and you can take the other one during alternate testing a week later. (I will be taking both chem and enviro this year and they are held at the same time)[/quote]

Hey aznphatso, if what you say is true and assuming I take both APs, does the college board have a rule saying you can't or your scores are invalid? Due to my schools slow science program, I wasted my freshman year in bio and now I want to fight back and get something done.

there is no rule against taking physics b and c in same year- it just would look weird but you definately can do it... why don't you take different subjects though, that would be more of a learning experience. many people in my school take physics b and c, though not in the same year, also some schools require phys b for c or calc ab for bc, so you can just take both, no problem!


So now that I actually understand what Cauchy's inequality is, I'm intrigued about using it to find the distance from a point to a line, or plane etc. Is there some general process for some point (a,b,c) and plane Ax+By+Cz+D=0?

Cauchy is basically derived from the dot product, which is the usual method of showing the below:

Let $x_{0}, y_{0}, z_{0}$ be a point on the plane, so that $Ax_{0}+By_{0}+Cz_{0}=-D$.  Then

$(A^{2}+B^{2}+C^{2}) \left( (x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}\right) \ge \left( A(x-x_{0})+B(y-y_{0})+C(z-z_{0}) \right)^{2}= \left( Ax+By+Cz+D \right)^{2}\implies$
$\sqrt{ (x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}}\ge \frac{ | Ax+By+Cz+D |}{ \sqrt{ A^{2}+B^{2}+C^{2}}}$


geometry
circumcircle
trigonometry
geometry unsolved

In a triangle ABC, the bisectors of <ABC and <ACB meet AC and AB at D and E respectively, and <ADE-<AED=60°. Prove that <ACB=120°.

Well, I can prove the converse, namely that $\angle ACB=120^{\circ}\Rightarrow \angle ADE-\angle AED=60^{\circ}$.

Let $F$ be the intersection between $BC$ and the circumcircle of $DEC$. It's clear that $D EF$ is equilateral, and from here and from the fact that $BD$ is the bisector of $\angle EBF$ we get $BE=BF$. $\angle ADE-\angle AED=\angle EFC-(120^{\circ}-\angle BEF)$, which is equal to $(180^{\circ}-\angle BEF)-(120^{\circ}-\angle BEF)=60^{\circ}$.

I think this converse can be used to prove the direct statement, but I don't have the actual argument right now.

[color=red][Moderator edit: the "DEF" string in TeX code is not compiled. Replaced by "D EF".][/color]

$1.\blacktriangleright\ B=2x$, $C=2y$ $\Longrightarrow$ $m(\widehat {CDB})=180^{\circ}-(x+2y)$, $m(\widehat {EDB})=y-30^{\circ}$, $m(\widehat {BEC})=180^{\circ}-(2x+y)$, $m(\widehat {DEC})=x+30^{\circ}$.
$2.\blacktriangleright\ \sin\widehat {BEC}\cdot \sin \widehat {CBD}\cdot \sin \widehat {DCE}\cdot \sin \widehat {EDB}=$
$\sin \widehat {DEC}\cdot \sin \widehat {EBD}\cdot \sin \widehat {BCE}\cdot \sin \widehat {CDB}\Longleftrightarrow$
$\sin (2x+y)\cdot \sin (y-30^{\circ})=\sin (x+30^{\circ})\cdot \sin (x+2y)\Longleftrightarrow$
$\cos (2x+30^{\circ})-\cos (2x+2y-30^{\circ})=\cos (2y-30^{\circ})-\cos(2x+2y+30^{\circ})\Longleftrightarrow$
$\cos (2x+2y-30^{\circ})-\cos (2x+2y+30^{\circ})=\cos (2x+30^{\circ})-\cos (2y-30^{\circ})\Longleftrightarrow$
$\sin (2x+2y)+2\sin (x+y)\cdot \sin (30^{\circ}+x-y)=0\Longleftrightarrow$
$\cos (x+y)+\sin (30^{\circ}+x-y)=0\Longleftrightarrow\cos (x+y)+\cos (60^{\circ}+y-x)=0\Longleftrightarrow$ $y=60^{\circ}\Longleftrightarrow \boxed {\ C=120^{\circ}\ }\ .$

[quote=Tsikaloudakis]stisiporsin[/quote]

What does this mean???

these are Greek words written using English letters ''στις υποψιν'' ,

it means ''to keep it in mind'' (in order to remember this later)


logarithms
trigonometry
LaTeX
calculus
quadratics
linear algebra

If $a^x=b^{x+1}$, then find $x$.

