function
algebra
functional equation
IMO
IMO 1992

[quote="orl"]Let $ \,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $ \,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that
\[ f\left( x^{2} \plus{} f(y)\right) \equal{} y \plus{} \left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.\]
[/quote]

Denote with $ P(x,y)$ the property $ f(x^2\plus{}f(y))\equal{}y\plus{}f^{2}(x)$

Let $ c\equal{}f^{2}(0)$. So we have
$ P(0,x) \Rightarrow f^{(2)}(x)\equal{}x\plus{}c\;\;\;x\in \mathbb{R} \: (*)$ where $ f^{(2)}(x)\equal{}f(f(x)).$

Consider $ a,b\in \mathbb{R}$ with $ a\ne b$ and $ f(a)\equal{}f(b)$. So we have
$ f^{(2)}(a)\equal{}f^{(2)}(b) \Rightarrow a\plus{}c\equal{}b\plus{}c \Rightarrow a\equal{}b.$ This is a contraddiction.
So it must be $ f(a)\ne f(b)$ when $ a \ne b$ and $ f(x)$ must be injective.

Consider now $ a\in \mathbb{R}$. From $ (*)$ we have $ f^{(2)}(a\minus{}c)\equal{}a.$
Therefore, $ \forall a\in \mathbb{R} \;\;\; \exists b\in \mathbb{R}, b\equal{}f(a\minus{}c),$ such that $ f(b)\equal{}a.$
So $ f(x)$ must be surjective.
Therefore $ f(x)$ must be bijective.
Then there must be $ u\in \mathbb{R}$ such that $ f(u)\equal{}0$ and this point is unique.

So we have
$ (1)$ $ P(u,u) \Rightarrow f(u^2)\equal{}u$
$ (2)$ $ P(\minus{}u,u) \Rightarrow f(u^2)\equal{}u\plus{}f^{2}(\minus{}u)$

From $ (1)$ and $ (2)$ we have $ u\equal{}u\plus{}f^{2}(\minus{}u) \Rightarrow f(\minus{}u)\equal{}0$

Since $ f(x)$ must be bijective we have $ u\equal{}0$. Therefore $ f(0)\equal{}0, \ c\equal{}0$

The equation $ (*)$ becomes $ f^{(2)}(x)\equal{}x\;\;\;\forall x\in \mathbb{R}\: (**)$

From $ (**)$ we derive that $ f^{(2)}(x)$ is strictly increasing.

So consider $ a,b\in \mathbb{R}$ with $ f(a)>f(b)$. Then $ f(f(a))>f(f(b))$ and this implies $ a>b.$

Consider now $ a,b\in \mathbb{R}$ with $ f(a)<f(b)$. Then $ f(f(a))<f(f(b))$ and this implies $ a<b.$

Therefore we derive that $ f(x)$ is also strictly increasing.

Consider $ a \in \mathbb{R}\;\;a\ne 0$. If $ f(a)>a$ then $ f(f(a))>f(a) \Rightarrow a>f(a).$
This is a contraddiction.

If $ f(a)<a$ then $ f(f(a))<f(a) \Rightarrow a<f(a).$
This is again a contraddiction.

The last two results say that $ \forall x\in \mathbb{R}, \;\;x \ne 0,$ it must be $ f(x)\equal{}x.$

Since $ f(0)\equal{}0$ we have that $ \forall x \in \mathbb{R}$ it must be $ f(x)\equal{}x.$

Therefore
$ \boxed{f(x)\equal{}x,\  \forall x \in \mathbb{R}}$ is a solution of the initial equation and it is the unique solution.

[quote="prester"]
So consider $ a,b\in \mathbb{R}$ with $ f(a) > f(b)$. Then $ f(f(a)) > f(f(b))$ and this implies $ a > b.$

Consider now $ a,b\in \mathbb{R}$ with $ f(a) < f(b)$. Then $ f(f(a)) < f(f(b))$ and this implies $ a < b.$

Therefore we derive that $ f(x)$ is also strictly increasing.
[/quote]

[b][i]The above steps are obviously wrong.[/i] [/b]

We cannot derive that $ f(f(a)) > f(f(b))$ because we do not know that $ a > b$ but we do know that $ f(a) > f(b)$ and then the hypothesys uses the assert we want to demonstrate. It is non-sense.

To demonstrate that $ f(x)$ is strictly increasing, I did not find a best way than previous post (by probability1.01).
Let $ a > b\;\;\; a,b \in \mathbb{R}.$ So $ \exists u \in \mathbb{R}$ such that $ a \equal{} u^2 \plus{} b$.
So $ f(a) \equal{} f(u^2 \plus{} b) \equal{} f^2(u) \plus{} f^{ \minus{} 1}(b) > f^{ \minus{} 1}(b) \equal{} f^{(2)}(f^{ \minus{} 1}(b)) \equal{} f(b)$ and therefore $ f(x)$ is strictly increasing.

Then the demonstration can proceed as before.

Sorry for mistake.

[quote="probability1.01"]Set x = 0 to get $f(f(y)) = y+f(0)^{2}$. We'll let $c = f(0)^{2}$, so $f(f(y)) = y+c$. Then

$f(a^{2}+f(f(b))) = f(b)+f(a)^{2}$
$f(a^{2}+b+c) = f(b)+f(a)^{2}$
$f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}$
$a^{2}+b+2c = b+(a+c)^{2}$
$2c = c^{2}+2ac$

Since this holds for all $a$, it follows that $c = 0$. Now we have $f(0) = 0 \implies f(f(y)) = y$. Note that f must be surjective since we may let y vary among all reals, and f must be injective since if $f(a) = f(b)$, then $a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}$. Finally, if $u > v$, then there is some $t$ s.t. $u = t^{2}+v$, and so $f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)$. Hence f is strictly increasing. It is now clear that since $f(f(y)) = y$, we must have $f(x) = x$ for all x.[/quote]
heloo,
can you explain me how you get $ f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2} $??

[quote="R.Maths"]I think it's easy to chek that $f(0)=0$

thereby $f(f(x))=x$

So , consider the sequence defined by : $f(f(f(....)))=x_{n} $ and $x_{0}=x$

So $f(f(x))=x \implies x_{n+2}=x_{n}$  so for all $n \in \mathbb N $ there existe

$a;b \in \mathbb R $ such that $x_{n}=a+b(-1)^{n} $

for $n=1$ we have $f(x)=a-b=a+b-2b=x-2b$ because $a+b=x$

replacing this values in the equation $ f(f(x))=x$ and we will merely find that $b=0$

then $ ( \forall x \in \mathbb R ) f(x)=x$

Wich is indeed a solution for the initial fonctioanl equation .

Redwane.[/quote]
helooo,
van you explian me why So f(f(x))=x \implies x_{n+2}=x_{n} so for all n \in \mathbb N there existe???

[quote="arshakus"][quote="e.lopes"]The first step is to establish that $f(0) = 0$. Putting $x = y = 0$, and $f(0) = t$, we get $f(t) = t^2$.
Also, $f(x^2+t) = f(x)^2$, and $f(f(x)) = x + t^2$.
We now evaluate $f(t^2+f(1)^2)$ two ways.
First, it is $f(f(1)^2 + f(t)) = t + f(f(1))^2 = t + (1 + t^2)^2 = 1 + t + 2t^2 + t^4$.
Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$. So $t = 0$, as required.

It follows immediately that $f(f(x)) = x$, and $f(x^2) = f(x)^2$.
Given any $y$, let $z = f(y)$. Then $y = f(z)$, so $f(x^2 + y) = z + f(x)^2 = f(y) + f(x)^2$.
Now given any positive $x$, take $z$ so that $x = z^2$.
Then $f(x + y) = f(z^2 + y) = f(y) + f(z)^2 = f(y) + f(z^2) = f(x) + f(y)$.
Putting $y = -x$, we get $0 = f(0) = f(x + -x) = f(x) + f(-x)$. Hence $f(-x) = - f(x)$.
It follows that $f(x + y) = f(x) + f(y)$ and $f(x - y) = f(x) - f(y)$ hold for all $x$, $y$.

Take any $x$.
Let $f(x) = y$. If $y > x$, then let $z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z$.
If $y < x$, then let $z = x - y$ and $f(z) = f(x - y) = f(x) - f(y) = y - x$.
In either case we get some $z > 0$ with $f(z) = -z < 0$.
But now take $w$ so that $w^2 = z$, then $f(z) = f(w^2) = f(w)^2 \ge 0$.
Contradiction. So we must have $f(x) = x$[/quote]
helooo,
can you explain me why Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$[/quote]
yes yes, I have just understand it

Here's my solution:

Taking $y = -f(x)^2$ we get $f(x^2 + f(y)) = 0$ i.e., $\exists b$ such that $f(b)=0$
Then $f(b^2+f(y))=y$ and so $f$ is bijective.

Then $f(b)^2= f(b^2+f(0))=f((-b)^2 +f(0)) = f(-b)^2$ so $f(-b)^2 = 0$
Hence $f(-b)=0=f(b) $ and so $-b=b$ since $f$ is injective, giving us $b=0$
So $f(0)=0$

Then $f(f(x))=x$
Suppose $x>y$ for some $x$ and $y$. Let $x=y+z^2$
Then $f(x)=f(z^2+f(f(y)))=f(y)+f(z)^2>f(y)$ so $f$ is strictly increasing.

It is then well known that the only increasing solutions to $f(f(x))=x$ is $f(x)=x$

So $f(x)=x$ which works.

hi guys,
I found this solution in shortlist, but I think that it is not full, can you look at this solution and explain me dear reader!!!

It is easy to see that f is injective and surjective. From $f(x^2 + f(y)) =$
$f((-x)^2 +f(y))$ it follows that $f(x)^2 = (f(-x))^2$, which implies $f(-x) =
-f(x)$ because f is injective. Furthermore, there exists $z∈R$ such that
$f(z) = 0.$ From $f(-z) = -f(z) = 0$ we deduce that $z = 0$. Now we
have $f(x^2) = f(x^2 + f(0)) = 0 + (f(x))^2 = f(x)^2$, and consequently
$f(x) = f(sqrt(x)^2) > 0$ for all $x > 0$. It also follows that $f(x) < 0$ for $x < 0$.
In other words, f preserves sign.
Now setting $x > 0$ and $y = -f(x)$ in the given functional equation we
obtain
$f(x - f(x)) = f(sqrt(x)^2+ f(-x)) = -x + f(√x)2 = -(x - f(x)).$
But since f preserves sign, this implies that $f(x) = x for x > 0. $Moreover,
since $f(-x) = -f(x),$ it follows that $f(x) = x$ for all x. It is easily verified
that this is indeed a solution.

[quote="orl"]Let $ \,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $ \,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that
\[ f\left( x^{2} \plus{} f(y)\right) \equal{} y \plus{} \left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.\]
[/quote]
Denote by $ P(x,y)$ the assertion $ f(x^2\plus{}f(y))\equal{}y\plus{}f^{2}(x)$.

[hide="1) $f(0) = 0$"]\[P(0,0) \implies f(f(0)) = f(0)^2 \\ \\
P(0,x) \implies f(f(x)) = x+f(0)^2 \implies f(x) + f(0)^2 = f(f(f(x))) = f(x+f(0)^2)
\]

\[P(x,0) \implies f(x^2+f(0)) = f(x)^2 \\ \\
P(-x,0) \implies f(x^2+f(0)) = f(-x)^2
\\
\\
\implies f(-x) = \pm f(x)\]
But,
\[f(-x) = f(x) \implies x+f(0)^2 = f(f(x)) = f(f(-x)) = -x+f(0)^2\]
But this is true only for $x=0$.

