# 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?

Last edited:

## 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##.

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:
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 $$(a^2+b^2) \geq (a+b)^2$$