Given the roots of $(x-m)^2+x=0$ are equal, find $m$.

If $\log_{10}2=0.3010$ and $\log_{10}3=0.4771$, then find the number of digits in $2^{12}*3^{10}*5^8$

Given $3^{x+1}-3^{x-1}=24$, find $(2x)^{2x}$. (Is there a way to do it without guess and check?)

If arctan(a)+arctan(b)=$45^\circ$, then $a+b=?$

[quote="eryaman"][hide="hints"]
1. I don't think there is enough information...are there some kind of conditions on $a$, $b$, and $x$?

2.  I haven't done this question but my first instinct is to expand and find the value of $m$ that will make the expression in the form $a^2+2ab+b^2$...

3. Notice that if $10^n=y$, then $y$ has $\lfloor{n}+1$ digits.

4. Make a substitution...say $y=3^{x-1}$?

5.  Not sure...maybe take the $\tan$ of both sides and use tangent sum/diff. identities?  Haven't worked this out so not sure...[/hide][/quote]
For #1, I think we are supposed to solve for $x$ in terms of $a$ and $b$.

For #5, yes, that is probably the easiest  approach.

[quote="Drunken_Math"]If arctan(a)+arctan(b)=$45^\circ$, then $a+b=?$[/quote]

[hide]$\arctan{a}=x\\
\tan{x}=a\\\\
\arctan{b}=y\\
\tan{y}=b\\\\
\arctan{b}=45^\circ-x$[/hide]

Okay, nevermind. :blush:

[quote="Drunken_Math"][quote="Silverfalcon"]Number 5 was 25 in mathfanatic's mock AMC.. and also MMPC.

Anyway..

[hide="Sol"]

Taking $\tan$ on LHS and RHS, we get:

$\frac{\tan(\arctan a) + \tan(\arctan b)}{1 - \tan(\arctan a) \tan(\arctan b)} = \frac{a+b}{1-ab} = \tan 45 = 1$.

Hence, $a+b = \fbox{1-ab}$. [/hide][/quote]

All of these were in the MMPC.

How'd you know about the MMPC?

Do you live in Michigan or something?  :?[/quote]

No, I do not live in Michigan. I just know these contests because I searched them on the Google (and still do for more math problems)

If a=b=1 in number 1, then x can be any number, same going with 0 (if x doesn't equal 0 or -1).

I found out the first one.

You're supposed to find it in terms of $a$ and $b$, using logarithms.  :)

[hide="Solution to #1"]
$x(lna) = (x+1)(lnb)$

$\frac{(x+1)}{x} = \frac{lna}{lnb}$

$1+\frac{1}{x} = \frac{lna}{lnb}$

$\frac{1}{x} =  \frac{lna-lnb}{lnb}$

$x = \frac{lnb}{ln(\frac{a}{b})}$
[/hide]

Please correct me if I'm wrong.

(BTW, sooooo sorry for not using LaTeX. Still have to learn >_<. I promise I will start soon :P)

EDIT: Converted it to LaTeX

[quote="thebrokenlight"]w00! I'm new here, so I was nervous that I would make a fool of myself :P. Some of these problems are a bit difficult for me, because (unfortunately) I've been concentrating my math education on higher level mathematics (calculus, linear algebra, etc) instead of honing my skills in problem solving. Which is, I guess, why I'm here :P.[/quote]

Well then, Welcome to AoPS! *offers handshake*  :)

*accepts handshake* Thanks for the warm welcome.