Now, take any non-zero real $a$.
We have,
\[-a +f(0)^2 = f(f(-a)) = f(-f(a)) = -f(f(a)) = -(a+f(0)^2) \\ \\
\implies f(0) = 0\][/hide]

[hide="2) Sign of $f(x)$ is the same as the sign of $x$. "]We have,
\[f(f(x)) = x+f(0)^2 \implies f(f(x)) = x\]

\[P(x,0) \implies f(x^2) = f(x)^2\]
Combining this with $f(-x) = -f(x)$, we get the desired.[/hide]

[hide="3) $f(x+1) = f(x) +1$"]
We have,
\[P(1,0) \implies f(1) = f(1)^2 \implies f(1) = 0  \ \text{or } f(1) = 1.\]
But, if $f(1) =0$, then
\[P(x,1) \implies f(x^2) = 1+f(x)^2\]
This is clearly not possible, because
\[P(x,0) \implies f(x^2) = f(x)^2 \not = 1+f(x)^2\]
Therefore, $f(1) =1$.
Now,
\[P(1,x) \implies f(f(x) + 1) = x+1 \implies f(x+1) = f(f(f(x) +1)) =f(x) +1\]
since $f(f(x)) =x$.[/hide]

[hide="4) $f(x)=x \ \forall x \in \mathbb{R}$"]
We have,
\[P(x,1-x^2) \implies f(x^2+ f(1-x^2)) = 1 + (f(x)^2 - x^2)\]
Now,
\[f(1-x^2) = 1+f(-x^2) = 1-f(x^2) \implies f(x^2+ f(1-x^2)) = f(1+(x^2 - f(x^2))) = 1+ f(x^2 - f(x^2))\]
Also, using $f(x)^2 = f(x^2)$, which has been proved earlier, we have
\[1+ f(x^2 - f(x^2)) =  f(x^2+ f(1-x^2)) = 1 + (f(x)^2 - x^2) = 1 + (f(x^2) - x^2)\]
Therefore, we have,
\[f(x^2 - f(x^2)) = f(x^2) - x^2\]
Therefore, from 2nd lemma,
\[x^2 - f(x^2) = 0 \iff f(y) = y\ \forall y \in \mathbb{R}^+ \]
Combining this with $f(-x) = -f(x)$, we get the desired.[/hide]

Let $P(x,y)$ be the assertion that $f(x^2 + f(y))=y+f(x)^2$. By $P(0,y)$, we have that $f(f(y))=y+f(0)^2$ which implies that $f$ is a surjective function. Furthermore, if $f(x)=f(y)$, then it follows that $x+f(0)^2=f(f(x))=f(f(y))=y+f(0)^2$ which implies that $x=y$ and hence that $f$ is injective. Now let $a \in \mathbb{R}$ be such that $f(a)=0$. By $P(a,f(y))$, it follows that $f(a^2+f(f(y)))=f(y)+f(a)^2=f(y)$ which implies that $y=f(f(y))+a^2$ since $f$ is injective. However, as previously established, $f(f(y))=y+f(0)^2$ which implies that $a^2 + f(0)^2=0$ and hence that $a=f(0)=0$ and $f(f(y))=y$. Now by $P(x,0)$, it follows that $f(x^2)=f(x)^2$ and furthermore that $f(x)^2=f(x^2)=f((-x)^2)=f(-x)^2$ which implies that $f(-x)=-f(x)$ since $f$ is injective. Now by $P(x,f(y))$, it follows that $f(x^2+y)=f(x^2+f(f(y)))=f(y)+f(x)^2=f(y)+f(x^2)$. Combining this with the fact that $f(-x)=-f(x)$ yields that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Now note that for all $a \ge 0$, $f(a)=f((\sqrt{a})^2)=f(\sqrt{a})^2 \ge 0$. Hence $f$ is bounded on an interval which implies by the Cauchy functional equation that $f(x)=cx$ for some constant $c \in \mathbb{R}$. Substituting this into the equation yields that $f(x)=x$.

[quote=jasperE3]There's a small detail which is easily fixed in your proof. What if $f(x)=-f(-x)$ for some $x$ but $f(x)=f(-x)$ for some other $x$?[/quote]

Is it correct now? I've edited it. Nice detail.

I think you fixed the wrong part. You are assuming that there exists $x$ such that $f(x)=f(-x)$, use $f(k)=0$ and try to prove that $x$ must be $0$.

[hide=storage]
$P(0,y) \implies f(f(y))=y+f(0)^2 \implies f$ is bijective.
So $\exists a$ such that $f(a)=0$
$P(a,0) \implies f(a^2+f(0))=0$
By injectivity, $a=a^2+f(0)$.
Similarly, $P(0,a) \implies f(0)=a+f(0)^2$.
so we have $f(0)=a-a^2=a+(a-a^2)^2$.
Solving this gives $a=0 \implies f(0)=0$
Then, $P(x,0) \implies f(x^2)=f(x)^2$. Similarly we get $f(f(x))=x$
notice that $f(x^2)=f(x)^2=f(-x)^2 \implies f(x)=-f(x) \forall x$ (by injectivity , $f(x) \ne f(-x)$ except when $x=0$).
Finally, $P(x,f(y)) \implies f(x^2+y)=f(y)+(f(x))^2=f(y)+f(x^2)$.
Now we prove, $f$ is additive. Notice that the above equation implies the following :
$f(a+b)=f(a)+f(b) \forall a,b \geq 0$
$f(a+b)=f(a)+f(b) \forall a \geq 0, b \leq 0$
Finally, choose $x,y >0$, and notice $f(x^2+y)=-f((-x^2)+(-y))=-f(-y)-f(-x^2) \implies f$ is additive over negative reals too.
Now notice that $f(x^2)=f(x)^2 \geq 0$, that is $f$ is bounded below over $[0, \text{infinity})$, so we must have that $f$ is linear.
Now, putting in $f=ax$ easily gives $a =  1 \implies f(x)=x$

[quote=orl]Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that \[ f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}. \][/quote]

Let $P(x,y)$ the assertion of the given F.E.
$P(0,x)$
$$f(f(x))=x+f(0)^2 \implies f \; \text{bijective}$$
Let $c$ be a real such that $f(c)=0$ (it exists becuase $f$ is bijective)
$P(c,f(x))$ and injectivity
$$y+f(0)^2=f(f(y))=y+c^2 \implies c=\pm f(0) \implies f(\pm f(0))=0$$
On the last equation if $y=-c^2$ then $f(-c^2)=c$ but by $P(c,c)$ we have $f(c^2)=c$ thus by injectivity $c=0$.
$P(0,x)$
$$f(f(y))=y \implies f \; \text{involution}$$
$P(x,0)$
$$f(x^2)=f(x)^2 \implies f(f(x)^2)=x^2$$
And now by replacing this on the original F.E.
$$f(x^2+f(y))=y+f(x^2)$$
Let $Q(x,y)$ the assertion of this F.E.
Choose $x>y$ so that $x=y+a^2$ where $a$ is a real number.
$Q(a,y)$
$$f(x)=f(y)+f(a^2)>f(y) \implies f \; \text{increasing}$$
Thus $f$ involutive+increasing means that $f$ is the identity.
Thus the only sol is $f(x)=x$ for every real $x$. Thus we are done  :D

We'll use $P(x,y)$ for the assertion $(x,y)$ into the FE:

$P(0,y)$ gives $f(f(y))=y+f(0)^2\Longrightarrow f$ is bijective (as $y$ runs from $-\infty$ to $+\infty$, $y+f(0)^2$ covers $\mathbb{R}$, so $f$ in surjective and if we assume that $\exists a\neq b\in\mathbb{R}$ such that $f(a)=f(b)$ then $b+f(0)^2=f(f(b))=f(f(a))=a+f(0)^2$, contradiction$\implies$ $f$ is bijective)

Therefore $\exists$ a unique $ z\in\mathbb{R}$ such that $f(z)=0$ since $f$ is bijective:
$P(0,0)\Longrightarrow f(f(0))=f(0)^2$ $(1)$
$P(z,f(0))\Longrightarrow f(z^2+f(f(0)))=f(0)+f(z)^2\overset{(1)}{\Longrightarrow} f(z^2+f(0)^2)= f(0)\Longrightarrow z^2+f(0)^2=0\Longrightarrow z=0, \boxed{f(0)=0}$

Now $P(0,y)$ gives $f(f(y))=y$ and $P(x,0)$ gives $f(x^2)=f(x)^2\implies f(x)=-f(-x)$. Using these two results we have that $P(x,f(y)):\quad f(x^2+y)=f(y)+f(x^2)$. This is the Cauchy equation and as $f(x^2)=f(x)^2\geq 0$ this means that $f$ is monotonic, which implies that $f(x)=cx$ for some constant $c\in \mathbb{R}$. As $c\times 1^2=f(1^2)=f(1)^2=c^2\times 1^2\Longrightarrow c\in\{0,1\}$. The function $f\equiv 0$ is not a solution as $f$ is bijective, so the only possible solution is $f(x)=x$. And indeed:
$$f(x^2+f(y))=f(x^2+y)=x^2+y=y+f(x)^2$$
Therefore $f(x)=x$ is the only solution to the FE.

[quote=orl]Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that \[ f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}. \][/quote]

$P(0;y)$ we have $f(f(y))=y+(f(0))^2, (1)$ so f is bijective
Exist a such that $f(a)=0$
$P(a;0):$ $f(a^2+f(0))=0=f(a)$ -> $a^2+f(0)=a, (2)$
Let $y=a$ in $(1):$ $f(0)=a+f(0)^2, (3)$
From $(2)$ and $(3)$ we have $f(0)=a=0$
We get $f(x^2)=f(x)^2, (*)$ and $f(f(y))=y$
$P(x;f(y)):$ $f(x+y)=f(x)+f(y), (4)$ for all $x \geq 0$ and all real $y$
$Q(x;y)$ in $(4)$
Let $x>0$ then from $(4)$ we have $0=f(0)=f(x)+f(-x)$ so f is odd
From $(*)$ we have $f(x) \geq 0$ for all $x \geq 0, (5)$
From $(4)$ and $(5)$ we have $f(x)=x$ , which is a familiar problem
Q.E.D

A rather lengthy solution IG ;-;(Pls correct if I am wrong at some point)
We claim that only $f(x)=x$ which can be easily checked.

Let the assertion be $P(x,y)$.
It is easy to see $f(x)$ is injective.
$P(x,-f(x)^2) \Rightarrow f(x^2+f(-f(x)^2)=0$ Since $f(x)$ is injective it follows that $ x^2+f(-f(x)^2)=t ...(1) $ for some unique $t$
Now, Plugging $x=t$ in $(1)$ we get $t^2+f(0)=t$
Again, $P(x,t) \Rightarrow f(x^2)=f(x)^2 +t \Rightarrow f(0)=f(0)^2+t$
Combining the above two expressions we get $f(0)=0$
Plugging $x=f(x)$ in $(1)$ we get $f(-x)=-f(x)$
Again,$P(x,-x^2) \Rightarrow f(x^2-f(x)^2)=f(x)^2-x^2$
Now lets assume $f(a)=-a$ for some $a \neq 0$ also assume WLOG that $a>0$ since $f(-x)=-f(x)$
Then $P(\sqrt{a},a) \Rightarrow f(\sqrt{a})^2=-a$ which is a contradiction. So, $a=0 \Rightarrow f(x)^2=x^2$
Now we fix the point wise error.
assume for $f(a)=a,f(b)=-b$
$P(a,b) \Rightarrow f(a^2 -b)=a^2+b$ Contradiction.
$f(x)=x$ is the only solution(since $f(x)=-x$ don't work).

$P(0,x): f(f(x))=x+f(0)^2$, so $f$ is bijective.

Claim: $f(0)=0$
$P(x,f(y)): f(x^2+y+f(0)^2)=f(x)^2+f(y)$.

$P(f(x),y): f(f(x)^2+f(y))=(x+f(0)^2)^2+y$.

Note that $f(f(x)^2+f(y))=x^2+y+2f(0)^2$. Suppose FTSOC that $f(0)\ne 0$.

So we get \begin{align*}
x^2+y+2f(0)^2=x^2+2xf(0)^2+f(0)^4+y \\
2f(0)^2=2xf(0)^2+f(0)^4 \\
2=2x+f(0)^2\forall x \\
\end{align*}

This is a contradiction because it implies that $f(0)^2+2\cdot 332.5=f(0)^2+665=2\implies f(0)^2=-663$ and also $f(0)^2+2\cdot 93.5=f(0)^2+187=2\implies f(0)^2=-185$.

So $f(0)=0$, which implies $f(f(x))=x$.

$P(x,f(y)): f(x^2+y)=f(x)^2+f(y)$.

$P(x,0): f(x^2)=f(x)^2$.

So $f(x^2+y)=f(x^2)+f(y)$.

Claim: $f$ is odd.
Proof: We have $f(x^2)=f(-x)^2=f(x)^2$. If $f(x)=f(-x)$ for some $x$, then we have $f(f(x))=f(f(-x))\implies x=-x\implies x=0$. So $f(x)=-f(-x)$ for all $x$ since $f(0)=0$. $\blacksquare$

For all reals $x$ and $y$ that are not both negative, we get $f(x+y)=f(x)+f(y)$. Now if $u$ and $v$ are negative, we get $f((-u)+(-v))=-f(u+v)=f(-u)+f(-v)=-(f(u)+f(v))$, so $f(u+v)=f(u)+f(v)$. Thus, $f$ is additive.

From $f(x^2)=f(x)^2$, $f$ is bounded over the interval (always nonegative) $[0,\infty]$, so $f(x)=kx$ for some constant $k$.

From involution we get $k=1$ or $k=-1$. From $f(x^2)=f(x)^2$, we get $k=1$. So $\boxed{f(x)=x}$ is the only solution.

Edit June 10: added a crucial word.

[b][u]Claim: $f(0)=0.$[/u][/b]
[i]Proof.[/i] Let $P(x,y)$ denote the original equation.
$P(0,y)\implies f(f(y))=y+(f(0))^2.$
$P(0,0)\implies f(f(0))=(f(0))^2.$
$P(0,1)\implies f(f(1))=1+(f(0))^2.$
Applying $f$ to both sides of $P(x,y)$ yields,
$x^2+f(y)+(f(0))^2=f(y+(f(x))^2).$
$P(1,f(0))\implies f(f(0)+(f(1))^2)=1+2(f(0))^2.$
$P(f(1),0)\implies f(f(0)+(f(1))^2)=(1+(f(0))^2)^2.$
From this the claim follows. And $f(f(x))=x.$ $\blacksquare$

---------

[b][u]Claim: $f$ is strictly increasing.[/u][/b]
[i]Proof.[/i] Let for any reals $a,b$, $a>b.$
Note that $f(f(\sqrt{a-b})) =\sqrt{a-b} \implies f(\sqrt{a-b}) \neq 0.$
$P(\sqrt{a-b}, b) \implies (f(\sqrt{a-b}))^2+f(b) =f(a).$
And thus our claim follows. $\blacksquare$

----------

[b][u]Claim: $f(x)=x,$ which in fact holds.[/u][/b]
[i]Proof.[/i] It's pretty well known and easy to show that the solution of any strictly increasing and involutive function is an identity function. This completes the proof. $\blacksquare$

[hide=2nd solution]
Let $P(x,y)$ denote the given assertion.

