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I'm approximating the second derivative of an arbitrary function.

Specifically, I'm asked to show that

$$ -\frac{d^2}{dt^2}\approx \frac{-u(t+h)+2u(t)-u(t-h)}{h^2} $$

for small $h$.

Intuitively, this seems like a reasonable approximation, the definition of the first derivative is

$$ \lim_{h\to0} \frac{u(t+h) - u(t)}{h} $$

There's a few parts that confuse me to get to the desired result... If we call that result $g(t)$ then we could apply the definition again.

$$ \lim_{h\to0} \frac{g(t+h) - g(t)}{h} = \lim_{h\to0}\left(\left[\lim_{h\to 0} (u(t+2h) - u(t+h))/h^2\right]-\left[\lim_{h\to0} (u(t+h) - u(t))/h\right]\right)/h^2 $$

Which doesn't actually make sense but that's where I'm at in my thought process.

financial_physician
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    Refer to https://math.stackexchange.com/questions/210264/second-derivative-formula-derivation – UNOwen Nov 09 '21 at 03:25

1 Answers1

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As you've taken let, $\begin{align}g(t) = \dfrac{du(t)}{dt} = \lim_{h\to0}\dfrac{u(t+h)-u(t)}{h}\end{align}$

Now let us define $\begin{align}\dfrac{dg(t)}{dt}\end{align}$ as,
$\begin{align} \dfrac{dg(t)}{dt} = \lim_{h\to0}\dfrac{g(t)-g(t-h)}{h}\end{align}$

So,

$\begin{align}&\dfrac{dg(t)}{dt} = \dfrac{d^2u(t)}{dt^2} = \lim_{h\to0}\dfrac{\dfrac{u(t+h)-u(t)}{h}-\dfrac{u((t-h)+h)-u(t-h)}{h}}{h}\\\\\Rightarrow& \dfrac{d^2u(t)}{dt^2}= \lim_{h\to0}\dfrac{u(t+h)-2u(t)+u(t-h)}{h^2}\end{align}$

19aksh
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