2
$\begingroup$

I have a function, $H$, which is a dependent on a number of parameters, $\theta_{i=1,\ldots,n}$ and a number of complex coefficients. The function hence gives a complex quantity. $H$ is relatively simple to differentiate w.r.t $\theta_i$.

What I am interested in finding is,

\begin{equation} \sum_i \left| \frac{\partial|H|}{\partial \theta_i} \right|^2 \end{equation}

Here is where I am unsure of myself: Because H is a function of theta

\begin{equation} \frac{\partial|H|}{\partial \theta_i} = \frac{\partial|H|}{\partial H}\frac{\partial H}{\partial \theta_i} \end{equation}

My question is: How do I calculate the differential of $|H|$ w.r.t $H$, given that H is complex (I think that this differential is undefined everywhere except zero?)

Is there another way of expanding this in a form where I can calculate the differential.

Note: although I can calculate dH/dtheta, it is difficult to directly calculate d(abs(H))/dtheta.

Many thanks in advance.

Edit

If you will indulge me for one more question,

Am I right in saying:

\begin{equation} \frac{\partial H_R}{\partial \theta_i} = 0.5 \left(\frac{\partial H}{\partial \theta_i} + \frac{\partial \overline{H}}{\partial \theta_i} \right) \end{equation}

So that I can use the partial differentials I have already calculated?

$\endgroup$
2
$\begingroup$

For a complex function $f(H)$, its derivative is defined as

$$f'(H)=\lim_{\Delta H\to 0}\frac{f(H+\Delta H)-f(H)}{\Delta H}$$

if the limit exists.

If $f(H)=|H|$, we have

$$\begin{align} f'(H)&=\lim_{\Delta H\to 0}\frac{|H+\Delta H|-|H|}{\Delta H}\\\\ &=\lim_{\Delta H\to 0}\frac{H \Delta \bar H+\bar H \Delta H+|\Delta H|^2}{\Delta H \left(|H+\Delta H|+|H|\right)}\\\\ \end{align}$$

which exists nowhere.

Therefore, we conclude that the function $|H|$ is nowhere differentiable.


Now, suppose that $H$ is a complex function of real variables, $\theta_i$, $i=1,\cdots,n$. Then, we can write

$$H(\vec \theta)=H_R(\vec \theta)+H_I(\vec \theta)$$

where $H_R$ and $H_I$ represent the real and imaginary parts of $H$, respectively. We can find the partial derivatives of $|H|$ as

$$\begin{align} \frac{\partial |H(\vec \theta)|}{\partial \theta_i}&=\frac{\partial \sqrt{H_R^2(\vec \theta)+H_I^2(\vec \theta)}}{\partial \theta_i}\\\\ &=\frac{H_R(\vec \theta)\frac{\partial H_R(\vec \theta)}{\partial \theta_i}+H_I(\vec \theta)\frac{\partial H_I(\vec \theta)}{\partial \theta_i}}{\sqrt{H_R^2(\vec \theta)+H_I^2(\vec \theta)}}\\\\ &=\frac{H_R(\vec \theta)}{|H(\vec \theta)|}\frac{\partial H_R(\vec \theta)}{\partial \theta_i}+\frac{H_I(\vec \theta)}{|H(\vec \theta)|}\frac{\partial H_I(\vec \theta)}{\partial \theta_i} \end{align}$$

$\endgroup$
  • $\begingroup$ If you could address my edit I would be very great-full. $\endgroup$ – Will Sep 28 '15 at 10:34
  • $\begingroup$ @will Yes. We can write $H_R=(H+\bar H)/2$ and proceed. Well done. $\endgroup$ – Mark Viola Sep 28 '15 at 13:50
  • $\begingroup$ Thanks for your help. It is much appreciated! $\endgroup$ – Will Sep 28 '15 at 14:58
  • $\begingroup$ You're welcome! My pleasure. $\endgroup$ – Mark Viola Sep 28 '15 at 14:59

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.