Note that $f$ is bijective. Comparing $P(x,y)$ with $P(-x,y)$ we conclude $f$ is odd. Note that this forces $f(0)=0.$ Then $P(0,x)$ gives $f$ involution. So we get $f$ is additive (e.g. $P(x,0)$ gives $f(x^2)=f(x)^2$ then $P(x,f(y)).$)

$P(x,0)$ implies $f$ is bounded. Hence $f(x)=cx$ and $P(x,y)$ yields $c=1$ so our answer is the identity, which fits. [/hide]

[hide = Solution] I claim that the only solution is $f(x) = x$.
Denote by $P(x, y)$ as the property pertaining to the original functional equation.

$\underline{P(0, y)}$: $\implies f$ is surjective.
$\underline{P(0, y) - P(0, x)}$: $\implies f$ is injective. (If we assumed $f(x) = f(y)$ to prove injectivity). Thus $f$ is bijective.
$\underline{P(x, x)}$: $\implies f(x^2 + f(x)) = f(x)^2$.
By the surjectivity of $f$, there exists some $\alpha \in \mathbb{R}$ such that $f(\alpha) = 0$.
$\underline{P(0, \alpha)}$: $f(0) = \alpha + f(0)^2$.
$\underline{P(\alpha, 0)}$: $f(\alpha^2 + f(0)) = 0$. Thus we have that as $f(\alpha) = f(\alpha^2 + f(0)) = 0$, then from the injectivity of $f$ $\alpha = \alpha^2 + f(0)$. Now using the above, we have that $\alpha - \alpha^2 = f(0) = \alpha + f(0)^2 \implies -\alpha^2 = f(0)^2 leq 0$. Hence $\alpha = f(0) = 0$.

Now the first equation ($P(0, y)$) becomes $f(f(y)) = y$.
$\underline{P(x, 0)}$: $\implies f(x^2) = f(x)^2$.
$\underline{P(x, f(y)}$: $\implies f(x^2 + f(f(y))) = y + f(x)^2 \Longleftrightarrow f(x^2 + y) = f(x^2) + f(y)$.
$\underline{P(x, -x^2)}$: $\implies f(x^2) = -f(-x^2)$. Thus $f$ is odd.
Now setting $x \mapsto \sqrt{x}$ in the second last equation yields Cauchy's functional equation. Now to show that it has "Cauchy's intended solution", let $a \geq 0, b \in \mathbb{R}$ be reals. Then $f(a + b) = f(a) + f(b)$. If $a < 0, b \in \mathbb{R}$, then $f(a + b) = -f(-a - b) = -(f(-a) + f(-b)) = f(a) + f(b)$. Thus $f(x) = cx$, for some constant $c$. Plugging this back into the original equation yields $c = 1$, so that $\boxed{f(x) = x}$ is the only solution. $\blacksquare$. [/hide]

Cool problem! I think I understand Cauchy Functional Equations a little bit now.

[i]Solution.[/i] The answer is $f \equiv \mathrm{id}$ which clearly works. We will proceed to show that this is unique. Denote
$P(x,y)$ as the assertion to the functional equation.

One can quickly see that $f$ is bijective from $P(0,x)$. Let $\alpha$ be "the zero" of $f$. Now we show that $\alpha = 0$. For this consider the following substitutions.
[list]
    [*] $P(\alpha, 0)$ gives us $f(\alpha^2 + f(0)) = 0 = f(\alpha)$. By injectivity, we have that $\alpha^2 + f(0) = \alpha$.
    [*] $P(0, \alpha)$ gives us $f(0) = f(0)^2 + \alpha$. Now solve the simultaneous equations to get $f(0) = 0$ as desired.
[/list]
$P(x,0)$ reveals $f(x^2) = f(x)^2 \ge 0$. So $f$ is positive over positive domain values. $P(0,x)$ gives that $f$ is an involution. Finally observe that $P(x,f(y))$ gives the following partial "Cauchy Condition"
    \[f(x^2 + y) = f(x^2) + f(y) \implies f(a+b) = f(a) + f(b)\]
for all reals $a$ and all $y \ge 0$. One can extend this partial condition to a full "Cauchy" easily. Observe that for some arbitrary $r \in \mathbb{R}$
    \[f(0) = f(r-r) = f(r) + f(-r) \iff f(r) = -f(-r)\]
which shows $f$ is odd. With the odd nature of $f$, we can now say that $f$ is completely additive over $\mathbb{R}$. We will now present two finishes.
[list]
    [*] Since $f(x) > 0$ for $x > 0$, $f$ must be strictly increasing. For strictly increasing additive function $f$, it is well-known that $f(x) = ax$. Plugging this in would finish the solution.
    [*] The second finish uses the involution. Since $f$ is an involution and strictly increasing, $f$ must be the identity function.
[/list]
So we're done. $\blacksquare$


If $ a,b,c>0$ such that $ abc\equal{}1$ , prove that :

\[ a^3\plus{}b^3\plus{}c^3\plus{}\frac{9}{a\plus{}b\plus{}c}\ge 6\]

$ (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c) \plus{} 9 \ge 6(a \plus{} b \plus{} c)$

Using the fact that $ a \plus{} b \plus{} c\ge 3(abc)^{1/3} \equal{} 3$, we have

$ (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c) \plus{} 9\ge 3(a^3 \plus{} b^3 \plus{} c^3) \plus{} 9$
$ \equal{} 3((a^3 \plus{} 1 \plus{} 1) \plus{} (b^3 \plus{} 1 \plus{} 1) \plus{} (c^3 \plus{} 1 \plus{} 1)) \minus{} 9\ge 3(3a \plus{} 3b \plus{} 3c) \minus{} 9$
$ \equal{} 9(a \plus{} b \plus{} c) \minus{} 9 \equal{} 6(a \plus{} b \plus{} c) \plus{} 3(a \plus{} b \plus{} c \minus{} 3)\ge 6(a \plus{} b \plus{} c)$

Cheers,

Rofler

[quote="Pain rinnegan"]If $ a,b,c > 0$ such that $ abc \equal{} 1$ , prove that :
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} \frac {9}{a \plus{} b \plus{} c}\ge 6\]
[/quote]

The inequality is equivalent with :

\[ \frac{a^3\plus{}b^3\plus{}c^3}{3abc}\plus{}\frac{3\sqrt[3]{abc}}{a\plus{}b\plus{}c}\ge 2\]

which is well-known .

[quote="Pain rinnegan"]If $ a,b,c > 0$ such that $ abc \equal{} 1$ , prove that :
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} \frac {9}{a \plus{} b \plus{} c}\ge 6\]
[/quote]
Another solution
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} \frac {9}{a \plus{} b \plus{} c}\ge\ (a \plus{} b \plus{} c)\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3} \plus{} \frac{9}{{a \plus{} b \plus{} c}}\ge 2\sqrt {3(a^2  \plus{} b^2  \plus{} c^2 )}  \ge 2\sqrt {9\sqrt[3]{{a^2 b^2 c^2 }}}  \equal{} 6$ :D


$ a,b,c\geq 0$ such that $ a\plus{}b\plus{}c\equal{}3$,prove that:
\[ \frac{13}{12}(ab\plus{}bc\plus{}ca)\plus{}3\geq \sqrt{ab}(a\plus{}1)\plus{}\sqrt{bc}(b\plus{}1)\plus{}\sqrt{ca}(c\plus{}1)\]
and when does the equal hold?

:maybe: No one interests  in this inequlity?


got 3 word probs i dont get :

1)

A jar 13 black and 12 red marbles. If one marble is drawn at random, what is the probability that it is black.

2)

If two marbles are drawn randomly from the jar in problem 1, what is the probability the first is red and second is black if the marbles drawn....

a) with replacment?                              b) without replacement?

3)

At 10 AM, Eve started out at 55 kilometers per hour to catch Adrienne, who had a one hour head start. Eve was still 100 kilometers behind at 6PM, how fast Adrienne driving?

Okay, these are relatively simple problems as long as you understand how probability works. Remember, probability is usually:

$\frac{number\ of\ desired\ outcomes}{total\ number\ of\ outcomes}$

And if you want two events to both occur, just multiply their individual probabilities.

[hide="Problem 1"]Simply:

$\frac{13}{25}$[/hide]
[hide="Problem 2"]This one is a bit longer. If each marble is replaced after recording the color:

$\frac{12}{25}*\frac{13}{25}=\frac{156}{625}$

Without replacement the number of marbles in the bag will decrease after each drawing:

$\frac{12}{25}*\frac{13}{24}=\frac{13}{50}$[/hide]

[hide="Problem 3"]$v_at_a-v_et_e=d\\ t_a=t_e+1\\ t_e=8\\ v_e=55\\ d=100$

Rearranged:
$v_a=\frac{d+v_et_e}{t_a}\\ v_a=\frac{100km+55kph8hrs}{9hrs}\\ v_a=60kph$[/hide]

You need to know that probability is $\frac{\text{number of working outcomes}}{\text{number of total outcomes}}$.
[hide="numero uno"]
There are $13$ black marbles and 25 total. So there are 13 working outcomes, and 25 total outcomes. The probability is thus $\frac{13}{25}$[/hide]
[hide="numero dos"]
For this you need to know multiplying probabilities.
For two events, you can multiply their individual probabilities to find the total probability of one following the other.
In this case, there is a $\frac{12}{25}$ chance that a red marble is drawn first.
Now, if the problem says with replacement, that means that you put the marble back after you draw it. If it says without replacement, that means you leave it out for the next time you pick a marble.
Let's deal with replacement first.
As we saw above, the probability that we draw a black marbe is $\frac{13}{25}$.
Multiplying that with the probability that we draw a red marble first, we get that it is
$\frac{12}{25}\times\frac{13}{25}$, or $\frac{157}{625}$.
For the without replacement, the probability of drawing a black marbe changes. Since we don't replace the red marble we drew first, there are now only 24 total outcomes for the second drawing. There are still 13 ways to draw a black marble.
So the probability is $\frac{12}{25}\times\frac{13}{24}=\frac{13}{50}$.\
[/hide]
[hide="numero tres"]
Let's say that Adrienne's speed is $x$.
That is how far she traveled in the one hour head start.
Then the distance she traveled is $d=9x$
The distance that Eve traveled is $d-100=8(55)$
So solving that equation, we get $540$, so she traveled 540 miles.
Plugging that into the second equation, we get
$540=9x$, and $x=60$[/hide]


algebra
polynomial
function
trigonometry
real analysis
real analysis unsolved

Given the nodes \[ {t_0} \equal{} 0,{t_1} \equal{} \frac{\pi }
{2},{t_2} \equal{} \pi ,{t_3} \equal{} \frac{\pi }
{4}\] and the function \[ f(x) \equal{} \sin x\]

How do I calculate:

a) The interpolation polynomial P_0 regarding the node t_0
b) The interpolation polynomial P_1 regarding the nodes t_0,t_1
c) The interpolation polynomial P_2 regarding the nodes t_0,t_1,t_2
d) The interpolation polynomial P_3 regarding the nodes t_0,...,t_3

What kind of interpolation?  Do you want [url=http://en.wikipedia.org/wiki/Lagrange_interpolation]Lagrange interpolation[/url]?  In the link I gave you, they work through an example for the tangent function, which is of course a trig function like sine.  The nodes are not the same, but the technique is illustrated.

[quote="JRav"]What kind of interpolation?  [/quote]

Hi!

Thank you for your answer. A Newton interpolation with divided differences is requested. I will take a look at your example.

Hi!

I have a problem. In case a), we have only one node. So we have the pair
\[ \left( {{t_0},\sin ({t_0})} \right) \equal{} (0,0)\]

Now, using divided differences, I would have to find a polynomial of the order 0 (zero, because we have only one pair), which goes through the pair (0,0). But this would be the constant 0...

In the cases where I have two or more pairs (cases b,c,d), I actually can construct a polynomial, but what happens with case a?

Yes, but is this how this problem should be approached? Meaning, is P_0 = 0 the interpolation polynomial for sin(x) regarding the node t_0 = 0?

If you want a $ \le$ zero-degree polynomial, the only thing you can do is match the value at one point. Yes.


Demonstrati ca (2^32+1) este divizibil cu 641.

Se spune ca Euler a gasit o solutie la asta in 5 minute :D
[Am rezolvarea intr-un caiet,cu solutia data de Euler]
E importanta relatzia,caci arata ca nu toate numerele Fermat(de forma $2^{2^{n}}+1)$[parca asa le zice] sunt prime.
Apropo,s-a demonstrat ca exista o infinitate de numere prime de tip Fermat?Sau o infinitate de numere compuse?

asa e!!!asa scrie si in enciclopedie (e una veche de prin 70').Si mai scrie ca probabil Fermat a verificat pana la n=4 si probabil ca mai departe (pentru n=5) nu a mai verificat.astfel , Euclid a gasit un contraexemplu pentru n=5. Practic problema se rezuma la demonstratia pe care Euclid a facut-o in 5 minute (asa cum spune Andrei Ciupan -in enciclopedie nu mai scrie si asta).

