Second Derivative for Non standard Calculus

Question

I am trying to understand how second derivatives work in Non standard calculus (NSA). The definition of the first derivative in NSA is $ st(\frac{ f(x + ε) - f(x) }ε) $ .

So if we have $f(x) = x^2 $, then the first derivative is $ st(\frac{ (x + ε)^2 - x^2 }ε) $ which becomes $ st(2x + ε) $ which is $2x$ as you'd expect.

Lets say we set $x = 8$. Here the first derivative is $ st(2 *8 + ε) $ which is 16 as you'd expect. Now, lets say we make $x = 8+ε_1 $. Then our first derivative at $8+ε_1 $ is $ st(2 (8+ε_1) + ε) $ which is also 16.

But as we move from $x = 8 $ to $x = 8+ ε_1$, then our first derivative does not change, so this implies that our second derivative is 0, when it should be 2 from standard analysis. I'd greatly appreciate anyone telling me where I'm wrong. Thanks

Answer 1

The reason why this doesn't work is simple: This way to derive the derivative only works at standard points. Observe what happens if you e.g. calculate $f'(\delta)$ for some infinitesimal $\delta$.

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(Edited as per comment discussion)

Since you haven't given us your exact process, I'll fill in what I think it is, to show the exact problem.

First we define for any $x_0\in D\subseteq ℝ$, where $x_0$ is an accumulation point of $D$, for a function $f:D\toℝ$ its derivative in $x_0$ as $f'(x_0):=st(\frac{^*f(x)-f(x_0)}{x-x_0})$, if this value is identical for all $x\approx x_0\land x\neq x_0$.

This is a text-book definition, but only valid for $x\inℝ$. So we can talk about $f'(x_0)$, but not about $f'(x_0+\delta)$ for $\delta\approx 0\land \delta\neq 0$.

What I think you now did was just plug in this definition twice to arrive at $f''(x_0)$: $$ f''(x_0)=st(\frac{\color{green}{^*f'(x)}-f'(x_0)}{x-x_0})\color{red} = st(\frac{^*\color{blue}{st(\frac{{^*f(x')}-{f(x)}}{x'-x})}-st(\frac{^*f(x'')-f(x_0)}{x''-x_0})}{x-x_0}) $$ You here end up with a wrong answer since at the green part, you're trying to use the definition on a non-standard-number. This then causes in the blue part both $x$ and $x'$ to be non-standard, so the $st$ smooths out more than you want it to.

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Trying to compute $^*f'$ at non-standard points, as far as I know, either uses $\epsilon$-$\delta$-method over the hyperreals, or computes $f'$ as standard function and then uses the transfer method to deduce $^*f'$.

There are, however, ways to directly compute $f^{(n)}(x)$ for standard $x$. In the end you simply need to find the non-standard expression that is equivalent to the stack of limits that you get when iteratively substituting the definition of a derivative in $f^{(n)}(x)$.

This however is a bit of a hassle, so in many cases it's skipped over. For example you need to differentiate between iterated limits and simultaneous limits.

Answer 2

$ \newcommand\st{\mathop{\mathbf{st}}} $If you take the derivative of $2x$ you of course get the right thing: $$ \st\frac{2(x+\epsilon)-2x}\epsilon = \st(2\epsilon/\epsilon) = 2. $$ The issue is you didn't compute the second derivative. Maybe you're saying "the second derivative is nonzero, but the derivative doesn't change when I add $\epsilon_1$." But it does change, you just deleted the change by taking the standard part. In general $$ f(x+\epsilon) \approx f(x) + f'(x)\epsilon. $$ The derivative is the coefficient of $\epsilon$. At the same time, $\st f(x+\epsilon) = f(x)$ because you only moved infinitesimally.

Answer 3

The definition of the first derivative via standard part of the differential ratio will always produce a standard number. Therefore that particular definition only works at standard points and cannot work at a nonstandard point. For example, for the function $x^2$ the derivative is $2x$, which is nonstandard if $x$ is infinitesimal. But in fact one can compute the second derivative via standard part. Indeed, if $f(x)$ is twice differentiable at a standard point $c$, then $f''(c)$ can be found as the standard part of $\frac{f(c+h)+f(c-h)-2f(c)}{h^2}$ where $h$ is infinitesimal. This produces the second derivative at all standard points. By the transfer principle, the resulting function will be naturally defined at all points.

Answer 4

The second derivative is the derivative of the derivative. $\DeclareMathOperator{st}{st}$ \begin{align*} f''(x) &= (f'(x))' \\ &= \st\left( \left. \frac{f'(x+u) - f'(x)}{u} \right|_{u=\varepsilon}\right) \\ &= \st \left( \left. \frac{\st \left( \left. \frac{f(x+u+v) - f(x+u)}{v} \right|_{v = \varepsilon} \right) - \st \left( \left. \frac{f(x+v) - f(x)}{v} \right|_{v = \varepsilon}\right)}{u} \right|_{u = \varepsilon}\right) \text{.} \end{align*}

When you say "move from $x=8$ to $x=8+\varepsilon_1$, then our first derivative does not change", this is because you've lost track of the three infinitesimal shifts we need, the one we need for $f'(x+\varepsilon) - f'(x)$, the one inside $f'$ evaluated at $x+\varepsilon$, and the one inside $f'$ evaluated at $x$.

A similar thing happens in normal calculus. "$f''(3)$" means to find the second derivative function, then evaluate it. When you say "let $x = 8 + \varepsilon_1$", you're evaluating $f'$ at $8 + \varepsilon_1$. What you are not doing is measuring the second derivative from the rate of change between $(x,f(x))$ and $(x+\varepsilon, f(x+\varepsilon))$ and the rate of change through $(x+\varepsilon,f(x+\varepsilon))$ and $(x+2\varepsilon,f(x+2\varepsilon))$.

For your example, $f(x) = x^2$, the second derivative is \begin{align*} f''(x) &= \st \left( \left. \frac{\st \left( \left. \frac{(x+u+v)^2 - (x+u)^2}{v} \right|_{v = \varepsilon} \right) - \st \left( \left. \frac{(x+v)^2 - (x)^2}{v} \right|_{v = \varepsilon}\right)}{u} \right|_{u = \varepsilon}\right) \\ &= \st \left( \left. \frac{\st \left( \left. \frac{2xv + 2uv + v^2}{v} \right|_{v = \varepsilon} \right) - \st \left( \left. \frac{2xv+v^2}{v} \right|_{v = \varepsilon}\right)}{u} \right|_{u = \varepsilon}\right) \\ &= \st \left( \left. \frac{\st \left(2x + \overbrace{2u}^{\text{here's your $\varepsilon_1$ and it's not just $u = \varepsilon$}} + \varepsilon \right) - \st \left(2x + \varepsilon \right)}{u} \right|_{u = \varepsilon}\right) \\ &= \st \left( \left. \frac{2x + 2u - 2x}{u} \right|_{u = \varepsilon}\right) \\ &= \st \left(\frac{2\varepsilon}{\varepsilon} \right) \\ &= 2 \text{.} \end{align*}

Admittedly, at the place we have "$2x+2u+\varepsilon$", we have $f'(x+u)+\varepsilon$ and we are going to specialize $u$ to $\varepsilon$. But we shouldn't be surprised that in this step, we are looking at $f'(x+\varepsilon) - f'(x)$, since that's what we'd want/expect to see in the numerator of the difference quotient for the second derivative.