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1=2
Birch & Swinnerton Dyer
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#41
Cube. I'm too lazy to go on.

not kidding!
Could someone cube (a+b+c)?
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PostPosted: Mon Jul 17, 2006 4:47 pm  Back to top 
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lotrgreengrapes7926
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#42
nutz_for2.718281828 wrote:
Cube. I'm too lazy to go on.

not kidding!
Could someone cube (a+b+c)?

(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3a^{2}b+3ab^{2}+3a^{2}c+3ac^{2}+3b^{2}c+3bc^{2}+6abc. Mr. Green
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PostPosted: Tue Jul 18, 2006 6:54 am  Back to top 
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lotrgreengrapes7926
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#43
Could someone provide a hint or something? maybe
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PostPosted: Tue Jul 25, 2006 11:18 am  Back to top 
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JBL
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#44
Sure -- once you've cubed, you have a whole bunch of terms that you want to turn into abc terms. You can use the case of AM-GM on 2 elements to convert some of those terms into abcs, but the a^{3} terms will need special treatment, namely a somewhat unusual factorization.

Alternatively, you could take the approach which is used to prove AM-GM in general and use the 2 element case to show the 4 element case and then use the 4 element case to show the 3 element case.
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PostPosted: Tue Jul 25, 2006 12:39 pm  Back to top 
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lotrgreengrapes7926
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#45
JBL wrote:
Sure -- once you've cubed, you have a whole bunch of terms that you want to turn into abc terms. You can use the case of AM-GM on 2 elements to convert some of those terms into abcs, but the a^{3} terms will need special treatment, namely a somewhat unusual factorization.

Click to reveal hidden content
a^{3}+b^{3}+c^{3}+3a^{2}b+3bc^{2}+3a^{2}c+3b^{2}c+3ab^{2}+3ac^{2}+6abc \ge 27abc.

\begin{eqnarray*}3a^{2}b+3bc^{2}& \ge &6abc\\3a^{2}c+3b^{2}c& \ge &6abc\\ 3ab^{2}+3ac^{2}& \ge & 6abc....

But isn't it not allowed to subtract inequalities?

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PostPosted: Tue Jul 25, 2006 2:15 pm  Back to top 
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JBL
Birch & Swinnerton Dyer
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#46
lotrgreengrapes7926 wrote:
a^{3}+b^{3}+c^{3}+3a^{2}b+3bc^{2}+3a^{2}c+3b^{2}c+3ab^{2}+3ac^{2}+6abc \ge 27abc.

\begin{eqnarray*}3a^{2}b+3bc^{2}& \ge &6abc\\3a^{2}c+3b^{2}c& \ge &6abc\\ 3ab^{2}+3ac^{2}& \ge & 6abc....

But isn't it not allowed to subtract inequalities?


Let's categorize your equations:

1] Things you would like to show:
a^{3}+b^{3}+c^{3}+3a^{2}b+3bc^{2}+3a^{2}c+3b^{2}c+3ab^{2}+3ac^{2}+6abc \ge 27abc.

2] Things you can already show:
\begin{eqnarray*}3a^{2}b+3bc^{2}& \ge &6abc\\3a^{2}c+3b^{2}c& \ge &6abc\\ 3ab^{2}+3ac^{2}& \ge & 6abc....
And, as a consequence, the sum of the things on the left is larger than the sum of the things on the right.
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PostPosted: Tue Jul 25, 2006 2:51 pm  Back to top 
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Rzeszut
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#47
amcavoy wrote:
Here's another easy one:

For n=3, prove the AM-GM inequality. That is, prove that \frac{a+b+c}{3}\geq\sqrt[3]{abc}.

nutz_for2.718281828 wrote:
Cube. I'm too lazy to go on.

not kidding!
Could someone cube (a+b+c)?

Are you joking? It's very easy.
Let a=e^{x},b=e^{y},c=e^{z}. Then the inequality rewrites as \frac{e^{x}+e^{y}+e^{z}}{3}\geq \sqrt[3]{e^{x}e^{y}e^{z}}\iff \frac{e^{x}+e^{y}+e^{z}}{3}\geq e^{\frac{x+y+z}{3}}, which is just Jensen for f(t)=e^{t}. I don't advice you using brute force and expanding. It isn't nice. nutz_for2.718281828, I'm sure that you just were lazy and you know how to use Jensen. BTW, try to prove \mathrm{AM-GM} for n variables. Mr. Green
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PostPosted: Wed Jul 26, 2006 12:46 am  Back to top 
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JBL
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#48
That's like saying, "Proving the medians are concurrent? That's easy, you just use Ceva's Theorem!" It works, but you're applying a bigger piece of machinery in order to derive a littler piece.
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PostPosted: Wed Jul 26, 2006 6:09 am  Back to top 
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#49
JBL wrote:
That's like saying, "Proving the medians are concurrent? That's easy, you just use Ceva's Theorem!" It works, but you're applying a bigger piece of machinery in order to derive a littler piece.