Welcome to AoPS Getting Started forum.  :)

[quote="Drunken_Math"]If $a^x=b^{x+1}$, then find $x$.[/quote]

[hide="solution to #1"]$a^x=b{x+1}$

$x \ln a=(x+1) \ln b$

$x=\frac{(x+1) \ln b}{\ln a}$

$\frac{x}{x+1}=\frac{\ln b}{\ln a}$

$\frac{x+1}{x}=\frac{\ln a}{\ln b}$

$1+\frac{1}{x}=\frac{\ln a}{\ln b}$

$\frac{1}{x}=\frac{\ln a/b}{\ln b}$

$x=\frac{\ln b}{\ln a/b}$[/hide]

This is what I got.

Masoud Zargar

EDIT:  I just noticed that it was already solved.

[quote="thebrokenlight"][hide="Solution to #2"]
In order for the roots of $(x-m)^2+x$ to be equal, then the discriminant must equal 0.
$x^2-2mx+m^2+x=0$
$x^2-(2m+1)x+m^2=0$
$x=\frac{(2m+1) \pm \sqrt{(2m+1)^2-4m^2}}{2}$
$\sqrt{(2m+1)^2-4m^2}=0$
$(2m+1)^2-4m^2=0$
$4m+1=0$
$m=-1/4$
[/hide]

[hide="Test"]
EDIT: Revised my work above. Made a careless error. Well, two careless errors. I got -1/4, then 1/4, and now -1/4 again. Here's a test (on the correct quadratic this time):

$x=\frac{(2m+1) \pm \sqrt{(2m+1)^2-4m^2}}{2}$
$x=\frac{(1/2) \pm \sqrt{1/4-1/4}}{2}=\frac{1}{4}$
[/hide]

Please correct me if I'm wrong.

EDIT: Made a few mistakes, which should now be fixed.[/quote]

Ummm accordin to ur calculations,whaen u reach
$(2m+1)^2-4m^2=0$
This is where u go wrong......over here.2m and 2m get cancelled,so is the answer 0?

[quote="shyong"]$m$ is $-\frac{1}{4}$ ...

we have $(2m+1)^2-4m^2=(4m^2+4m+1)-4m^2=4m+1=0$

so how you guys got $0$ ??  :huh:[/quote]
 :oops:  I thought... nevermind.

[quote="shyong"]$m$ is $-\frac{1}{4}$ ...

we have $(2m+1)^2-4m^2=(4m^2+4m+1)-4m^2=4m+1=0$

so how you guys got $0$ ??  :huh:[/quote]

Shyong,if u transfer $4m^2$ to r.hs and do sqroot,u'll find that 2m amd 2m get cancelled.....this is weird....so what can u conclude by this?

[quote="Drunken_Math"]If $a^x=b^{x+1}$, then find $x$.

[hide]
$a^x=b^{x+1}x\ln {a}=(x+1)\ln{b}x\ln{a}=x\ln{b}+\ln{b}$
$\displaystyle x\ln{a}-x\ln{b}=\ln{b}x(\ln{a}-\ln{b})=\ln{b}x=\frac{\ln{b}}{\ln{\frac{a}{b}}}$

[/hide]

Given the roots of $(x-m)^2+x=0$ are equal, find $m$.
[hide]
$(x-m)^2+x=0x^2-2mx+m^2+x=0x^2+(-2m+1)x+m^2$
This is a perfect square, thus
$\displaystyle(\frac{-2m+1}{2})^2=m^2\\\frac{-2m+1}{2}=m\\2m=1-2m\\4m=1\\m=\frac{1}{4}$
[/hide]

[/quote]

[quote="IntrepidMath"][quote="Drunken_Math"]If $a^x=b^{x+1}$, then find $x$.

[hide]
$(x-m)^2+x=0\\x^2-2mx+m^2+x=0\\x^2+(-2m+1)x+m^2$
This is a perfect square, thus
$\displaystyle(\frac{-2m+1}{2})^2=m^2\\\frac{-2m+1}{2}=m\\2m=1-2m\\4m=1\\m=\frac{1}{4}$
[/hide]

[/quote][/quote]

Try testing that solution. I think you're off by a factor of $-1$.