Faza cu 5 minute ne-a spus-o un profesor la pregatirea lotului,atunci cand ne-a dat si rezolvarea.Noi toti am fost uimiti ca a facut-o asa de repede,iar profesorul ne-a replicat:Euler era un tip destept :D

Eu am pus problema pe site si pentru faptul ca e interesanta, dar mai ales pentru povestea ei.

folosind ordinul subgrupului generat de $2$ in grupul multiplicativ $(\mathbb{Z}_{p},\cdot)$, unde $N=2^{32}+1$ si $p|n$, $p$ prim, se arata ca orice divizor prim al acestui numar este congruent cu $1$ modulo $64$ :
vorbind mai elementar :
fie $p$ un divizor prim al lui $N$.
fie $s$ cel mai mic numar astfel incat $2^{s}\equiv 1\pmod{p}$. atunci, deoarece $2^{32}\equiv-1\pmod{p}$, rezulta ca $2^{64}\equiv 1\pmod p$. daca $s<64$, atunci $s$ nu este putere a lui $2$ (altfel am avea $2^{32}\equiv 1\pmod{p}$, si nu e bine), si deci $(s,64)=1$, si asta ar insemna ca $2\equiv 1\pmod{p}$, care iar nu e bine. deci $s=64$. cum $2^{p-1}\equiv 1\pmod{p}$, rezulta ca $2^{(s,p-1)}\equiv 1\pmod{p}$. dar cum $s$ este cel mai mic numar cu aceasta proprietate, rezulta ca $(s,p-1)=s$, adica $s|(p-1)$. cum $s=64$, rezulta ca $p\equiv 1\pmod{64}$, adica ce vroiam sa arat. de aici, verificarea este manuala, deoarece se fac doar vreo 7-8 impartiri.


Find the last two digits of the following sum: \[ 5! \plus{} 10! \plus{} 15! \plus{} \cdots \plus{} 100!\]

Since three numbers in $ 10!$ are $ 2$, $ 5$, and $ 10$, the last two digits will be $ 00$.  This holds true for $ n \ge 10$.

Since the last two digits of $ 5!$ are $ 20$, our answer is simply $ \boxed{20}$.

Ahhhh I sillied this one by looking at "last digits"
and my second chance i just typed the last digit again
notice every factorial num more or equal to 5! ends in 0....
and any factorial num more or equal to 10! ends in 00...
also i mean greater not more


inequalities
function
inequalities proposed

Nice inequality, Reyes! :lol:
I obtained some similar nice inequalities:

$\frac{1}{22a^2+5bc}+\frac{1}{22b^2+5ca}+\frac{1}{22c^2+5ab} \geq \frac{1}{(a+b+c)^2}$

and

$\frac{1}{2a^2+7bc}+\frac{1}{2b^2+7ca}+\frac{1}{2c^2+7ab} \geq \frac{1}{ab+bc+ca}$.

[b]Remark. [/b]The second ineq. is wrong (see below)  ;)

obtained?

Can you obtain these inequalities from something?

[quote="Vasc"]Nice inequality, Reyes! :lol:
I obtained the similar nice inequalities:

$\frac{1}{22a^2+5bc}+\frac{1}{22b^2+5ca}+\frac{1}{22c^2+5ab} \geq \frac{1}{(a+b+c)^2}$

and

$\frac{1}{2a^2+7bc}+\frac{1}{2b^2+7ca}+\frac{1}{2c^2+7ab} \geq \frac{1}{ab+bc+ca}$.   ;)[/quote]
Can you show me proofs for them,Vasc  ?

The second ineq is wrong. I made a miscalculation. The corrected form is:
$\frac{1}{8a^2+bc}+\frac{1}{8b^2+ca}+\frac{1}{8c^2+ab} \geq \frac{1}{ab+bc+ca}$.

I say to wait first some solutions, Reyes.

[quote="georgesand"]Can you obtain these inequalities from something?[/quote]
Of course, from a more general ineq.

I solve first problem of Vasc:
[tex] \sum \frac {1}{22a^2+5bc}\geq\ \frac{1}{(a+b+c)^2}[/tex]
<=>[tex] \sum \frac{1}{22a^2+5bc}-\frac{9}{22(a^2+b^2+c^2+5(ab+bc+ca)}\geq\ \frac{1}{(a+b+c)^2}-\frac{9}{22(a^2+b^2+c^2)+5(ab+bc+ca)}[/tex]
We uses the equality: [tex] \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{9}{x+y+z}=\frac{\sum x(y-z)^2}{xyz(x+y+z)}[/tex] for [tex]x=22a^2+5bc ,y=...[/tex].Then,after some calculations,our inequality becomes:[tex] \sum (b-c)^2Ma\geq\ 0[/tex]. [tex]Ma=[2(a+b+c)^2(22b+22c-5a)^2-13(22b^2+5ca)(22c^2+5ab)][/tex]...
now assume that [tex]a\geq\ b\geq \ c.[/tex].We have [tex] (a-c)^2Mb+(b-c)^2Ma\geq\ (b-c)^2(Mb+Ma)[/tex]
[tex] Mc\geq 0[ /tex] and [tex] Mb+Ma\geq 0[/tex].And we are done.
The second problem can be solved in the same way

My proof is similar spdf's proof .
[tex]<=>\sum\frac{\sum a^2 +\sum bc}{2a^2+bc}\geq 6 <=>\sum\frac{b^2+c^2+ab+ac-3a^2-bc}{2a^2+bc} \geq 0[/tex]
Since : $-bc \geq -\frac{b^2+c^2}{2}$ ( By AM-GM) ; We try to prove that :
[tex]\sum\frac{b^2+c^2+2ab+2ac-6a^2}{2(2a^2+bc)}\geq 0 (1)[/tex].
We have equality :
[tex]\frac{b^2+c^2+2ab+2ac-6a^2}{2a^2+bc} = \frac{(b-a)(b+3a)+(c-a)(c+3a)}{2a^2+bc}[/tex] .Thus :
[tex](1) <=> \sum (a-b).(\frac{3b+a}{2b^2+ac}-\frac{b+3a}{2a^2+ab})\geq 0[/tex]
[tex]<=>\sum (a-b)^2.M_c \geq 0[/tex] .Then assume $a\geq b\geq c $ , and solve similarly spdf's solution . :)

There are some typist's errors in spdf's proof, namely

$ \sum \frac{1}{22a^2+5bc}-\frac{9}{22(a^2+b^2+c^2+5(ab+bc+ca)}\geq\ \frac{1}{(a+b+c)^2}-\frac{9}{22(a^2+b^2+c^2)+5(ab+bc+ca)}$,

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{9}{x+y+z}=\frac{\sum x(y-z)^2}{xyz(x+y+z)}$,

$Ma=(22a^2+5bc)[2(a+b+c)^2(22b+22c-5a)^2-13(22b^2+5ca)(22c^2+5ab)]$.

I think the ineq $Mb+Ma \ge 0$ is more difficult than the original ineq. ;)

Reyes, the ineq. (1) in your proof is not true. Put $a=0$ and $b=c=1$.  :P

Are you sure Reyes your solution is OK?

I have made all computation and I have found for $M_a +M_b$ the expression

$M_a+M_b=4a^2b^2 +12b^2c^2 +12c^2a^2 +13a^2bc+13ab^2c-6abc^2-4b^3c-6b^3a-6a^3b-4a^3c+2c^3a+2c^3b$

which cannot be positive.

Am I wrong? Can you please post a more detailed solution?

Thank you very much.

Thanks .You're right .I 'll try to prove it .This problem is difficult for me . :(

Reyes, which is the source of the inequality you posted ?

It's belong to a friend of mine .I had a proof but i don't like it .It isn't nice .

What do you mean when you say it is not nice ?

It's so bad for me .I have checked it .It isn't true .Sorry for my stupidity. :(

Sorry for my mistake :) .The rest is not as easy as I think.To prove [tex]\sum (b-c)^2Ma\geq\ 0[/tex],I think we can use the substitution [tex] b=c+x;a=b+y=c+x+y[/tex](assume that [tex]a \leq\ b\leq\ c[/tex])),x,y are positive real numbers and replace c by a function of x and y.I will try to complete the solution.Can you post your proof,Vasc?

Only an indication for the moment.
Use Cauchy-Schwarz ineq. for the original ineq.

can you tell me your more generalized ineq, vasc??? :)

how can we apply cauchy for this one??
 :D

I think the ineq. holds

$\sum\frac{1+p}{a^2+pbc} \ge \frac{9(1+q)}{\sum{a^2}+q\sum{bc}}$
for $p>0$ and $q>-1$ such that $(1+p)(2+1/p) \ge \frac{9(1+q)}{2+q}$.
I haven't a complete solution to this general ineq.
For $p=1/2$ and $q=1$ we get the original ineq. To prove this one with Cauchy,
use $\sum{(b+c)^2(2a^2+bc)}$.

[quote="Reyes"]I don't think your inequality can prove my ineq .Because, as you said :
$ \sum \frac{1}{2a^2+bc}\geq \frac{9}{2\sum a^2 +\sum ab }$ . Thus we need to prove that :
$ \frac{3}{2\sum a^2+\sum ab } \geq \frac{2}{\sum a^2 +\sum ab}$ .
$\sum ab \geq \sum a^2$
But this ineq isn't true .Am i wrong ?[/quote]

why?
Vasc used cauchy in this way,
\[
\sum {\frac{1}{{2a^2  + bc}}}  \ge \frac{{4\left( {a + b + c} \right)^2 }}{{\sum {\left( {b + c} \right)^2 (2a^2  + bc)} }}
\]

I understand you .I  believed it's :
$\sum\frac{1+p}{2a^2+pbc}\geq \frac{9(1+p)}{2\sum a^2 +p\sum ab }$ .It's true but it can't prove my ineq . :D .Sorry Vasc and thanks you because of your proof .

If I apply Vasc method at

$\displaystyle \frac{1}{8a^2+bc}+\frac{1}{8b^2+ca}+\frac{1}{8c^2+ab} \geq \frac{1}{ab+bc+ca} $

I obtain

$3\sum (a^2+b^2)ab +4abc\sum a \geq 10 \sum a^2b^2 $

But this is not true, Vasc?

[quote="manlio"]If I apply Vasc method at

$\displaystyle \frac{1}{8a^2+bc}+\frac{1}{8b^2+ca}+\frac{1}{8c^2+ab} \geq \frac{1}{ab+bc+ca} $

I obtain

$3\sum (a^2+b^2)ab +4abc\sum a \geq 10 \sum a^2b^2 $[/quote]

This inequality is wrong (take a = 1, b = 2, c = 2 for a counterexample).

But proving the inequality

$\frac{1}{8a^2+bc}+\frac{1}{8b^2+ca}+\frac{1}{8c^2+ab} \geq \frac{1}{bc+ca+ab}$

for positive a, b, c is actually easy:

We have

$\frac{1}{8a^2+bc}+\frac{1}{8b^2+ca}+\frac{1}{8c^2+ab}=\frac{\left(bc\right)^2}{\left(bc\right)^2\left(8a^2+bc\right)}+\frac{\left(ca\right)^2}{\left(ca\right)^2\left(8b^2+ca\right)}+\frac{\left(ab\right)^2}{\left(ab\right)^2\left(8c^2+ab\right)}$.

After [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]the Cauchy-Schwarz inequality in the Engel form[/url], we have

$\frac{\left(bc\right)^2}{\left(bc\right)^2\left(8a^2+bc\right)}+\frac{\left(ca\right)^2}{\left(ca\right)^2\left(8b^2+ca\right)}+\frac{\left(ab\right)^2}{\left(ab\right)^2\left(8c^2+ab\right)}\geq\frac{\left(bc+ca+ab\right)^2}{\left(bc\right)^2\left(8a^2+bc\right)+\left(ca\right)^2\left(8b^2+ca\right)+\left(ab\right)^2\left(8c^2+ab\right)}$,

and on the other hand,

$\left(bc\right)^2\left(8a^2+bc\right)+\left(ca\right)^2\left(8b^2+ca\right)+\left(ab\right)^2\left(8c^2+ab\right)\leq\left(bc+ca+ab\right)^3$

(that's because

$\left(bc+ca+ab\right)^3-\left(\left(bc\right)^2\left(8a^2+bc\right)+\left(ca\right)^2\left(8b^2+ca\right)+\left(ab\right)^2\left(8c^2+ab\right)\right)$
$=3abc\left(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\right)$,

what is clearly $\geq 0$). Thus,

$\frac{1}{8a^2+bc}+\frac{1}{8b^2+ca}+\frac{1}{8c^2+ab}=\frac{\left(bc\right)^2}{\left(bc\right)^2\left(8a^2+bc\right)}+\frac{\left(ca\right)^2}{\left(ca\right)^2\left(8b^2+ca\right)}+\frac{\left(ab\right)^2}{\left(ab\right)^2\left(8c^2+ab\right)}$
$\geq\frac{\left(bc+ca+ab\right)^2}{\left(bc\right)^2\left(8a^2+bc\right)+\left(ca\right)^2\left(8b^2+ca\right)+\left(ab\right)^2\left(8c^2+ab\right)}$
$\geq\frac{\left(bc+ca+ab\right)^2}{\left(bc+ca+ab\right)^3}=\frac{1}{bc+ca+ab}$,

and the inequality is proven.

