So, we have two equations...ok? One equation tells us the relationship between the two Lagrange multipliers In particular, it tells us "Z" is a function of Lambda 1 Ok? And the second equation here two equations, two unknowns will allow us to solve for Lambda 1 itself So, "Z" here (unintelligible) But if you remember how your geometric series work, you can actually write this out because this, the first couple terms just to drive your intuition, look like one plus 'e' to the negative lambda one plus 'e' to the negative 2 lambda one and so forth. And so we know the geometric series of that form by a very nice argument that we won't do here. Sum to one over 1 minus 'e' to the negative lambda 1. So we know already how to reduce this infinite sum to a very elegant form. The next thing I'll point out, because I'm an expert at this...I'm always, I'm always gonna...going to, uh, eventually (unintelligible) from this We like to call 'z' the Partition Function Ok? 'z' here is the sum over these terms here. And so one thing you notice about 'z' about the partition function is if you take the derivative of 'z' with respect 'i', look what happens. The derivative of 'z' - sorry, we just got to lambda 1 - derivative of 'z' with respect to lambda 1 what happens is the following: Well, you get a negative sign and you get the sum. And now what you've done is pulled down a factor of 'i'. So, now all of a sudden, you have the same infinite sum you have in the other equation appearing. This sum here you can't do immediately, by the geometric series trick because instead of it being 1 plus 'p' plus 'p' squared plus 'p' to the fourth...plus 'p' cubed and so forth, it now has these funny terms in the front here hanging off. But notice that if you the derivative of the partition function, you pull down that factor of 'i'. So, in fact, we can now rewrite this second constrained equation here as 1 over 'z', minus, 'd' 'Z' 'd' lambda 1 equals 4. So, first thing you have to do is take a derivative of 'z' with respect to lambda 1 and then we have to divide by 'z'. And so we can do that quite easily. We have a factor 'e' to the negative lambda 1 on the top, 1 minus 'e' to the negative lambda 1 squared - that's the derivative of 'z' with respect to lambda 1...minus sign. And then we have a factor of 1 over 'z', and all that does is pull one of these out. So 'z' is 1 over 1 minus 'e' to the negative lambda 1. Ok? So if we divide by that...we just lose one of these. And so now, we've turned this equation here, which we would have trouble solving, because it's an infinite sum, and you know, putting an infinite sum into MatLab tends to take a long time to solve. But instead, we have a analytic form, a functional form that we can put in right away. Ok? So now, all we have to do is solve this equation here. Find the value of lambda 1 that sets this term here equal to 4. Once we find that value of lambda 1, we can then plug into 'z' here, ok? And once we know both 'z' and lambda 1, we can recover not just the functional form of the probability of waiting for 'x', but actually we'll be able to compute the waiting time as a function of 'x', ok? Er, the probability of a waiting time for a particular value, ok? So, we're left with this equation here and instead of solving that exactly, because we have some logarithms, I can tell you the answer that lambda 1 is equal to roughly 0.22. And from there, we can also compute the value of 'z', and we can now write out the proportionality, roughly speaking, ok? So this distribution here, with appropriate normalization, will be the exponential distribution, it's an exponential distribution, linear in the waiting time 'x'. And what I've shown to you, is this distribution has the following properties, ok? It satisfies this linear constraint. It's normalized. Ok? And it has the maximum entropy of all the distributions with those two previous properties.