So then we get that the Laplace transform of cosine of a t is equal to s over s squared plus a squared.

We've shown is equal to well I'll write the notation it's equal to some function of s.

Now, the great thing is that we can make a small modification to our Q learning algorithm where, when we were updating, the Q of S, A got updated in terms of a small change to the existing Q of S, A values.

Laplace transform of cosine of t, we know that this is equal to s over s squared plus 1, which this kind of looks like if this was an s and this was an s squared plus 1.

Then if you take the Laplace transform of that, that means that F of s is equal to s minus 1 over s minus 1 squared plus 1.

Then this is going to be f of s minus 2.

So times, 5 times 10 to the second seconds, seconds, I'll just use that with an s.

So let's just ignore that for a second.

In some notations, you'll see

It's 1 over s squared. y is equal to 1 over s squared, times s squared, over s squared plus a squared.

Plus f of 0 is equal to s times the Laplace transform of f.

If we call this expression right here F of s, then what's this expression?

So s minus 3, which is equal to s minus 3 over s minus 3 squared plus 4.

Let's multiply s times r.

That one is expanded by looking at the different conditions that can make us happy.

We'll initialize a value m equals minus infinity, and then we'll say for all pairs of actions and successors states in successors of S, we'll say the value is and let's call this S prime so we don't get confused the value of S prime and M for keeping track of the maximum so far and the new value.

So what's anti derivative of e to the minus st with respect to dt?

Remember that SDS PAGE is for small DNA or protein cells.

Let's go back to this.

We said, look, if I take the Laplace transform of cosine of t, I'd get s over s squared plus 1.

The Laplace transform of 1 we just did it at beginning of the video was equal to 1 s, if we assume that s is greater than zero.

And I'll give you a hint.

And they both have s plus 5 as a common factor, so we can now factor out s plus 5.

So I have just dividing both sides of this equation by a minus 1, I get s is equal to a to the n plus 1 minus 1 over a minus 1.

This expression right here is F of s minus 2.

Cosine of 0 is 1, so we have minus 1 over s squared, minus 1 over s squared, times 1.

Jei, Chulsu, Chaek sang, sae chaeksang. Jei

But this isn't s over s squared plus 1.

Bart, what's new in the shoe world?

So this is going to be the limit as A approaches infinity of minus 1 s e to the minus sA minus minus 1 s.

Yes, there is. There is still one more thing on the to do list.

The Laplace transform of f is equal to 1 s.

So now we've got , the string, Month which is what it said before, and our field and I've added this new value parameter for (month), and I've added another one for (day), and another one for (year).

Because if a is equal to 0, then the Laplace transform of e to the 0 is just 1 s minus 0.

But that's the same thing as s squared over s squared, plus a squared over s squared.

And that's just 4 s times this.

Well, you still have an s left over, you just have an s there.

P of happiness given S and R times P of S and R, which is of course the product of those 2 because they are independent, plus P of happiness given not S R, probability of not as R plus P of H given S and not R times the probability of P of S and not R plus the last case,

If you don't believe me, take the derivative of this.

Well, we just use this formula up here.

And I did that for a reason because now I'm going to add both sides of this equation.

So that was this term and that term are real.

This gives us a new particle, and this gives us a new non normalized importance weight.

It's some old man who's heard about him.

And actually, remember, we even made the condition when s is greater than 0, right?