Ceva (it has four-line proof using areas) and Jensen are simple theorems. Jensen for n variables isn't hard to using 2-variable Jensen. In our case we only need convexity of e^{t}. Proof without deriviatives is 2-line:
\frac{e^{x}+e^{y}}{2}\geq e^{\frac{x+y}{2}}\iff e^{x}+e^{y}\geq
\geq 2\cdot e^{\frac{x}{2}}e^{\frac{y}{2}}\iff e^{x}-2\cdot e^{\frac{x}{2}}e^{\frac{y}{2}}+e^{y}\iff \left( e^{\frac{x}{2}}-e...
Now my proof of \mathrm{AM-GM} is complete.
In general, the idea of creating theorems is to prove some general theorem and then use it in particular case in some problem just saying "from ... theorem". It's one of the main ideas of mathematics.
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PostPosted: Wed Jul 26, 2006 6:30 am  Back to top 
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lotrgreengrapes7926
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#50
Problem 9:
darkdevil wrote:

f=x^{3}-x-1
Prove that f has only one real root r \in \left(1, \frac{3}{2}\right)
Prove that if a is a root and a \notin \mathbb{R} then |a| \in \left( \sqrt{\left(\frac{2}{3}\right)}, 1\right)

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PostPosted: Wed Jul 26, 2006 8:22 am  Back to top 
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#51
I think I can do the first part (if the lower bound thing works when a coefficient is 0)

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The decartes rule of signs says that there is 1 positive root and 0 or 2 negative roots.

Dividing the polynomial by (x-1) we get x^2-x+0-1/x so there is no root below 1.

There is no root between -1 and 0 because x^3-x=1 and x^3 and x cannot differ by more than or equal to 1. Hence there is 1 positive root.

Since f(1) is negative and f(3/2) is positive there is a root between the two.


but at the moment I don't see how to prove the absolute value of the complex roots are between sqrt(3/2) and 1 (if that is what the question is asking
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PostPosted: Wed Jul 26, 2006 9:23 am  Back to top 
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#52
When is the next time the new problem come out/due?
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PostPosted: Wed Jul 26, 2006 9:34 am  Back to top 
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lotrgreengrapes7926
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#53
Problem 10:
Find, with proof, |z^{i}| when

a) z is a positive real;

b) z is a negative real;

c) z is in the form a+ai for positive real a;

d) z is in the form a+ai for negative real a;

e) z=a+bi, in terms of a and b.

Hint
\sin^{2}{x}+\cos^{2}{x}=1

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PostPosted: Sun Sep 10, 2006 6:57 am  Back to top 
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Farenhajt
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#54
lotrgreengrapes7926 wrote:
Problem 10:
Find, with proof, |z^{i}| when

a) z is a positive real;

b) z is a negative real;

c) z is in the form a+ai for positive real a;

d) z is in the form a+ai for negative real a;

e) z=a+bi, in terms of a and b.

Hint
\sin^{2}{x}+\cos^{2}{x}=1


Solution
If z=re^{i(\varphi+2k\pi)}, r,\varphi\in\mathbb{R}, k\in\mathbb{Z}, then

\begin{eqnarray*}|z^{i}| &=& |r^{i}e^{-\varphi-2k\pi}|\\ &=& e^{-\varphi-2k\pi}|e^{i\ln r}|\\ &=& e^{...

where n=-k

Hence

a) |z^{i}|\in\{e^{2n\pi}|n\in\mathbb{Z}\}

b) |z^{i}|\in\{e^{(2n-1)\pi}|n\in\mathbb{Z}\}

c) |z^{i}|\in\{e^{-{\pi\over 4}+2n\pi}|n\in\mathbb{Z}\}

d) |z^{i}|\in\{e^{{5\pi\over 4}+2n\pi}|n\in\mathbb{Z}\}

e) |z^{i}|\in\{e^{-\arctan{b\over a}+2n\pi}|n\in\mathbb{Z}\}
Last edited by Farenhajt on Sat Sep 16, 2006 7:16 pm; edited 1 time in total 
PostPosted: Sun Sep 10, 2006 7:18 am  Back to top 
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lotrgreengrapes7926
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#55
Farenhajt wrote:
lotrgreengrapes7926 wrote:
Problem 10:
Find, with proof, |z^{i}| when

a) z is a positive real;

b) z is a negative real;

c) z is in the form a+ai for positive real a;

d) z is in the form a+ai for negative real a;

e) z=a+bi, in terms of a and b.

