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March 24th, 2018, 10:58 AM   #1
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God's Book of Proofs

A diverting article on an interesting book.

https://www.wired.com/story/in-searc...ect-proofs/amp
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March 24th, 2018, 12:41 PM   #2
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Let's show one of them.
Prove that $\displaystyle a_1 ^2 +a_2 ^2 +...+a_n^ 2 \geq \frac{1}{n} for \; \; a_1 + a_2 + ... + a_n = 1$
$\displaystyle a_1 , a_2 ,..., a_n $ real numbers

Last edited by skipjack; March 24th, 2018 at 12:54 PM.
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March 24th, 2018, 01:17 PM   #3
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Beautifully stated, yet I have no idea how to prove it.

Ramanujan?

Last edited by Loren; March 24th, 2018 at 01:25 PM.
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March 24th, 2018, 01:40 PM   #4
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Start from assuming all $a_k=\frac{1}{n}$. Then sum $=\frac{1}{n}$. Show that any deviation increases the sum. You might be able to use mathematical induction starting from n=2.
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March 25th, 2018, 06:14 AM   #5
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I will show at least one method ...
$\displaystyle (1^2 + 1^2 +... + 1^2)\sum a_\omega ^2 \geq \sum a_\omega$
$\displaystyle \Rightarrow n\sum a_\omega ^2 \geq ( \sum a_\omega )^2 = 1^2 = 1 $ proved
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March 25th, 2018, 08:15 AM   #6
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Quote:
Originally Posted by idontknow View Post
$\displaystyle (1^2 + 1^2 +... + 1^2)\sum a_\omega ^2 \geq \sum a_\omega$
Perhaps I'm being a bit slow, but it's not immediately clear to me why this is true.

It's not the nicest, but here's what I came up with: for each $i$, we have $(\frac{1}{n} - a_i)^2 = \frac{1}{n} (\frac{1}{n} - 2a_i) + a_i^2$, so

$\begin{align*} \sum_{i=1}^{n} {a_i^2}
&= \sum_{i=1}^{n} \left(\frac{1}{n} - a_i \right)^2 + \frac{1}{n} \sum_{i=1}^{n} \left( 2a_i - \frac{1}{n} \right) \\
&\geq \frac{1}{n} \sum_{i=1}^{n} \left( 2a_i - \frac{1}{n} \right)
\qquad \qquad & \text{since $\left(\frac{1}{n} - a_i \right)^2 \geq 0$ for all $i$} \\
&= \frac{1}{n} \left( 2\sum_{i=1}^{n} a_i - \sum_{i=1}^{n} \frac{1}{n} \right) \\
&= \frac{1}{n} \left(2 \times 1 - n \times \frac{1}{n}\right)
& \text{since $\sum_{i=1}^{n}a_i = 1$} \\
&= \frac{1}{n} \end{align*}$
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January 10th, 2019, 12:05 AM   #7
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Quote:
Originally Posted by cjem View Post
Perhaps I'm being a bit slow, but it's not immediately clear to me why this is true.

It's not the nicest, but here's what I came up with: for each $i$, we have $(\frac{1}{n} - a_i)^2 = \frac{1}{n} (\frac{1}{n} - 2a_i) + a_i^2$, so

$\begin{align*} \sum_{i=1}^{n} {a_i^2}
&= \sum_{i=1}^{n} \left(\frac{1}{n} - a_i \right)^2 + \frac{1}{n} \sum_{i=1}^{n} \left( 2a_i - \frac{1}{n} \right) \\
&\geq \frac{1}{n} \sum_{i=1}^{n} \left( 2a_i - \frac{1}{n} \right)
\qquad \qquad & \text{since $\left(\frac{1}{n} - a_i \right)^2 \geq 0$ for all $i$} \\
&= \frac{1}{n} \left( 2\sum_{i=1}^{n} a_i - \sum_{i=1}^{n} \frac{1}{n} \right) \\
&= \frac{1}{n} \left(2 \times 1 - n \times \frac{1}{n}\right)
& \text{since $\sum_{i=1}^{n}a_i = 1$} \\
&= \frac{1}{n} \end{align*}$
How did you get this:

$ \displaystyle \sum_{i=1}^{n} \frac{1}{n} = n \times \frac{1}{n} $


Isn't $ \displaystyle \sum_{i=1}^{n} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} $ ?

Last edited by pnerd; January 10th, 2019 at 12:08 AM.
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January 10th, 2019, 06:43 AM   #8
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No $$ \sum_{i=1}^{n} \frac{1}{i} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} $$
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January 10th, 2019, 07:36 AM   #9
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Quote:
Originally Posted by v8archie View Post
No $$ \sum_{i=1}^{n} \frac{1}{i} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} $$
Oh yes. I had a brain fart.
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January 10th, 2019, 08:19 AM   #10
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It is solved but symbols are not used well .

Last edited by idontknow; January 10th, 2019 at 08:22 AM.
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