Vasc's other two inequalities seem to be less simple. ;)

  Darij


calculus integration function geometry conics hyperbola

Does anybody have a way to prove that e is convergent (besides plugging in big numbers and showing that it doesn't get much bigger each time)?

If you don't know, the definition of e is

lim x-> :inf:  (1 + 1/x)^x

and is approximately 2.71828.

You might be able to use Taylor Series to prove it, since :Sigma: 1/n! would be e^1 or e when n goes from 0 to infinity. However, to get that series, I think you assume that e is convergent in the first place (d/dx (e^x) = e^x), although I'm not positive.

[hide]

With basic calculus (I know JBL likes this expression  )

f(x)=(1+1/x)^x

log (f(x)) = x log(1+1/x) = x (1/x - 1/2x^2 + o(1/x^2)) (little 'o' notation)

So log (f(x)) -> 1 when x -> +oo and f(x) -> exp(1)

It's also possible to show that f(x) is increasing and bounded.

 [/hide]

Whoa, whoa, explain o notation. I've heard of Big-O notation in comp sci, but that's about it.

We say f(x)=o(g(x)) if f(x)/g(x) --> :inf: as x gets big.

Compare this with big O notation: We say f(x)=O(g(x)) if there exists a constant C so that |f(x)|<=C*|g(x)| for all (suitable) x. (The suitable part is necessary if g(x)=0 for some small x.)

Belenos:

OK, that's a good proof. But there seems to be a reasoning error (I haven't "officially" done calculus yet, so I could be wrong.)

You use log without defining it rigorously, which is a problem when you're trying to prove that e isn't infinite (and therefore show that log is a function.)

I could be missing something here (for example, I don't know if you can prove that the expression for log defining it as the integral of 1/x is equivalent to its other definition regardless of whether e is finite). Tell me how this hole is patched, please.

I think he's okay -- we can prove plenty of stuff about the log function without knowing whether or not that particular sequence converges to e.

[quote="rlee"]Does anybody have a way to prove that e is convergent (besides plugging in big numbers and showing that it doesn't get much bigger each time)?[/quote]

The Wolfram Mathworld Web site states:  "e is the unique number with the property that the area of the region bounded by the hyperbola, y=1/x, the x-axis, and the vertical lines x=1 and x=e is 1."

There is much more about e at their web-site at [url]http://mathworld.wolfram.com/e.html[/url]

Yeah, and I can prove plenty of stuff about solutions of a^n+b^n=c^n for n>2, but that doesn't mean those solutions exist. (Of course, we now know that no solutions exist.)


conics
parabola
geometry
perpendicular bisector
calculus
calculus computations

Take the parabola $ y\equal{}x^{2}$, and some point P on it. The point Q is where the perpendicular bisector of OP intersects the y-axis, where O is the origin. As P approaches the origin along the parabola, what happens to Q? Does it have a limiting position?

I can help you with your first question: From your description it seems like the perpendicular bisector can be seen as a normal line to the parabola at some point on the graph of x^2 between the origin and P. Since the normal line is perpendicular to the tangent, it will coincide with the y-axis when P reaches the origin because at that point the tangent line will be horizontal(x-axis)


I can solve this problem by trigonometry, but I think this problem can solve by complex number, too. who can help me :)
Let $ ABCD$ be a quadrilateral with $ \widehat{CBD} \equal{} 2\widehat{ADB}$, $ \widehat{ABD} \equal{} 2\widehat{CDB}$, and $ AB \equal{} CB$. Prove that $ AD \equal{} CD$.  thanks

[geogebra]b5a6b525529c243287fa885e67fe6e33d76f5ea5[/geogebra]
$ BM$ and$ BN$ are angle bisectors of $ \angle CBD,\angle ABD$ respectively

$ \implies \angle MBK = \angle ADB = b$ and
$ \angle NBD = \angle MDB = a$

$ \implies \angle CMB = a + b$
$ \angle C = 180^\circ - \left( {\angle CMB + \angle CBM} \right) = 180^\circ - \left( {a + b + a} \right) = 180 - 2a - b$

$ \& \because AB = BC \\
\& \implies \angle BCA = \angle BAC = \frac {{180 - \left( {2a + 2b} \right)}}{2} = 90 - a - b$

$ \& \angle C = \angle BCA + \angle DCK = 90 - a - b + \angle DCK = 180 - 2a - b \\
\& \implies \angle DCK = 90 - a$

Similarly, we get $ \implies \angle DAK = 90 - a$

$ \& \therefore {\text{ In }}\triangle CKD, \\
\& \angle CKD = 180 - \left( {a + 90 - a} \right) = 90^\circ$

$ \& \implies \boxed{\triangle CKB \cong \triangle AKB}{\text{ By R.H.S Congruence}} \\
\& \implies 2a = 2b \implies a = b \\
\& \implies 90 - a = 90 - b \implies \angle DCK = \angle DAK$

$ \therefore \text{ In } \triangle ADC \text{ we have } \boxed{AD = DC}$

oh, it's nice (but longer than my solution by trigonometry, but do you have a complex number solution? can you post for me please :) thanks so much :D

Another approach

since its given that $\angle CBD= 2\angle ADB$ and $\angle ABD= 2\angle CDB$
let $BI$ and $BH $ be angle bisectors of $\angle ABD $and $\angle CBD  $respectively.
Now in $\triangle ABD,$
$BL$ is bisector of$ \angle ABD$.
Thus, $ AB/BD$= $AL/LD$
and in $ \triangle CBD$,
$BH$ is bisector of $\angle CBD$.
Thus,$ BC/BD$=$CM/MD$.
But , $AB=BC$.
hence we get$ AL/LD=CM/MD$.
$ \implies ML \parallel  AC.$

$\angle ADB=\angle MBD$
$\implies AD\parallel BM,$
similarly,$ CD \parallel BL $
$\implies BLDM , ALMW, CMLV $are parallelograms.
$\implies ML=CV $and $ML=AW$.
$\implies ALMW\cong CMLV $
thus $\angle VCM=\angle WAL$
$\implies AD=CD $
Proved.

Q.E.D.


geometry
circumcircle
search
trigonometry
geometry proposed

$ ABC$ a triangle,such that $ \angle ABC\equal{}60^\circ.$$ O$ is the circumcenter of $ ABC$ and $ H$ is the orthocenter.$ (OH)$ meet $ [AB]$ at $ P$ and $ [AC]$ at $ Q$.Prove that $ PO\equal{}QH$

rather easy.
B,O,H,C are concyclic
 => ∠APQ=∠PBH+∠PHB=∠OBC+∠OCB=60
 ∠AQP=∠QHC+∠QCH=∠OBC+∠OCB=60
AO=AH=R
done!

[quote="rachid"]$ ABC$ a triangle,such that $ \angle ABC \equal{} 60^\circ.$$ O$ is the circumcenter of $ ABC$ and $ H$ is the orthocenter.$ (OH)$ meet $ [AB]$ at $ P$ and $ [AC]$ at $ Q$.Prove that $ PO \equal{} QH$[/quote]
[hide="Solution"]
Let the midpoint of $ AC$ be $ U$ let $ CH$ meet $ AB$ at $ Z$.  Also, let the centroid of $ \triangle ABC$ be $ G$.  Now, $ \angle BHC \equal{} 180 \minus{} \angle ABC \equal{} 120 \equal{} 2\angle ABC \equal{} \angle BOC$, so $ BHOC$ is cyclic.  Now, $ \angle ABH \equal{} 90 \minus{} \angle BAC \equal{} 30$ and $ \angle BZH \equal{} 90$ since $ H$ is the orthocenter of $ \triangle ABC$.  Since $ \angle HBZ \equal{} 30$, we have that $ HZ \equal{} \frac {HB}{2}$.  We have that $ G$ lies on $ HO$, $ OU\perp AC$ since $ U$ is the midpoint of $ AC$, and $ BH\perp AC$, so $ \triangle BHG\sim \triangle UOG$, so $ \frac {BH}{OU} \equal{} \frac {BG}{GU} \equal{} 2$.  This means that $ OU \equal{} \frac {HB}{2} \equal{} ZH$.  Now, $ \angle ZHP \equal{} \angle OHC \equal{} \angle OBC \equal{} 90 \minus{} \angle ABC \equal{} 30$ since $ O$ is the circumcenter of $ \triangle ABC$ and $ BHOC$ is cyclic.  This implies that $ \angle ZPQ \equal{} 90 \minus{} \angle ZHP \equal{} 60$.  Since $ \angle PAQ \equal{} 60 \equal{} \angle APQ$, we have that $ \angle UQG \equal{} 60$.  Therefore, since $ \angle OUQ \equal{} 90\equal{}\angle HZP$, $ HZ \equal{} OU$, and $ \angle ZPH \equal{} 60 \equal{} \angle UQG$, we conclude that $ \triangle HZP\cong \triangle OUG$, so $ OQ \equal{} HP\implies OP \equal{} HQ$, as desired. [/hide]

[quote="rachid"]$ ABC$ a triangle,such that $ \angle ABC \equal{} 60^\circ.$$ O$ is the circumcenter of $ ABC$ and $ H$ is the orthocenter.$ (OH)$ meet $ [AB]$ at $ P$ and $ [AC]$ at $ Q$.Prove that $ PO \equal{} QH$[/quote]

 We had solved this problem in the greek forum in January at this post:

 http://www.mathlinks.ro/viewtopic.php?search_id=1067233330&t=184329

 The solution by Bilot uses a little but basic trigonometry and is very short.

 I had proposed there the problem , but I do not remember the source.

 Babis


ABCD - square. AB = 12, AЕ = 6, CG = 3, DL = 4, $ LM \bot GE$. AM = ?

I think $ E$ is the midpoint of $ AD$ and $ LM$ perpendicular with $ EG$!
Let $ H,K$ be the perpendicular feet from $ G,L$ to $ AD,CD$ respectively!
It easy to check that $ 2$ triangle $ GHE,LKM$ are equal!
Then $ AM\minus{}DL\equal{}KM\equal{}HE\equal{}DE\minus{}CG$.
Last $ AM\equal{}7$


$ \sum \frac{a^{2}\plus{}b^{2}}{c}\geq \frac{2}{3}(a\plus{}b\plus{}c)^{2}\sqrt{\frac{a\plus{}b\plus{}c}{3abc}}$

The first ineq we can prove it by using the following ineq:
  $ \sum\frac{a^2}{b}\geq\frac{\sum a\sum a^2}{\sum ab}$

[quote="iqtimo"]$ a,b,c \succ 0$ . prove that
$ \sum \frac {a^{2} \plus{} b^{2}}{c}\geq \frac {2(a \plus{} b \plus{} c)^{3}}{3ab \plus{} 3bc \plus{} 3ca}$[/quote]
$ \Leftrightarrow (\sum a^{2})(\sum \frac {1}{a})\geq \frac {7}{3}(a \plus{} b \plus{} c) \plus{} \frac {2}{3}\frac {(a \plus{} b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2})}{ab \plus{} bc \plus{} ca}$
we have:
$ (\sum a^{2})(\sum \frac {1}{a})\geq 3(a \plus{} b \plus{} c)$
and
$ (ab \plus{} bc \plus{} ca)^{2}\geq 3abc(a \plus{} b \plus{} c)\Leftrightarrow \sum \frac {1}{a}\geq \frac {3(a \plus{} b \plus{} c)}{ab \plus{} bc \plus{} ca}\Leftrightarrow (\sum a^{2})(\sum \frac {1}{a})\geq \frac {3(a \plus{} b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2})}{ab \plus{} bc \plus{} ca}$
Done ! :D

[quote="iqtimo"]$ \sum \frac {a^{2} \plus{} b^{2}}{c}\geq \frac {2}{3}(a \plus{} b \plus{} c)^{2}\sqrt {\frac {a \plus{} b \plus{} c}{3abc}}$[/quote]
We have :
$ (ab\plus{}bc\plus{}ca)^{2}\geq 3(a\plus{}b\plus{}c)abc\Rightarrow \sqrt{\frac{a\plus{}b\plus{}c}{3abc}}\leq \frac{ab\plus{}bc\plus{}ca}{3abc}$
We need prove :
$ 3(a^{2}\plus{}b^{2}\plus{}c^{2})(ab\plus{}bc\plus{}ca)\geq \frac{2}{3}(a\plus{}b\plus{}c)^{2}(ab\plus{}bc\plus{}ca)\plus{}(ab\plus{}bc\plus{}ca)^{2}$
it is true because $ a^{2}\plus{}b^{2}\plus{}c^{2}\geq ab\plus{}bc\plus{}ca$ and $ 3(a^{2}\plus{}b^{2}\plus{}c^{2})\geq (a\plus{}b\plus{}c)^{2}$
Done ! :lol:

from Holder,
${ (\sum \{\frac{a^2}{b}+\frac{a^2}{c}\})(\sum\{ab+bc})\times 6\geq 8(a+b+c)^3$
$ \Rightarrow \sum \frac{a^2+b^2}{c} \geq \frac{2(a+b+c)^3}{3(ab+bc+ac)}$


number theory
inequalities solved
inequalities

For which positive integers n is

             1^n+2^n+3^n+4^n

divisible by 5.

