Given a photon dropping from $\lambda_1$ to $\lambda_2$, its energy will drop from $\frac{hc}{(\lambda_1)}$ to $\frac{hc}{(\lambda_2)}$. However, I was wondering if there is any significance in the energy of the change in wavelength itself. For $\Delta \lambda = \lambda_2 - \lambda_1$, this change in wavelength has an energy $\frac{hc}{(\Delta \lambda)}$, yet this value does not correspond at all (as far as I can see) with $\Delta E = \frac{hc}{(\lambda_1)} -\frac{hc}{(\lambda_2)}$. Is this just a subtlety in the math, or is there actually meaning behind the value $\frac{hc}{(\Delta \lambda)}$?


Question: Interpolated FIR filter
Answer: The interpolation lowpass filter computes a weighted sum of input samples, which results in the zero input samples being interpolated using the non-zero samples of the input signal. The $*$ sign is NOT multiplication but convolution!

EDIT: I'm adding a simple example to clear things up a bit. Let's consider the impulse response $h = [0.5, 1, 0.5]$ and a zero-padded input signal
$x=[1,0,3,0,5,0,3,0,1]$. The time index is denoted by $n$, and $n=0$ corresponds to the left-most values of $h$ and $x$. The output $y$ is the convolution of $(x*h)(n)$:

$$y(0) = h(0)x(0) = 0.5\\
y(1) = h(0)x(1) + h(1)x(0) = 1\\
y(2) = h(0)x(2) + h(1)x(1) + h(2)x(0) = 2\\
y(3) = h(0)x(3) + h(1)x(2) + h(2)x(1) = 3\\
y(4) = h(0)x(4) + h(1)x(3) + h(2)x(2) = 4\\
y(5) = h(0)x(5) + h(1)x(4) + h(2)x(3) = 5\\\vdots$$

This impulse response obviously performs linear interpolation of the zero-padded input signal.