Fix doc

units/en/unit4/pg-theorem.mdx

@@ -21,18 +21,18 @@ So we have:
 
 We can rewrite the gradient of the sum as the sum of the gradient:
 
-\\( =  \sum_{\tau} \nabla_\theta (P(\tau;\theta)R(\tau)) = \sum_{\tau} \nabla_\theta P(\tau;\theta)R(\tau) \\) as \\(R(\tau)\\) is not dependent on \\(\theta\\)
+\\(\nabla_\theta J(\theta) =  \sum_{\tau} \nabla_\theta (P(\tau;\theta)R(\tau)) = \sum_{\tau} \nabla_\theta P(\tau;\theta)R(\tau) \\) as \\(R(\tau)\\) is not dependent on \\(\theta\\)
 
-We then multiply every term in the sum by \\(\frac{P(\tau;\theta)}{P(\tau;\theta)}\\)(which is possible since it's = 1)
+We then multiply every term in the sum by \\(\frac{P(\tau;\theta)}{P(\tau;\theta)}\\)(which is possible since it's = 1):
 
-\\( = \sum_{\tau} \frac{P(\tau;\theta)}{P(\tau;\theta)}\nabla_\theta P(\tau;\theta)R(\tau) \\)
+\\(\nabla_\theta J(\theta) = \sum_{\tau} \frac{P(\tau;\theta)}{P(\tau;\theta)}\nabla_\theta P(\tau;\theta)R(\tau) \\)
 
 
 We can simplify further this since \\( \frac{P(\tau;\theta)}{P(\tau;\theta)}\nabla_\theta P(\tau;\theta) =  P(\tau;\theta)\frac{\nabla_\theta P(\tau;\theta)}{P(\tau;\theta)}  \\). 
 
 Thus we can rewrite the sum as
 
-\\( P(\tau;\theta)\frac{\nabla_\theta P(\tau;\theta)}{P(\tau;\theta)}= \sum_{\tau} P(\tau;\theta) \frac{\nabla_\theta P(\tau;\theta)}{P(\tau;\theta)}R(\tau) \\)
+\\(\sum_{\tau} P(\tau;\theta) \frac{\nabla_\theta P(\tau;\theta)}{P(\tau;\theta)}R(\tau) \\)
 
 We can then use the *derivative log trick* (also called *likelihood ratio trick* or *REINFORCE trick*), a simple rule in calculus that implies that \\( \nabla_x log f(x) = \frac{\nabla_x f(x)}{f(x)} \\)
 
@@ -70,10 +70,8 @@ We also know that the gradient of the sum is equal to the sum of gradient:
 \\( \nabla_\theta log P(\tau^{(i)};\theta)=\nabla_\theta log\mu(s_0) + \nabla_\theta \sum\limits_{t=0}^{H} log P(s_{t+1}^{(i)}|s_{t}^{(i)} a_{t}^{(i)}) + \nabla_\theta \sum\limits_{t=0}^{H} log \pi_\theta(a_{t}^{(i)}|s_{t}^{(i)}) \\)
 
 
-Since neither initial state distribution or state transition dynamics of the MDP are dependent of \\(\theta\\), the derivate of both terms are 0. So we can remove them:
-
-Since:
-\\(\nabla_\theta \sum_{t=0}^{H} log P(s_{t+1}^{(i)}|s_{t}^{(i)}  a_{t}^{(i)}) = 0 \\) and \\( \nabla_\theta \mu(s_0) = 0\\)
+Since neither initial state distribution or state transition dynamics of the MDP are dependent of \\(\theta\\), the derivate of both terms are 0 
+(\\(\nabla_\theta \sum_{t=0}^{H} log P(s_{t+1}^{(i)}|s_{t}^{(i)}  a_{t}^{(i)}) = 0 \\) and \\( \nabla_\theta \mu(s_0) = 0\\)). So we can remove them:
 
 \\(\nabla_\theta log P(\tau^{(i)};\theta) =   \nabla_\theta \sum_{t=0}^{H} log \pi_\theta(a_{t}^{(i)}|s_{t}^{(i)})\\)