Vtyjkyuk

The second-derivative test states that if $x$ is a real number such that $f'(x)=0$, then:

1. If $f''(x)>0$, then $f$ has a local minimum at $x$.


2. If $f''(x)<0$, then $f$ has a local maximum at $x$.


3. If $f''(x)=0$, then the text is inconclusive.

I’m wondering whether non-standard analysis, involving the hyperreal numbers, allows for a more powerful version of the second-derivative test, one that has fewer or no inconclusive cases. In terms of nonstandard analysis, the second derivative of $f$ is defined as the standard part of $fracf’(x+epsilon)-f’(x)epsilon$ for any infinitesimal $epsilon$. But what if we omit the bit about the standard part? Fixing some infinitesimal $epsilon$, let $g(x)=fracf’(x+epsilon)-f’(x)epsilon$.

Then can we define a more powerful second derivative test as follows?

1. If $g(x)>0$, then $f$ has a local minimum at $x$.


2. If $g(x)<0$, then $f$ has a local maximum at $x$.


3. If $g(x)=0$, then the text is inconclusive.

Is this valid, and would this test apply to a greater class of functions than the regular second derivative test?

Let's take 1:

If $fracf’(x+epsilon)-f’(x)epsilon > 0$ then $f$ has a local minimum at $x$.

False in general.¹

This one works:

If $fracf’(x+epsilon)-f’(x)epsilon > 0$ for all positive infinitesmial $epsilon$, then $f$ has a local minimum at $x$.

But (by the transfer principle) that is the same as this one, using only ordinary real numbers $epsilon$:

If $fracf’(x+epsilon)-f’(x)epsilon > 0$ for all sufficiently small positive real $epsilon$, then $f$ has a local minimum at $x$.

¹ Example: $f(x) = int_0^x t sin frac1t;dt$, so $fracf’(0+epsilon)-f’(0)epsilon = sinfrac1epsilon$, which is positive for some positive infinitesimal $epsilon$ and negative for some other positive infinitesimal $epsilon$.