The derivative operator $D$ is given by $\frac{\ln(\varepsilon)}{h}$, where $\varepsilon$ is the shift operator.

Using the Taylor Expansion and the relation $\varepsilon = I + \Delta _{+}$, the derivative operator can be also written as $\frac{1}{h} (\Delta _{+} - \frac{1}{2}\Delta _{+}^2 + \frac{1}{3}\Delta _{+}^3 - \cdots)$.

One of my homework problem is to find a sequence of coefficients $\{a_i\}$ such that $D = \frac{1}{h} (\beta \Delta _{-} + \sum\limits_{i=1}^{\infty} a_i \Delta _{+}^i)$ for any given $\beta$. $a_i$ can be $\beta$-dependent (of course it should).

I tried using relation $\varepsilon = (I - \Delta _{-})^{-1}$ and ended up with $D = \frac{1}{h} (\Delta _{-} + \frac{1}{2}\Delta _{-}^2 + \frac{1}{3}\Delta _{-}^3 - \cdots)$. How can I use only one backward shift operator?


  • $\begingroup$ Can you give a little more background? Is your derivative a discrete or a numerical one instead of the usual derivative in analysis? Where are the functions defined that you want to differentiate? $\endgroup$ – Joonas Ilmavirta Feb 13 '15 at 22:13
  • $\begingroup$ It's concrete. But I think it converges to the real derivative by taking infinite sum. This D operator applies to all functions. In this class we use derivative approximation to solve some simple types of PDE using Matlab. $\endgroup$ – Isomorphism Feb 13 '15 at 22:17

You have,

$$D=\frac{1}{h} (\Delta _{+} - \frac{1}{2}\Delta _{+}^2 + \frac{1}{3}\Delta _{+}^3 - \cdots) $$ and, $$ D = \frac{1}{h} (\beta \Delta _{-} + \sum\limits_{i=1}^{\infty} a_i \Delta _{+}^i) $$

Thus, $$ \Delta_{-} = \sum\limits_{i=1}^{\infty} \dfrac{1}{\beta} \left( \dfrac{(-1)^{i-1}}{i} - a_{i}\right)\Delta_{+}^{i} $$

You also have,

$$ \Delta_{-} = \dfrac{\Delta_{+}}{I+\Delta_{+}} $$

Write the Taylor series for $\Delta_{-}=f(\Delta_{+})$. Set the coefficients of the two series to be equal and you'll find $a_i$.


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