0
$\begingroup$

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?

Thanks!

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $a_i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.