Well, as a Flanders I guy I want to mention that this is really not a final question :)
That would give a very bad image of our olympiad :)
Ok, our olympiad is easy, but not that easy :D
Actually, this problem was originally multiple choice (selection rounds).
And it is number theory, no algebra.

[u]Solution[/u] :D:

f(n) = 1n + 2n + 3n + 4n
f(n + 4) = f(n) [mod 5] by Fermat's little theorem (for example).
f(1) = 0 [mod 5] => f(4k + 1) = 0 [mod 5] for all k.
f(2) = 0 [mod 5] => f(4k + 2) = 0 [mod 5] for all k.
f(3) = 0 [mod 5] => f(4k + 3) = 0 [mod 5] for all k.
f(4) =/= 0 [mod 5] => f(4k) =/= 0 [mod 5] for all k.
There are lots of other methods, for example simple factorisations.
Too simple actually :)

But in 1991, there was a [b]very[/b] interesting problem in our final round:

[url]http://www.mathlinks.ro/phpBB/viewtopic.php?t=836[/url]

I really recommend this one :)


Give x, y, z are real numbers, we have:
2xyz \geq  2*x+4*y+7*z.
find the smallest value of A=x+y+z.

Sorry Kikicoco,but i think you have a mistake here :? .A can't have the smallest value.For example
x=1, y---> - \infty ,z--> - \infty .
I know this problem,it might be:
Find min(x+y+z)  when we have x,y,z>0 and 2xyz  \geq   2x+4y+7z :P

Oh, i'm sorry very much about my crazy mistake and i think Anh Cuong is right. :(

I think i can solve my problem by B.C.S Inequality. Please wait!

[quote="bozuy87"]Hello.
I don't know if it is true,but :
using AM-GM   2*x+4*y+7*z>=3(56xyz)^1/3,hence xyz>=3*(21)^1/2
so x+y+z>=3(3*(21)^1/2)^1/3[/quote]
A very deceiving mistake here. First application of the AM-GM inequality refers to 2x=4y=7z but the second application of the AM-GM inequality refers to x=y=z for the minimums, but they are not necessarily the same value!!! Be careful when you guys apply inequalities multiple times.

and even if it were correct...

even if you would find a minimal value for it, as Anh Cuong said, whichever you say is minimal, you can always go smaller!!

like if I'd say a>1, find the minimal value of a :maybe:

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

but let's change our problem a bit: x,y,z are [b]positive[/b] reals...
maybe that will work better :)

Hi.
I'think that we will if we will find a positive integer solutions of

   (4b+7c)/a=(2a+7c)/b=(2a+4b)/c     (at least one one solution).

                                               :?  :?  :?

sorry can anybody show me the complete solution?


Support articles Harvard college Columbia geometry search

I, personally, don't really see what McCain was thinking. Did he think an ardently anti-abortion, pro-gun candidate would win over Hillary supporters? Yes, it solidifies the base, but Pawlenty or Romney would have made a better play for independent voters - someone who could actually be trusted to take over if anything happened to McCain.

I dislike her more and more the more I hear - she supports "teaching the controversy," does not believe global warming is human-caused, etc. She is involved in an ethics scandal in her own state.

Thoughts from others?

To be a realist, McCain is quite old. I think the question should be raised on how prepared the VP is to take over in case of an emergency. I'm not saying this as an attack on McCain, but with health, you just never know, and to deny that possibility is to be irresponsible. Biden on the other hand, I would trust to take over in a heartbeat.

I think McCain's choice clinches the deal in my mind. There is absolutely no way I'd vote for him now. That's how I feel about his choice.

Maybe he was balancing the ticket a bit? I dunno.

but yes i agree it was a poor choice on his part.

Balances? I don't see how Palin can be equated with balance in any circumstance XD

I was seriously scared that McCain could win the presidency despite the failure of Republican politics for the last 8 years, so I'm somewhat relieved to see him shoot himself in the foot.

Basically, the Palin pick was a huge gamble, the reaction mentioned above is the 1 he was hopeing to avoid  :D

Here is an interesting article:

[url]http://www.politico.com/news/stories/0808/12997.html[/url]

I think that Palin was a great choice.
Even if McCain died and Palin had to take over as President, I still think she's better qualified than Obama is at being President.

In addition, she is a real conservative, which really helps McCain in getting votes.

McCain's plan seems pretty shallow and see through for me. Hes obviously trying to get women to vote for him by picking Palin. It was a terrible idea for John McCain and I think it'll probably cost him the white house.

Obama on the other hand made a very wise decision in picking Biden as his VP. I say this because Biden has life experiences, work experience and is a real family man.

Really the only positive thing for McCain in picking Palin is he's probably going to win Alaska.

Btw, according to a debate of some sort in 2006, she's a creationist who supports "teaching the controversy". That'll really help McCain with the conservative votes, considering that a significant portion supports teaching the controversy.

And the strongly Pro-life view will attract Evangelicals...especially with her story of choosing not to abort a child that she learned has down syndrome.

It seems pretty ridiculous to even venture that Palin would be better prepared than Obama for the job of president. Both are first term, but Obama clearly had the political strength to make a name for himself nationally whereas clearly, nobody has the least idea of who Palin is. (Also I think Palin's term started later.)

I think it's a very exciting pick and am pretty happy about it. I doubt it's going to work, but still it's a really ballsy move that really livens things up. Obviously, McCain needed to pick someone more socially conservative than he in order to retain a hold on his base but he still will attract some Hillary supporters (or simply female independents) this way -- it sounds stupid, but there will probably be a fair number of votes going this way. By the way, note that there's nothing secret about this -- McCain and Palin are obviously and openly trying to attract women; it's not like it's some clever underhanded thing which you alone figured out.

Also by the way, a lot of people are going to be like "Palin was a horrible pick because she is so socially conservative and blah blah blah". Well obviously, McCain wasn't trying to get your vote with that pick. You would have been pretty unlikely to vote for any Republican ticket in the first place, right? Whether or not the pick is good depends on how many votes it manages to attract, not whether people who wouldn't have voted for the ticket in the first place approve or disapprove.

I'm also a bit disappointed in everybody suddenly bringing up McCain's age. When he first publicly released all his health documents, everybody kind of gathered around and said "oh wow huh looks like you are actually pretty healthy" and there was a tacit assumption to drop the issue. Now, just because he appointed Palin, everybody's like "HE'S OLD AND GONNA DIE OMG". That's a bit silly. Certainly, you can justifiably blast Palin on being far too inexperienced but it'd be nice to do it in a way that doesn't immediately hark back to an already settled point of discussion.

mccains not gonna, hes perfectly healthy right now and is given the best medical care this nation can offer

picking palin wasn't a terrible move, its gonna make anything worse, but i'm not sure if it'll help much either.

The play for Hillary supporters won't work. It's that simple. The media has seriously created a conflict where there is none for the purpose of having a good story. The Democrats are in fact more unified (79% definitely for Obama) behind Obama than they were behind either Kerry or Gore (77% definitely for each).

Yes, she solidifies his base. But honestly, it just makes me question his judgment. Why would you choose someone with absolutely no experience? It's an incredibly political move that I think will backfire eventually. She's already under investigation in her home state (which has very few non-corrupt politicians, see Murkowskis, Stevens, Young, etc.). She has absolutely no foreign policy experience - the claim that she is more qualified than Obama is, frankly, weak at best. Obama has been in the Senate for three and a half years and currently serves on the Senate Foreign Relations committee - he's the chair of the European Affairs subcommittee. He also majored in Political Science, specializing in internation relations. When Palin lists being on her local PTA as a credential, it's really quite frightening.

And honestly, being governor of Alaska is really far less challenging than being that of a more popular state. They get huge amounts of tax revenue from oil companies as well as the federal government. When she supposedly said "thanks, but no thanks" to the Bridge to Nowhere, she just kept the money and used it on other stuff in the state, after changing her position from one of initial support.

On the other hand, Biden would obviously be able to take over and do a great job running the country - I place a lot of trust in him. And in terms of simple intelligence, we know that Barack Obama is incredibly intelligent - something we lacked in the last administration. He was *president* of the Harvard Law Review, a pretty impressive feat. He's willing to surround himself with smart advisers, like Biden, and learn from them. Palin, in contrast, has a minor in political science from some relatively unimpressive schools. While I'm not generally one to ascribe much value to a degree from the generally pretentious elite schools, it really takes quite a bit more to make it into Columbia than North Idaho University.

Edit: as an aside, Palin has only motivated me to sign up to volunteer for Obama.

Edit 2:
Note that McCain's campaign did a terrible job of vetting Palin. Nobody in his campaign went to Alaska, no one checked the archives of the Wasilla paper, and the decision was rushed and last-minute.

Well obviously it was rushed. Acctually it almost seems plausible that he put every single major republican into a hat and just randomly picked one. The fact that she has even less experience than Obama makes John McCain a hypocrite for even using the "experience attack".

Palin was a good choice.

As Mystic points out McCain is a significantly different position that Obama.  The goals of the two VP choices are very different.

Unless the feminists try the "she's not a woman because she doesn't have my political beliefs," if they want to be true to their goal of moving a woman into office, this has to make the McCain ticket more attractive than it was before.

Worth's comment on Democrat support being behind Obama is irrelevant.  The question is not percentage of supporters who are Democrat across the country.  Even though Al Gore lost the popular vote, as he found out, it is the electorial votes that determine the election.  Given that, there are only a minority of voters that are relevant to the question who will be the next US President.  Voters in solid states like Texas don't count.  Another 1,000 or 10,000 voters for either side there won't make a difference.  Voters in states like Michigan, where Obama didn't even run and which poll close to even are critical.

A state Governor is better experience than being in the legislature.  Being part of a group that works together to, at best make recommendations and review issues is a lot different from making decisions and running an administration.

While she is the VP candidate, comparing her experience to Obama' s doesn't work for one really important reason: he is the Presidential candidate, not the VP candidate.  Second, she has much more experience that he does.  Before everyone responds with the wrong information, Obama has spent most of his efforts since he was elected running for President, flying across the nation, since he is not held to any standard as to putting time in to represent the citizens of his state (as is true for any member of Congress).   However, in the two years since both were elected, Palin has been focusing on running the government of the State of Alaska, gaining at least a bit of experience that may be relevant to being President.  Obama has been spending his time flying about the country, making speeches and running for President.  This does not gain him any experience in foreign affairs or dealing with the armed services.  Both of those would at least have some relevance to being President.

The next thing that is overlooked by many is that Obama has the single most liberal voting record in the Senate, at least by some counts.  This means that he personally represents what his party regulars want more than what is electable.  McCain is from the liberal wing of the Republican party.  He represents what is electable more than what his party regulars want.  McCain needed to use his VP choice, not to get the votes of his party regulars, but to get them enthused.

Crazy comparing the experience level of the top of one ticket with the bottom of the other isn't a fair comparison.

Worth questions selecting a VP with Palin's experience level.  Did Edwards and others have more experience.  Let's look at Ike, Bill Clinton, Jimmy Carter, Reagan.  I don't think that a great deal of experience with federal or international issues has been needed for President, let along VP.

Palin is not running as one of the members of the US self appointed elite, like Obama, she is running as a real American, who just doesn't have someone tell them what a small business need to thrive, she has lived it.  Obama probably has never attended a PTA meeting and doesn't personally know the issues that most face daily.  This is the point, that seems to have been overlooked in this discussion.

I'm glad that Crazy has decided to get involved in the campaign.  Things are better with more involved, regardless of which side they support.

The lack of folks going to Alaska was probably to keep it a surprise, which was a success.  I'm sure that the politicians that live and work in Alaska with her were doing the vetting of Palin.  Just because there were no fingerprints doesn't mean that there wasn't any background checking.

Mystic - I don't understand why having internal political name recognition across the US prepares one to be President.  I do understand how it helps get elected, but not to actually be President.  As I suggested above, I think giving Obama credit for the time since he was sworn into office as being a Senator is very misleading.  He simply hasn't had the experience, he has been running instead.

I agree with you on many points, such as the need to address the conservative base and the foolishness of the age thing coming up again.

Protestan - she actually is beginning to raise the child you are talking about, it wasn't aborted and was delivered.  Just a bit more than not aborting the child.

Crazy - Obama lacks experience, so he picked a fellow Democrat that has been rejected consistently by his party for the office of President.  He has racist issues, saying on the floor of the Senate something like you have to have at least an Indian accent to enter a 7-11.

I disagree that Biden has work experience.  He was one of the youngest elected to the Senate from the smallest state and has just sat there ever since.  He has not run a company.  He has not run a state.  His experience is that of being a Senator.  No real boss, no real job responsibility.  No real threat for re-election.  HIs sons work for the government, both lawyers, on in the armed services the other as attorney general for Delaware.  Where is he getting real life experiences similar to what the vast majority of Americans have? He isn't.  I don't know what you mean by a real family man.  If this means that his sons have grown up, while Palin's children have not, this isn't true.  I understand that her oldest son is preparing to go overseas into Iraq, just like Biden's oldest son.