[quote="IntrepidMath"][quote="Drunken_Math"]

[hide]
$(x-m)^2+x=0\\x^2-2mx+m^2+x=0\\x^2+(-2m+1)x+m^2$
[/hide]

[/quote][/quote]

[hide="Corrections"]
I think your error occurs when you factor out the $x$:
$x^2-2mx+x+m^2\\x^2-(2m+1)x+m^2$
I believe that's how it's supposed to be. I think you incorrectly distributed the negative to the factor $-(2m+1)x$ to make it $+(-2m+1)x$. Correct me if I'm wrong.
[/hide]

hey guys , shoudnt it be $(2m-1)^2$ instead of $(2m+1)^2$ in the 2nd one??? :huh:

[quote="shyong"]if you take square root both side , you forget to put a " plus and minus " sign ...

i.e $(2m+1)^2=4m^2  \implies 2m+1=2m$ or $2m+1=-2m$ . Obviously the first cant hold since you will get $1=0$ so it only  remains $2m+1=-2m$ .....

Hope that clears your doubt .... :)[/quote]

Ohhhyeahhh,thanks shyong,pretty dumb of me not to see that :D

[quote="eminem"]can some one please clear my doubt  in  my above post ??

how did you guys get $(2m+1)^2$??? shouldnt it be $(2m-1)^2$???, in that case, the answer is coming 1/4......not -1/4  :huh:  :huh:  :huh:[/quote]

yes eminem , you are right for that $2m-1$ . It is NOT $2m+1$

whew! thats a relief! i was thinking like " am i going crazy or something, that i cant understand this" ??  :)  ;)


inequalities
trigonometry
geometry unsolved
geometry

Prove that in a $ \triangle ABC$ , where angle $ B$ is obtuse and $ AB \equal{} \frac {1}{2} AC$ . Then angle $ C > \frac {A}{2}$

Well, by mistake I have posted it in the unsolved section. Anyway , a nice and simple problem to try.
I have done it using Appolonius Theorem and Trigo {cosine rule}.
Construction : Drop the median from B to AC.

\[C>\frac{A}{2}\Longleftrightarrow 1-cos^2 C=\sin^2 C>\sin^2\frac{A}{2}=\frac{1-\cos A}{2}\]
Using the cosine rule and simplifying and substiting the fact that $b=2c$ provides us that we have to prove
\[3c^2> a^2\]
This is straightforward from nothing that since $\angle B$ is obtuse, we have $a^2+c^2<b^2=4c^2$ and hence the result.


Confronted with the box shown in the diagram, you wish to discover something about its internal workings. You have no tools and cannot open the box. You pull on rope B, and it moves rather freely. When you pull on rope A, rope C appears to be pulled slightly into the box. When you pull on rope C, rope A almost disappears into the box.  Describe the box's internal workings.

no diagram and this should be in the puzzles and [b]brainteasers[/b] forum.

Guess: Rope C and rope A are such that rope A is curved and rope C is straight. Rope B is not part of the system.

Something like this:

[code]
+++++++++++
+.........+
+.......b.+
+.........+
aaaa......+
+....a....+
+...c.a...+
+.....a...+
+.....a...+
+.....a...+
++++++a++++[/code]

where b and c ropes are positioned in the viewing axis.


Determine a pair of whole positive numbers $ a$ and $ b$ such that $ ab(a\plus{}b)$ is not a multiple of $ 7$ and $ (a\plus{}b)^7 \minus{} (a^7 \plus{} b^7)$ is a multiple of $ 7^7$

This is a current problem:
[url]http://mathcentral.uregina.ca/mp/current/[/url]


<Summons up best impression I can muster>

"YAY!"

Good job to all of our sixes, the A team for our decent placing, and the B team for having lots of fun!

Next year... NYC will be at least 6 points behind us!!

