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I was going through the chapter on "The Principle Of Least Action" of The Feynman Lectures, and came across a property of minimum for a well-behaved function dependent on a single-variable (as I have deduced by merely looking at the figures given in the book), which he puts forth thus: "If we go away from the minimum in the first-order, the deviation of the function from its minimum value, is only second-order." After a few lines, he again states: "At any place else on the curve, if we move a small distance, the value of the function changes in the first-order also."

Using basic concepts from calculus, I tried to prove these two statements on my own. I succeeded in proving the latter, but got stuck on the former.

Here's how I proceeded:

I assume a twice-derivable(of course, a special case, but simple to analyse) function $f(x)$, dependent on a single variable $x$, and emphasize, that within a certain interval,f(x) is well-behaved. I don't need to know its behavior out of that interval.

At any general point, $f'(x)=k_1(x)-------(1); $

$ f"(x)=k_2 $(independent of x)$-------(2); k,c$ belongs to the set of reals.

so, starting from equn.. (2), I get:

$f(x)=(k_2/2)(x^2) + ax + b$, where, a,b are arbitrary constants, and, $k_1(x)=cx + a$

let $(x_1,f(x_1)), (x_2,f(x_2))$ be two co-ordinates of interest.

Then,

$f(x_1)=(k_2/2)(x_1^2) + ax_1 + b$,

and,

$f(x_2)=(k_2/2)(x_2^2) + ax_2 + b$,

hence, $f(x_2) - f(x_1) = (c/2) [(x_2)^2 - (x_1)^2] + a(x_2 - x_1)$,

, or,

change in $f(x)$= linear combination of second and first-order changes in $x$,

which duly proves the second part.

What I find troublesome is to prove the first part, as well as the second part for a general function (dependent on single-variable only). As you have noticed, I have assumed $f(x)$ to be well-behaved, "twice"-derivable function of $x$.

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At a minimum point in a function, $f'(x) = 0$ but $f''(x) > 0$. The fact that there is a nonzero second derivative means that the dominant (lowest-order) term in $f$ at that point is from the $x^2$.

For the second statement, any function can be approximated as a straight-line tangent to the function. This is the first-order term in the Taylor expansion:

$$f(x) = f(a) + f'(a)(x-a) + ...$$

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  • $\begingroup$ Of course, the first statement is only true if we assume that $f''(x)$ exists. $\endgroup$ Sep 29, 2014 at 18:21
  • $\begingroup$ @John : Still unclear to me. Please elaborate your answer. $\endgroup$
    – abstract
    Sep 29, 2014 at 18:27
  • $\begingroup$ It may also be probable that $f'(x)$ doesn't exist. $\endgroup$
    – FreeMind
    Sep 29, 2014 at 18:29
  • $\begingroup$ @abstract Use Taylor series and study why it works, think of it's geometric approach, that's your answer. $\endgroup$
    – FreeMind
    Sep 29, 2014 at 18:31
  • $\begingroup$ @ John,Freemind: Thanks for the tip. Moving on to Apostol----Taylor Series... $\endgroup$
    – abstract
    Sep 29, 2014 at 18:34

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