Fixed issue with vertical scrollbar and inline math

1 files changed, 4 insertions, 4 deletions

diff --git a/content/blog/fourier-noise.md b/content/blog/fourier-noise.md
index e17c119..8e1a114 100644
--- a/content/blog/fourier-noise.md
+++ b/content/blog/fourier-noise.md
@@ -83,11 +83,11 @@ $$\begin{aligned}dx_rdx_i &= \|{J_{x_r, x_i}(\alpha, \theta)}\| d\alpha d\theta
 
 Therefore, the integral now becomes the following. Note that for the distribution of $A$ we allow $\theta$ across its entire range. The value of $a$ also has to remain positive, therefore, the lower limit is changed from $- \infty$ to $0$.
 
-$$\begin{equation}P_A(a) = \int_0^{2\pi} \int_0^a \frac{1}{2\pi\sigma_n^2} exp(\frac{-\alpha^2}{2\sigma_n^2}) \alpha d\alpha d\theta\end{equation}$$
+$$P_A(a) = \int_0^{2\pi} \int_0^a \frac{1}{2\pi\sigma_n^2} exp(\frac{-\alpha^2}{2\sigma_n^2}) \alpha d\alpha d\theta$$
 
 I leave the simplification of this to the reader but this results in the following distribution:
 
-$$\begin{equation}P_A(a) = 1 - exp(\frac{-a^2}{2\sigma_n^2}) \end{equation}$$
+$$P_A(a) = 1 - exp(\frac{-a^2}{2\sigma_n^2})$$
 
 This makes the Probability Distribution Function (PDF) of $A$ to be:
 
@@ -99,10 +99,10 @@ This is a well know distribution called the [Rayleigh Distribution](https://en.w
 
 The similar derivation for the phase can be done and shows that the Cummulative Distribution Function (CDF) is:
 
-$$\begin{equation}P_\Phi(\phi) = \frac{\phi}{2\pi}\end{equation}$$
+$$P_\Phi(\phi) = \frac{\phi}{2\pi}$$
 
 which makes the Probability Distribution Function (PDF):
 
-$$\begin{equation}p_\Phi(\phi) = \frac{1}{2\pi}\end{equation}$$
+$$p_\Phi(\phi) = \frac{1}{2\pi}$$
 
 This means that the phase is uniformly distributed across the the interval $(0, 2\pi)$.