Hint
\sin^{2}{x}+\cos^{2}{x}=1


Solution
If z=re^{i(\varphi+2k\pi)}, r,\varphi\in\mathbb{R}, k\in\mathbb{Z}, then

\begin{eqnarray*}|z^{i}| &=& |r^{i}e^{-\varphi-2k\pi}|\\ &=& e^{-\varphi-2k\pi}|e^{i\ln r}|\\ &=& e^{...

where n=-k

Hence

a) |z^{i}|\in\{e^{2n\pi}|n\in\mathbb{Z}\}

b) |z^{i}|\in\{e^{(2n-1)\pi}|n\in\mathbb{Z}\}

c) |z^{i}|\in\{e^{-{\pi\over 4}+2n\pi}|n\in\mathbb{Z}\}

d) |z^{i}|\in\{e^{{3\pi\over 4}+2n\pi}|n\in\mathbb{Z}\}

e) |z^{i}|\in\{e^{-\arctan{b\over a}+2n\pi}|n\in\mathbb{Z}\}


OK, give the answer that your calculator would give. Mr. Green
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PostPosted: Sun Sep 10, 2006 7:46 am  Back to top 
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Farenhajt
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#56
lotrgreengrapes7926 wrote:
OK, give the answer that your calculator would give. Mr. Green


My calculator is not as smart as I am (though that could be disputed) Very Happy

PostPosted: Sun Sep 10, 2006 10:05 am  Back to top 
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1=2
Birch & Swinnerton Dyer
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#57
Rzeszut wrote:
nutz_for2.718281828, I'm sure that you just were lazy and you know how to use Jensen. BTW, try to prove \mathrm{AM-GM} for n variables. Mr. Green


first of all, what the heck is Jensen?

second of all, I can't prove what you tole me to prove.
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PostPosted: Tue Sep 19, 2006 2:33 pm  Back to top 
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Rzeszut
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#58
nutz_for2.718281828 wrote:
Rzeszut wrote:
nutz_for2.718281828, I'm sure that you just were lazy and you know how to use Jensen. BTW, try to prove \mathrm{AM-GM} for n variables. Mr. Green


first of all, what the heck is Jensen?

second of all, I can't prove what you tole me to prove.

Jensen is a simple, but very strong and useful inequality and I do not agree that it's a heck.
Let f be a convex function defined at interval [a,b], x_{1},\ldots,x_{n}\in [a,b] and p_{1},\ldots,p_{n} be nonnegative reals whose sum is 1. Then: p_{1}f(x_{1})+\ldots+p_{n}f(x_{n})\geqslant f(p_{1}x_{1}+\ldots+p_{n}x_{n}). If f is concave, the sign of inequality should be reversed.
If you want to check if a function is convex or concave, just calculate the second deriviative: if f''\geqslant 0, then f is convex, if f''\leqslant 0, then f is concave.
Using Jensen you can easily prove \mathrm{AM-GM} for n variables: just substitute a_{i}=e^{x_{i}} in \frac{a_{1}+\ldots+a_{n}}{n}\geqslant \sqrt[n]{a_{1}\cdot\ldots\cdot a_{n}} and then it becomes Jensen for convex function f(t)= e^{t}.
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PostPosted: Wed Sep 20, 2006 5:03 am  Back to top 
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1=2
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#59
Since this is just not going anymore, I'll start it again.

Prove or disprove that the area of a regular n sided polygon with side length s is

\frac{ns^{2}\tan{\left(90-\frac{180}{n}\right)}}{4}
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Last edited by 1=2 on Mon Oct 30, 2006 10:21 am; edited 1 time in total 
PostPosted: Sun Oct 29, 2006 5:14 pm  Back to top 
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Rzeszut
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#60
nutz_for2.718281828 wrote:
Prove or disprove that the volume of a regular n sided polygon with side length is

\frac{ns^{2}\tan{\left(90-\frac{180}{n}\right)}}{4}

See the attached picture.
Let the regular n-gon A_{0}\ldots A_{n-1} be inscribed in a circle whose center is O and let's consider a triangle A_{i}OA_{i+1}. Of course \left|\measuredangle A_{i}OA_{i+1}\right|=\frac{2\pi}{n} and d=\frac{s}{2}\cot\left(\frac{\pi}{n}\right), so S_{A_{0}\ldots A_{n-1}}= \sum_{0\leqslant i\leqslant n-1}S_{A_{i}OA_{i+1}}= \sum_{0\leqslant i\leqslant n-1}\frac{d\cdot s}{2... n\cdot \frac{\frac{s}{2}\cot\left(\frac{\pi}{n}\right)\cdot s}{2}= n\cdot \frac{s^{2}\cdot\cot\left(\frac{\pi}{n}\right)}{4}=..., so the formula you wrote is correct.
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PostPosted: Mon Oct 30, 2006 12:41 am  Back to top 
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