# Is ${e^{ikx}} \to {x^2}$ when $k \to 0$

in an online video lecture,(around 36min, where the exactly statement is at 36min33secs.) i got one question, suppose we have a system of $$N$$ particles, $$\left\{ {{{\vec r}_i}(t)} \right\}i = 1, \cdot \cdot \cdot ,N$$ are the position vectors of the particles. I was told in the lecture that the so-called self intermediate scattering function is defined as. $${F_s}(k,t) = \frac{1}{N}\left\langle {\sum\limits_{i = 1}^N {{e^{i\vec k \cdot [{{\vec r}_i}(t) - {{\vec r}_i}(0)]}}} } \right\rangle.$$ (for homogeneous system, it only depends on the absolute value of $$\vec k$$.)

Furthrmore, it is said by the lecturer that when $$k \to 0$$, $${F_s}(k,t) \to \frac{1}{N}\left\langle {\sum\limits_{i = 1}^N {{{[{{\vec r}_i}(t) - {{\vec r}_i}(0)]}^2}} } \right\rangle$$

but i can't see why. Could anybody give me some help on it.

Start from this expression :

$$e^{ikx}=\cos kx + i \sin kx$$

As $$kx \to 0$$ we get $$\sin kx \to kx$$ and $$\cos kx \to 1-\frac {(kx)^2} 2$$

Now from listening to that section of the video I do not think that the lecturer is saying $$e^{ix}\to x^2$$ but is saying that there is a contribution in the low $$kx$$ range that includes a squared factor, which is coming from the expansion of $$\cos kx$$ in the low $$kx$$ range.

The squared quantity expectation value is referred by him as the mean squared displacement, but it is not equated to $$F_s(k,t)$$ (at least not in the section of video you refer to).

• Thanks StephenG! I think I misunderstood what the lecturer said.