Man... I should start posting here more often......

since this thread is titled "discussion", i guess i'll throw in a random comment or two.  :lol:

i will try making a "math league trading card" prototype in my "spare time" and then "email" it to "anyone" who's "interested".  that would be amusing.  i have to figure out if we're going for a baseball-card idea or a pokemon-card idea.  the difference being [insert random remarks about pokemon here].

Wow-I have not visited this Upstate NY forum in like 2 years...

But anyway, I concur: Yay!

One should also note that we did totally awesome on the power round, coming in fifth (and beating NYC)!


conics
parabola
calculus
algebra
polynomial
geometry
geometric transformation

I'm not entirely sure where this should go, as it might involve calculus content, but there's probably a more elementary way that I can't find...

i) Find the second-degree polynomial parabola $ g(x)$ orthogonal to $ f(x)\equal{}x^2$ at $ x\equal{}1,\minus{}1$.
ii) Find a general formula for the second-degree polynomial parabola $ g(x)$ orthogonal to $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ at $ x\equal{}x_1,\minus{}\frac{b}{a}\minus{}x_1$ (the latter value of $ x$ is the reflection of $ x_1,f(x_1)$ about the vertex line...  I think)

[hide="You can use an analogous technique for the general case"]First of all, by symmetry, you expect this parabola to be of the form $ f(x) \equal{} ax^2\plus{}b$, since it needs the same axis of symmetry as the first parabola.

Also, $ f(\pm 1)\equal{}1$, so $ a\plus{}b\equal{}1 \implies b \equal{} 1\minus{}a \implies f(x) \equal{} 1\plus{}a(x^2\minus{}1)$.

Now, you want $ f'(x) (x^2)' \equal{} \minus{}1$ at $ x\equal{}1$ ($ x\equal{}\minus{}1$ has been taken into account through symmetry).

Hence, $ (2a)(2) \equal{} \minus{}1 \implies a \equal{} \minus{} \frac{1}{4}$, and so $ f(x) \equal{} 1 \minus{} \frac{1}{4}(x^2\minus{}1)$.[/hide]

[hide="My solution was even more calculus heavy"]We have $ f'(x) \equal{} 2x$. Thus, the tangent lines to $ f(x)$ at $ x \equal{} \pm 1$ are $ \minus{} 2,2$. The perpendicular tangent lines would have slopes $ \frac {1}{2}, \minus{} \frac {1}{2}$. Thus,
$ g'(1) \equal{} \minus{} \frac {1}{2}$
$ g'( \minus{} 1) \equal{} \frac {1}{2}$.
The derivative of a second degree polynomial that a parabola is is obviously linear, so we easily find that $ g'(x) \equal{} \minus{} \frac {x}{2}$. Taking the integral of this, we have that $ g(x) \equal{} \minus{} \frac {1}{4}x^2 \plus{} c$. We know that it's orthogonal to $ f(x) \equal{} x^2$ at $ x \equal{} \pm 1$, so it passes through $ f(x)$ at $ (1,1),( \minus{} 1,1)$. These give basically the same equation, and we find that $ c \equal{} \frac {5}{4}$. Thus, $ g(x) \equal{} \minus{} \frac {1}{4}x^2 \plus{} \frac {5}{4}$.[/hide]


[size=200][color=red]FACT[/color][/size]: this forum is really pathetic and we need more ppl posting.

[size=200][color=red]WHAT WE (OR YOU) ARE GOING TO DO ABOUT IT[/color]: [/size]write something in ur sig encouraging ppl to at least look at the indiana forum.

P.S. i might get bored of this idea, so don't call me a hypocrite if i don't have it in my sig (although i will sometime or the other!)

[hide][hide][hide]look at my signature[/hide][/hide][/hide]

[quote="indianamath"]look at my signature[/quote]

cool! i hope this works *crossing fingers*!!

[quote="mathcrazed"]Yuqing, you have too much time on your hands[/quote]

amen

oookay, it said my message was too short.

[quote="mathcrazed"]Yuqing, you have too much time on your hands[/quote]

hmmm....

1. edit out my name or i will call you by your name, ------(yup, 6 letters)
2. it took me around 40 seconds.