How does the author obtain formula (4)? From formula (2), I only get that $u\left(\frac{k}{n}+\frac{1}{n},\cdot\right)=\left(1-\frac{c}{n}\partial_x\right)u\left(\frac{k}{n},\cdot\right)$ but I don't see how the exponent $k$ appears as in (4).
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for one step backward it is what you said: $$ u(t + \frac{1}{n},.) \approx (1-\frac{c}{n} \partial_x)u(t,.) $$ so if you start from $t=0$ and use this approximation $k$ times each time with step size $\Delta t = 1/n$ you get: $$ u(0+\frac{k}{n},.) \approx (1-\frac{c}{n} \partial_x)u(\frac{k-1}{n},.) \\ \approx (1-\frac{c}{n} \partial_x)(1-\frac{c}{n} \partial_x)u(\frac{k-2}{n},.) \approx \dots \approx (1-\frac{c}{n} \partial_x)^k u(0,.) = (1-\frac{c}{n} \partial_x)^k g(.) $$
Abolfazl Chaman motlagh
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Suppose that for some $k\ge 1$, we have $$ u(\frac kn, \cdot) \approx (1-\frac{c}{n}\partial_x)^kg(\cdot).\tag{$\ast$} $$ The case $k = 1$ holds by (2). Again by the approximation (2) and our assumption $(\ast)$, \begin{align*} u(\frac kn + \frac 1n,\cdot) &\approx (1-\frac cn\partial_x)u(\frac kn,\cdot)\\ &\approx (1-\frac{c}{n}\partial_x)^{k+1}g(\cdot). \end{align*} By induction (4) holds for each $k$.
Alex Ortiz
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