So, we have this expression here for the probability of measuring any particular y. I'm going to introduce the change of variable. So, instead of this expression i, this change of variable, I will write just e in Theta_ye. By this index, y, we recognize that this Theta_y depends on y. So, this variable change, everything will become shorter. Now, we are going to distinguish some y's which are in some sense better than the others. For those y's which satisfy this equation, we will call them good y's. We are going to estimate the probability of measuring good y's. For this expression here, if we rewrite it with the variable change, let's do it. We can notice that it is sum of geometric progression, and we know the formula of the sum and if you don't, we are still lucky to have it right here. So, this sum equals to e_Theta_y i minus 1 divided by e_Theta_yi minus 1. So, this is what we have here instead of this sum. Okay. So, instead of these here, you're going to write this. As I'm going to estimate this fraction for good y's. I'm going to do this separately for numerator and denominator. For the numerator I'm going to write this, that's e power A Theta_y i minus 1. I'm going to prove that this greater or equal to 2A absolute value Theta_y divided by Pi for good y's. To do this, I'm going to introduce the variable change to write less, so that X be A Theta_y, and let's take a closer look on this expression. So, it's absolute value e_xi minus 1. I'm going to use the Euler's formula. So, this cosine x minus 1 plus i sine x. Now, instead of this absolute value, I'm going to write square root. So, this cosine x minus 1 squared plus sine squared x. When we opened this square, I get square of cosine which among with sign gives us one, and another one we get from here. So, it will be 2 minus 2 cosine x which goes from the scope. Does not look good yet, so, we're going to employ, and the formula for cosine of double angle. Instead of this two here, I'm going to write 2cosine squared x divided by 2 plus sine squared x divided by 2. Instead of 2cosine, we're going to write 2cosine squared x divided by 2 plus 2sine squared x divided by 2. So here I employ the formula for the cosine of double angle. These loops now whether because these cosines, they go away, and there on the think left here is four squared sines. So, we can write that it equals to 2sine modular x divided by 2. So, this inequality here, I can rewrite now as 2sine_A absolute value Theta_y divided by 2 is greater or equal to A Theta_y divided by Pi. We can now remove this 2's. Let's take a look at these four good y's. So, this is the linear function on the interval from zero to Pi. Here, we have zero, here we have Pi, and zero which equals zero and Pi. When A Theta_y equals to Pi, this Pi divided by Pi, so it is one. Linear function. This sine here, when A Theta_y goes to zero also equals to zero, and when it equals to Pi, then sine of Pi divided by 2 is also one. So, in the ends of the intervals these two expressions are equal. On this integral sine is a convex function, so it goes like this, and on the wall integral, it is bigger than this linear function. We proved our inequality for good y's. So, we have estimated the numerator. Okay. We put this achievement here. We are going now to estimate the denominator. I'm going to prove that this denominator Theta_y i minus 1 is less or equal to Theta_y absolute value. So, if we estimated that numerator is bigger than we solve them, then we have to prove that the denominator is less than we solve them. How we are going to prove this? Well, let's see this geometrically. Here's the complex plane. So here, we have one, here we have one, and this is unitary circle. Not very pretty on the complex plane. We have this angle Theta_y here. This expression, this absolute value, is here we have e and power Theta_y i minus 1. So, this expression on the left is the length of this chord. This expression on the right is the length of this arc. So, what I am trying to say is that the arc is longer than the chord. It is always true, so I believe I proved this. Now, we have both estimations here, and we are ready to estimate the whole thing, this probability. So, it's going to be greater or equal to one divided by A. Instead of the numerator over here take this. So, it's 4A squared. So, it really Theta_y squared. Here, we have Pi squared, and here we have just Theta_y squared. This goes away, this goes away, and we have these. This expression here, it's approximately one divided by R as we remember, so I'm going to write it. So, this is approximately 4 divided by Pi squared R. This is our probability to measure any of good y's. Why these y's are good and what to do with this probability, we are going to discuss in the next episode.