I'm a bit tired of this "failed Republican politics" story the press is pushing.  Abraham Lincoln was elected with 30 % of the vote, was widely viewed as not very bright and led us through a very divisive civil war, that many Democrats of their day said were wasting solder's lives and arguing that we should settle with the South.  He didn't have a strategy how to occupy the South after military victory.  There were gangs causing violence and attacking the established governments and there was violence directed against members of the South society.  (This was the period of the wild west, when the soldiers of the war ran wild across the south and west for some 30 years after the war.  This was also the time for the establishment of the Klu Klux Klan, that were criminals - before the term terrorists was cast - that carried out the same type of attacks that are occurring in the middle east today.)

If Iraq and the other areas end up stable, which will be determined by future events, Bush will be viewed by history as a great president.  He is carrying on with what was started with the full support of the Senate (Biden and Clinton voting to go to war), just not giving up when the going got tough, both in the field and in the polls.

Anyone willing to direct concern against McCain because of his age, based on actuarial tables, which I think is wrong, should also be willing to look at actuarial tables for black males, living in DC for Obama.  Neither comparison is appropriate and valid, but if one candidate is questioned on that basis, shouldn't both have the same question?

If nothing else, McCain's choice and timing moved Obama off the center stage in the political news, which must have been a major goal.

As an aside, I bet the Alaskan government's web page set a record for number of hits and downloads in the day or two after the announcement.

hm. on what basis do you label Obama as a member of the elite? I thought his acceptance speech at the Democratic convention would have quashed those rumors pretty effectively.

Your introduction of Lincoln is a bit surprising since most Democrats today would probably have been Republican in that time period. That he was elected with 30% of the vote is a bit irrelevant since there were at least four candidates each with a sizable part of the vote, something that clearly does not hold with the present political picture. To be honest, I'm really not sure what the rest of the paragraph is about either -- are you saying that because Lincoln is viewed as a great president, Republican politics have been and will be infallible forever more?

I'm also a bit surprised by your characterization of United States senator as "not a real job". Incidentally, Biden probably has the closest to real life experiences of just about anybody recognizable in politics, having a net worth of, I believe, 150000 USD.

I doubt that Bush will be viewed warmly in retrospect; if anything, I suspect that when more (any) documents of this time period become declassified, people will simply become more cynical about Rumsfeld and Cheney's true motives in starting this war in the first place. I won't seriously argue the point, though, as I obviously don't have access to the evidence necessary.

The reported lack of vetting by the presidential nominee himself seemed odd, as McCain reportedly only talked to Palin once before offering her the vice presidential slot on the ticket, but I'm not going to use this in any argument over whether or not Palin was a good pick as McCain and his campaign probably did know what they were trying to accomplish rather than simply picking haphazardly; I'm still pretty happy over what this pick will do for both campaigns and the political season in general.

You are right; simply having name recognition is certainly not sufficient experience to run this country. I will cede this point as I did not formulate it well (and I am sure other people would be far happier to debate on the relevant experience levels of Obama and Palin).

[quote="wizard2378"]... Obama has the single most liberal voting record in the Senate, at least by some counts.[/quote]This is based on a very suspect methodology- it ignores bills on which the legislator didn't vote, and presidential candidates miss a lot of votes. The main candidates are likely to only show up for tightly contested partisan "important" bills, which naturally biases them to the extremes by this measure.
By measures which either adjust for this effect or aren't based on campaign years, Obama is near the middle of the Senate Democrats.

I'm not sure what these would say about McCain, who basically hasn't shown up at all for months in the Senate.

[quote="wizard2378"]Unless the feminists try the "she's not a woman because she doesn't have my political beliefs," if they want to be true to their goal of moving a woman into office, this has to make the McCain ticket more attractive than it was before. [/quote]I've been reading quite a bit of stuff posted by feminists. Their general position is that (1) McCain is anti-woman. (2) Palin is anti-woman. There was never a chance of them voting for McCain; it's only a question of voting Obama or voting other/not voting.
Along those lines, the only possible threat here is a backlash against sexist attacks on Palin convincing them not to vote for either major-party candidate. The Green Party candidate Cynthia McKinney is still out there for those who feel wronged.

[quote="wizard2378"]I'm a bit tired of this "failed Republican politics" story the press is pushing.[/quote]And I've got a collapsed bridge in Minnesota to sell you (see [url=http://www.nytimes.com/2007/08/07/us/07highway.html]VP contender Pawlenty[/url]). It's not so much "failed politics" as politics that creates failure.
What about the levees in New Orleans? Far less repair work has been done than was initially prepared, and people don't trust them.
These are not accidents. The "conservative" ideology is openly hostile to government, and seeks to destroy services whether they are effective or not.

That's the legacy of Bush and the modern Republicans- people no longer trust the government to do good.

You have got to love her:
http://caffertyfile.blogs.cnn.com/2008/09/26/is-palin-qualified-to-be-president/

Hi JoeBlow,

I'm unfortunately not going to give your post quite the quality of response it deserves.  For one thing, my knowledge of mid-20th century American history is sufficiently weak that I can't provide the kinds of analysis that would make this discussion really interesting.  For example, my guess (and I emphasize that this is just a guess) is that the Nixon/Ford number can be largely explained in light of the fact that Nixon followed LBJ than in terms of Nixon's own presidency, but I really don't know enough to make a compelling case around that guess.

To the extent that your point is that there are forces beyond the presidency that influence inequality (whether measured by the take of the top 1% or some other way), I certainly agree, as I've already said.  (In fact, if we omit Reagan from our data-set, the numbers you provided suggest strongly that the extent of Democratic control of congress dominates the party of the president: the four presidents with the lowest numbers in your sample all had Democratic congresses for their entire presidencies, while the two with the largest numbers suffered under Republican control for most of their terms.)  However, I think it's indisputable on the evidence that the Democratic party supports policies that tend to counteract inequality by relatively favoring the lower 20 or 50 or 90% of the population as measured by wealth or income while the Republican party tends to support policies that relatively favor the top 50 or 10 or 1%.  This (and my earlier comment) contains no guess or judgement on my part about [i]why[/i] the Republican party favors such policies.  I don't think that I've claimed or suggested here that the well-being of the upper class is an explicit goal of these policies or even that these policies are morally wrong (let alone that the people supporting them are sinister), just that they are what they are and that I think they are bad policy.  Thus, I think the portion of your post quoted below is rather much of a non sequitur.*
[quote="JoeBlow"]While I agree that the effect of Republican policy is to favor accumulation of wealth by the upper classes, I do not agree that this is the goal of the policy, if that is what you are suggesting. If the two ends of the political spectrum are to engage in any kind of constructive discussion, I think it's useful to recognize that the other side is, too, motivated by what they perceive is a noble cause. In the case of conservatives it's defense of individual freedom and economic self-determination, in the case of liberals it's the defense of those that cannot defend themselves. I don't think it's very fair to imply the other side is just a sinister group whose only purpose is to thwart that which you believe is righteous. [/quote]

As evidence like daboa's link show, Palin is going to be viewed as a first-class disaster as VP nominee.  Not only is she clearly unprepared to offer any coherent thoughts about anything relating to national or international politics, but her attempts to do so are just deeply embarrasing.  I think, though, that for humor value nothing beats wizard's claim in post 56 in this thread that Sarah Palin is a good VP pick because even if McCain dies after a couple of years in office, how much damage could she possibly do?  That one's going to keep me smiling through November.

* Some further comments on this: I have no doubt that Republicans believe they are acting towards some vision of the "Good Society."  It's not clear to me why you think it is useful to recognize this (or why you suppose I haven't done so).  Given [i]my[/i] vision of the Good Society, the Republican party supports policies that are almost uniformly destructive and counter-productive.  To recognize that Republican politicians support what they do despite this merely raises the question of whether they are incompetent, misguided, or evil.  (In the case of Sarah Palin, for example, I think we have a little from column A and a little from column B.)  To raise the question is to require me to explain why I think Republicans are doing bad things, a conversation that is not very likely to lead to compliments or collegiality -- to [i]not[/i] raise the question is to allow me to oppose those bad things without having to impugn anyone's honesty, goodness or motives.

[quote="JBL"]To the extent that your point is that there are forces beyond the presidency that influence inequality (whether measured by the take of the top 1% or some other way), I certainly agree, as I've already said[/quote]My assertion was a few notches stronger than this: Not only are there forces beyond the Presidency that influence inequality, but moreover these forces are significantly more important than who is President.  I assert that who happens to be President (and even who happens to control Congress) has, at least in the last several decades, had a lesser impact on inequality than have changing economic conditions.  I agree, however, that I haven't provided enough evidence so far to justify much more than the weaker version which you have stated.

Tangential: What I feel has been and continues to be [i]the[/i] most significant factor driving the increase in inequality is the shift from a manufacturing-driven to a service economy. But I guess that would be going even further off topic.[quote="JBL"]However, I think it's indisputable on the evidence that the Democratic party supports policies that tend to counteract inequality by relatively favoring the lower 20 or 50 or 90% of the population as measured by wealth or income while the Republican party tends to support policies that relatively favor the top 50 or 10 or 1%.[/quote]I agree.  What I do not agree with: (1)That the [i]intent[/i] of Republican policy is to favor the top whatever percent, and perhaps more importantly (2) that the policies pursued by one party or the other have had a significant impact on inequality.  You made clear that you haven't really implied (1) nor (2), so I guess we don't disagree. Still, let me elaborate why it seemed to me that we did.

In the original quote to which I was originally objecting, you state that "Republican domestic policy amounts to a giant redistribution of wealth in favor of economic elites."  While it's true that this statement, strictly speaking, contains no claim about Republicans' motivations, I find the wording and context to be very suggestive of both motivation and your judgment about it.  Moreover, you proceeded to defend a related statement by offering historical evidence that a powerful redistributive process has been occurring in the last three decades.  Again, though you make no explicit connection between this observation and Republican policy, because of the context and in light of your previous comment about Republican support for redistribution, it's hard to read this without concluding that you believe said redistribution is in large part the result of Republican policy.

Since we don't disagree on these two points after all (right?), I'm inclined to drop them as it seems our statements aren't at odds.
[quote="JBL"]It's not clear to me why you think it is useful to recognize this (or why you suppose I haven't done so). Given my vision of the Good Society, the Republican party supports policies that are almost uniformly destructive and counter-productive.[/quote]It seems useful to me because I believe anyone that doesn't understand others' motivations is much less capable of engaging in constructive discussion with them.  I feel this is useful in every relationship in everyday life, but it is particularly important in politics.  Insofar as our entire political system is hinged on the ability of people with opposing views to engage in compromises, I believe it is not merely useful, but absolutely necessary, that people involved in politics have a good understanding of the oppositions' motivations and intentions.  While you may retort that the average citizen is not directly involved in politics, everyone is a constituent of someone who is.  Thus, if many citizens hold the view that the entire opposing political party holds 'stupid', 'backwards', 'elitist', or 'deluded' views (words I've seen used to describe both liberal and conservative positions by otherwise intelligent people on these very fora), they will tend to elect more representative who hold, or at least espouse, the same intolerant views.  Invariably this leads to increased animosity between the two sides and, just because of the compromising nature of our legislative system, to less efficient lawmaking.

Incidentally, I think it's this very quality of being able to understand the concerns of your opposition that will allow Obama to pass more bipartisan legislation during his term(s) than any Presiden since LBJ (if only he didn't hold so many views with which I disagree ;)).

As for the Republican VP nominee: I agree that she is inexperienced, and I wouldn't be comfortable with the possibility of having her as president sometime in the next four years.  But I wouldn't place much weight on the particular quote alluded to in daboa's link; the statements were made on a liberal show aimed at a liberal audience in a scrambled effort to defend a position that both she and McCain would've been vehemently opposed to even a month ago.  It's sad that politicians have to defend a nonsensical position they probably don't approve of merely because their President pushed for it, but it's the reality of our system.

nick42: since Obama polls better among women than men and polls better than McCain overall, it is not possible for McCain to have a lead (1% or otherwise) among female voters.  (I also don't think there has been any time since the nominees were decided that McCain has held a lead over Obama among female voters, though I could be wrong.)  So your second parenthetical is either complete nonsense or made sense in the original context that you've stripped it from -- which is it?

[quote="JoeBlow"]Not only are there forces beyond the Presidency that influence inequality, but moreover these forces are significantly more important than who is President.  I assert that who happens to be President (and even who happens to control Congress) has, at least in the last several decades, had a lesser impact on inequality than have changing economic conditions.  I agree, however, that I haven't provided enough evidence so far to justify much more than the weaker version which you have stated.

Tangential: What I feel has been and continues to be [i]the[/i] most significant factor driving the increase in inequality is the shift from a manufacturing-driven to a service economy.[/quote]  Let me continue to agree-but-not-entirely, starting off from this tangent.  (Not much interesting discussion is happening on the original topic, anyhow.  Perhaps I should just claim moderator's priviledge to have off-topic discussions? :)) The shift from manufacturing to service economy is undoubtedly a major factor in many other large-scale changes in the economy, no doubt inequality among them.  But one reason for this is that this shift has been accompanied by a serious and steady decline in union strength in the country.  This is partly due to the differing nature of the economy, but it is also largely due to regulatory and legislative changes that have led to a vastly more hostile environment for labor than in the past.  Anti-union sentiment is (unsurprisingly) mostly the realm of the Republican party, and the Republican party has been the driver behind anti-union legislation and regulation.  Meanwhile, unions are probably the most significant institutions counteracting increasing inequality.  This is just one example (though I think a particularly strong one) of how changes associated with changing economic conditions are or might be partially explained by political changes, as well.  Thus, it's not at all clear to me that explaining something in terms of changing economic conditions implies that it can't equally well be explained in terms of political changes.  And I guess the strong version of this claim is that one really can identify a huge chunk of Republican domestic policy as "policies whose effect will be to increase inequality," regardless of the stated intent of said policies.

