The principle of least Action proof of minimum

[shakaman2]

Homework Statement

Reading Feynman The Principle of Least Action out of The Feynman Lectures on Physics, Vol 2. Link to text http://www.feynmanlectures.caltech.edu/II_19.html
So I'm having a problem proving that, section 19-2 5th paragraf, that
"Now the mean square of something that deviates around an average, as you know, is always greater than the square of the mean; ". The purpose of this was to illustrate the principle of least action. Then proving that the path that is taken by a particle is the one with Actions = 0, and the way to get to that path. Below is a comment that is underlined(sp?) made while discussing how to do this.

Homework Equations

So definitions:
Numbers: x_1+...x_n
Mean: (x_1+...x_n)/n
Mean square: ((x_1+...x_n)/n)^2
Square mean: (x^2_1+...x^2_n)/n
Kinetic Energy: KE
Potential Energy: PE
Action = S = \int_{t_1}^{t_2} (KE-PE) dt

The Attempt at a Solution

So i can prove for myself that the following is true:
((x_1+...x_n)/n)^2 \geq (x^2_1+...x^2_n)/n \quad \forall x_i \geq 0

The problem here s that it is not always that x_i \geq 0. Since we have v \in \mathbb{R}, and therefore can be less then 0. So when we later go to find the path that is right we can't minimize with respect to the action.
What am I missing here?

 

[Ray Vickson]

Homework Statement

Reading Feynman The Principle of Least Action out of The Feynman Lectures on Physics, Vol 2. Link to text http://www.feynmanlectures.caltech.edu/II_19.html
So I'm having a problem proving that, section 19-2 5th paragraf, that
"Now the mean square of something that deviates around an average, as you know, is always greater than the square of the mean; ". The purpose of this was to illustrate the principle of least action. Then proving that the path that is taken by a particle is the one with Actions = 0, and the way to get to that path. Below is a comment that is underlined(sp?) made while discussing how to do this.

Homework Equations

So definitions:
Numbers: x_1+...x_n
Mean: (x_1+...x_n)/n
Mean square: ((x_1+...x_n)/n)^2
Square mean: (x^2_1+...x^2_n)/n
Kinetic Energy: KE
Potential Energy: PE
Action = S = \int_{t_1}^{t_2} (KE-PE) dt

The Attempt at a Solution

So i can prove for myself that the following is true:
((x_1+...x_n)/n)^2 \geq (x^2_1+...x^2_n)/n \quad \forall x_i \geq 0

The problem here s that it is not always that x_i \geq 0. Since we have v \in \mathbb{R}, and therefore can be less then 0. So when we later go to find the path that is right we can't minimize with respect to the action.
What am I missing here?

I hope you can't prove that
((x_1+...x_n)/n)^2 \geq (x^2_1+...x^2_n)/n \quad \forall x_i \geq 0 \; \Leftarrow \:\text{FALSE}
because the exact opposite is true:
\frac{1}{n} \sum_{i=1}^n x_i^2 \geq \left( \frac{1}{n} \sum_{i=1}^n x_i \right)^2 \;\; \forall x_1, x_2, \ldots, x_n \in \mathbb{R}
This follows from expansion and simplification of
\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \geq 0,
where $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is the average of the $x_i$.

 

[shakaman2]

I hope you can't prove that
((x_1+...x_n)/n)^2 \geq (x^2_1+...x^2_n)/n \quad \forall x_i \geq 0 \; \Leftarrow \:\text{FALSE}
because the exact opposite is true:
\frac{1}{n} \sum_{i=1}^n x_i^2 \geq \left( \frac{1}{n} \sum_{i=1}^n x_i \right)^2 \;\; \forall x_1, x_2, \ldots, x_n \in \mathbb{R}
This follows from expansion and simplification of
\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \geq 0,
where $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is the average of the $x_i$.

Yes sorry the (...)^2 should be before the n \quad (x_1+...x_n)^2 /n \geq (x^2_1+...x^2_n)/n \quad \forall x_i \geq 0 \;
Which I get from (a^2+b^2) \geq (a+b)^2
But when I try to edit I get:
The following error occurred:
Your content can not be submitted. This is likely because your content is spam-like or contains inappropriate elements. Please change your content or try again later. If you still have problems, please contact an administrator.

Was trying to wait it out.

 

[Ray Vickson]

Yes sorry the (...)^2 should be before the n \quad (x_1+...x_n)^2 /n \geq (x^2_1+...x^2_n)/n \quad \forall x_i \geq 0 \;
Which I get from (a^2+b^2) \geq (a+b)^2
But when I try to edit I get:
The following error occurred:
Your content can not be submitted. This is likely because your content is spam-like or contains inappropriate elements. Please change your content or try again later. If you still have problems, please contact an administrator.

Was trying to wait it out.

You keep writing false things. It is not true that $a^2 + b^2 \geq (a+b)^2$. Try putting $a = b = 1$ and see what happens.