I took out the constant terms there, and it's going to be a delta of t minus c dt.

So we could write average rate of change or delta

You're already hopefully reasonably familiar with it.

Let's say I wanted to graph let me do it in another color.

I was going to write an x again.

Let me write it all at once.

If we take out the constants from inside of the integral, we get e to the minus sc times f of c times the integral from 0 to infinity of f of t minus c dt.

Well, I would go to t minus 2.

So it's this constant times my Dirac delta function, times delta of t minus c.

delta

default

And so this is what I meant by pseudoinfinity, because if I have 2 times the Dirac delta function, and if I'm taking the area under the curve of that, of 2 times the Dirac delta function t minus c dt, this should be equal to 2 times the area of just under the Dirac delta function 2 times from minus infinity to infinity of the delta function shifted by c dt, which is just 2 times we already showed you, I just said, by definition, this is 1, so this will be equal to 2.

All of these are just constants, so this might as well just be 5.

Delta beta two at six, delta gamma three at five.

That's actually the capital Greek alphabet delta .

delta

Delta Row

And then we're going to use the back propagation algorithm to compute delta L minus 1, delta L minus 2, and so on down to delta 2 and once again there is now delta 1 because we don't associate an error term with the input layer.

This thing, if I were to draw my x axis like that, and then right at t equals 0, my Dirac delta function pops up like that.

Well, it's going to be the value of the Dirac delta function.

And I'm going to shift it and multiply it times some arbitrary function f of t.

Delta Row Compression

So e to the minus sc f of c times my delta function t minus c dt.

Delta

Delta Junction

I have to be very careful.

Concretely, if you think of delta ij as a matrix, indexed by subscript ij.

Compressed Replacement Delta Row Encoding

Attack pattern Delta. Go now.

Then, if delta L is a matrix we can rewrite this as delta L, gets updated as delta L plus

And I'll assume that c is greater than zero, that the delta function pops up someplace in the positive t axis.

So let's say my first parametrization, I have x of t is equal to t.

So this is going to be the integral from t equals a to t equals b, of p of p of x, really, instead of writing x, y, it's x of t, right? x as a function of t, y as a function of t.

Here's a formula for computing delta 3 is delta 3 is equal to theta 3 transpose times delta 4.

No, I'm defining it.

I'm running out of space.

So if we take from zero to infinity, what I'm saying is taking this integral is equivalent to taking this integral.

Everywhere, when t equals anything other than c, the

This is t equal to 3.

Delta Six Actual, this is Murphy.

Let me do it on the side.

I want to put out a tracer on Eleanor Burnett.

I'll do it in another color, just to ease the monotony.

I'm claiming that this is equivalent to this.

But you actually put an arrow there, and so this is your