Right after Palin's selection McCain did indeed gain a short-lived lead among female voters. Identity politics is alive and well, and we can't forget that.

[quote="JBL"]But one reason for this is that this shift has been accompanied by a serious and steady decline in union strength in the country.  This is partly due to the differing nature of the economy, but it is also largely due to regulatory and legislative changes that have led to a vastly more hostile environment for labor than in the past.[/quote]I agree that the decline of unions has played a role in the process.  I do not agree that federal regulatory attacks on labor unions -- the most significant of which occurred in the 40's and 50's -- can account for increasing inequality that began for the most part in the mid to late 70's.  Moreover, I don't believe said federal legislation can even account for the decline in union strength itself; while union membership did decline during 1947-1975, this decline was quite small compared to what followed.  The major change occurred, once again, only after the mid 70's.  Because of the observation that the major effect of federal legislation of the 40's and 50's was delayed for two decades, it seems to me that while the politics of the mid-century played a role in [i]enabling[/i] an attack on labor unions, the actual cause of the decline could not have been primarily political.  It also seems to me that even in the absence of previously established legislation, the conditions of the 80's would have eventually necessitated similar legislation coming into existence anyway.  For these reasons, I believe the decline in equality (and the decline in labor union strength as a particular instance of this more general phenomenon) is a process better-explained by economic changes, and that federal-level politics is a mostly helpless bystander rather than a co-conspirator.
[hide="note"]The claims are only made for politics and legislation at the federal level.  I do agree that state and local politics can have a much more direct and profound impact on the conditions of labor, but it seems our discussion was mostly focused on the effect of which party holds the Presidency and Congress.

Incidentally, the two major pieces of federal legislation often cited as the cause of unions' eventual decline are Taft-Hartley (1947) and Landrum-Griffin (1959).  Note that the first was passed under a Democratic president (though I concede he opposed it, and the Congress was Republican), and the second was passed during one of the most overwhelmingly Democratic Congresses ever, and co-sponsored by a fairly liberal Democrat.[/hide]

How do people think she did at the debate?

I liked her flag pin  :) It definitely made up for John McCain's failure to wear one at his debate.

[b]Other comments:[/b]

[b]0.[/b] She sounded much more intelligible and intelligent than she was in her interview. I was impressed.
[b]1.[/b] I thought it was hilarious when she randomly changed the topic, especially when she changed the topic back to energy -- twice. It just screamed, "I don't know how to respond, so I'm going to say what they taught me in debate camp!"
[b]2.[/b] She didn't really have an effective defense when she threw out the misleading statistics. Biden just ran her over. That's a relief in a sense -- at least she's not a good liar when she's put on the spot.
[b]3.[/b] On the whole, she didn't screw up or visually get stuck [i]too [/i]much. When she did, she didn't seem to get flustered or angry. I'm talking about responses here, not mistakes like saying General McClellan. Part of this was that when she was directly attacked, she sometimes just changed the subject.
[b]4.[/b] She said "maverick" and "Main St" so much I wanted to go to Wash U and poke her with the pointy end of a pencil.
[b]5. [/b]In general, a lot of her answers sounded incredibly canned.
[b]6.[/b] The last question, about differences between her and McCain, she started off answering that well (with the exception of using "maverick" yet again), and I was ready to hear a good answer from her. But then it went downhill.
[b]7.[/b] Her closing was decent, given the body of the debate.
[b]8.[/b] She was at least entertaining, and seemed much friendlier than John McCain.

I don't think she hurt herself with the performance, but I don't think it will make too much of a difference either.  Her refusal to answer Ifill's questions, as TZF pointed out, was irritating.  I really liked the fact that Biden had a good counter to almost everything that Palin said.  My favorite line of the night was Biden's "bridge to nowhere" quip, to which Palin couldn't counter due to time.  You could hear the audience "ooh" and "aah" at that line.

[url=http://www.salon.com/news/feature/2008/10/10/palin_chryson/]Here[/url]* is a nice article on Palin's links to paranoid secessionists.

*You may have to go through the front page of Salon to actually load the article.

Well her husband is/was a member.  Also note that her husband has a DUI under his belt, didn't register to vote until he was 25, and Sarah Palin didn't get a passport until last year.

My grandma thought she did great :rofl: and she said they edited it to show her in her worst light and I'm like I don't think they edit those things..... I would like to see an intelligence comparison of Obama supporters vs. McCain supposrters. THAT would be funny if it were like who I encounter....

Related to this topic, here is an e-mail that has recently been sent around. It mentions a recent PBS poll that asks whether Palin is qualified to become VP. If you happen to support it (or even if you feel the other way), you might as well vote.

--------------------------------------------------
PBS has an online poll posted asking if Sarah Palin is qualified. Apparently the right wing knew about this in advance and is flooding the voting with YES votes.

The poll will be reported on PBS and picked up by mainstream media. It can influence undecided voters in swing states.

Please do two things -- takes 20 seconds.

1) Click on link and vote yourself.

Here's the link:

            http://www.pbs.org/now/polls/poll-435.html

2) Then send this to every single Obama-Biden voter you know, and urge them to vote and pass it on. Note:  Only one vote/computer will be counted.

The last thing we need is PBS saying their viewers think Sarah Palin is qualified.
--------------------------------------------------

Here is a nice blog about how out of touch with reality she is: [url=http://mudflats.wordpress.com/2008/10/12/through-the-looking-glass-with-sarah-palin/]Through the Looking Glass[/url]

She's in complete denial-similar to John McCain.


Hello everyone i am new here to the forums and i need alittle bit of help in my math class. I need help using my CASIO Graphing calculator. Every one in my math class has Texas Instruments calculators. My teacher never told me i had to get a TI because i arrived to class1 week late due to scheduale problems. The problem i have is that he teaches everyone TI method and i cant keep up. Heres what we are learning now and what i need to know to do my home work.

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

--------------------------------------------------------------------------------------------------------------------------------------
-The Commission rate, the price of each item, and the number of items sold are given at the bottom.
-------------------------------------------------------------------------------------------------------------------------------------
a. Calculate the commissions earned by selling the first number of items sold ( for example, 10 items in Excercise 1).

b. Write an equation to show the commission y earned by selling x number of items.

c. Graph the euqation on your graphing calculator. Then use the TRACE function to find the commissions for all four amounts. Remember to use an integer function or a "friendly window."

=====================================================================
Now here is the chart with commission rate and number of items sold and price of each item.

1. Price- $16.75                       Rate-9%                       Number sold-10;18;24;32
=====================================================================
-OK thats the problem. This is what i have figured on my own.
you must multiply 16.75*.09*10     -16.75 For the cash-  -.09 for the Rate-  -10 for the first amount of items sold-
After doing that i come up with
$15.08. But i must now make it a Y for my calculator.    $15.08= y=  "1.508x"

Now heres where the trouble starts. What i must do is plug what i have into the calculator and then find the other 3 prices of the number sold. The prices for 18,24 and 32.

What i do is open up my calculator and go to graph then i change the scales to as follows

xmin:0
max:60
Scale:10
yMin:-10
Max:10
Scale:10

After that i go to Y= and type in "1.508"

Now what i must do is find out where to plug in 18, 24 and 32! But i dont know how to use this thing. I have read the Graphing section for this thing front to back and im clueless! I am guessing that i plug in the other 3 in x intercepts some where.

Everyones Help Will Be Greatly Appreciated, Please Feel Free To Make Any Suggestions And Thank You For Even Taking The Time To Read My Post If You Couldn't Figure It Out.

You double posted.  ;)

And, I don't know. I own a TI myself.

I apologize for double posting it was an honest error. If any one can please help me i would appreciate.

I got the same calculator for Alg2/Geom.   :)   TI really has a monopoly on the graphing calc market.

Instructions
Go to [b]MENU[/b]
Go to [b]TABLE (7)[/b]
Enter the equation [b]Y1:15.08X[/b] (use the X right under the red alpha button)
Press [b]EXE[/b]
Goto [b]RANG (F5)[/b]
Since you want to find the values 10-32, make the start 10, and the end 32.  Ignore the pitch.
Go back to the equations.  Press [b]TABL (F6)[/b].
And you're done.  :)


vector
linear algebra
matrix
analytic geometry
combinatorics unsolved
combinatorics

We name a set like $T$ even if the number of elements of $T$ be a positive even number.
Let $n$ be a positive even number and $S_1,S_2,S_3,...,S_n$ be even sets such that
$S_i \subset S(S=$set of  first n natural numbers$)$ .Show that there exist one $i$ and one $j$ such that $1 \leq i < j \leq n$ and
$S_i \cap S_j$ be an even set.

[quote="lomos_lupin"]We name a set like $T$ even if the number of elements of $T$ be a positive even number.
Let $n$ be a positive even number and $S_1,S_2,S_3,...,S_n$ be even sets such that
$S_i\subset S=\{1,2,3,...,n\}$ .Show that there exist one $i$ and one $j$ such that $1 \leq i , j \leq n$ and
$S_i\cap S_j$ be an even set.[/quote]

I like it, because it's a bit trickier than all the other (and there were soo many :)) combinatorial problems which appeared on the forum and which can be solved by algebraic arguments :).

Identify each set $S_i$ with its characteristc vector (which we also name $S_i$) regarded as an element of the vector space $\mathbb Z_2^n$ over the field $\mathbb Z_2$. For vectors ${\bf x}=(x_1,\ldots,x_n),{\bf y}=(y_1,\ldots,y_n)$, define the bilinear form $<{\bf x},{\bf y}>=\sum x_iy_i$ (reduced modulo $2$, of course, because this is the field we're working over).

Assume the contrary of what we must show. We have $<S_i,S_i>=0,\ \forall i,<S_i,S_j>=1,\ \forall i\ne j$. Because $n$ is even, the vectors $S_i$ form a linearly independent system (all you need to prove is that an $n\times n$ matrix having $0$ on the diagonal and $1$ everywhere else has odd determinant when $n$ is even). This means that if $e_i,\ i \in\overline{1,n}$ is the standard basis of this vector space, we can find subsets $\mathcal I_k\subseteq\{1,2,\ldots,n\}$ s.t. $\sum_{i\in\mathcal I_k}S_i=e_k,\ \forall k\in\overline{1,n}$.

Let's look at the vectors $S_i,\ i\in\mathcal I_1$. If we assume WLOG that these vectors are $S_1,\ldots,S_m$, then $S_1+\ldots+S_m=(1,0,\ldots,0)$. At least one of $S_1,\ldots,S_m$ (say it's $S_1$) has $1$ on the first coordinate, and assume there's also one vector $S_t,t>m$ with $1$ on the first coordinate. Then, by computing the bilinear forms $1=<S_1,e_1>=<S_t,e_1>=\sum_{i=1}^m<S_1,S_i>=\sum_{i=1}^m<S_t,S_i>$, we get different results, so we have a contradiction. This means that all the vectors which have $1$ as the first coordinate are among the $S_1,\ldots,S_m$. On the other hand, if one of $S_1,\ldots,S_m$ does not have $1$ on the first coordinate, then, by applying a similar argument we get, once more, a contradiction.

We have thus shown that $S_1,\ldots,S_m$ are precisely those vectors which have $1$ as the first coordinate. This shows that $m$ is odd, because $1$ is the sum of the $m\ 1$'s which are the first coordinates of $S_1,\ldots,S_m$. Now, $1=<S_1,e_1>=\sum_{i=1}^m<S_1,S_i>=m-1=0$, because, by assumption, exactly one of the $<S_1,S_i>$ is $0$, namely $<S_1,S_1>$, and $m$ is odd, so $m-1$ is even, meaning that it's $0$ when regarded as an element of $\mathbb Z_2$. We thus have a contradiction.

I hope it's Ok.

Still the same problems...
This is point b) here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=caruso&t=13941

There is an attached file with all the proofs.

Pierre.


I took this exam last February without fully understanding the last third of the AP Physics C Mechanics curriculum, and now I was wondering what I should do in order to attain a semi-finalist spot...I had no problems with the AP exam, I just need to  stay sharp from now till then. Any good books I should try? (I used Halliday for the exam, and did over half of its problems) I got a book on dynamics (covers all of mechanics but with more math) and I was wondering what else I needed...

Barron's AP PHYSICS C would'nt hurt as review.


function
real analysis
real analysis unsolved

Let $ f: [a, b] \rightarrow \mathbb R$ be a continuous function. Call a point $ x \in [a, b]$ a special zero of $ f$ if $ f(x) \equal{} 0$ and in any neighborhood $ (x \minus{} \epsilon, x \plus{} \epsilon) \cap [a, b]$ of $ x$ the function $ f$ attains both positive and negative values. Let $ S_f$ be the set of special zeroes of $ f$. Are the following claims true (prove them or provide a counter-example):

a) $ S_f$ has measure zero.
b) $ S_f$ is countable.