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In the last video, we started to explore
the notion of an error function.

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Not to be confused with the expected value

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because it really does reflect the same
notation.

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But here E is for error.

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And we could also thought it will

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some times here referred to as Reminder
function.

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And we saw it's really just the difference
as we,

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the difference between the function and
our approximation of the function.

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So for example, this, this distance right
over here, that is our error.

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That is our error at the x is equal to b.

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And what we really care about is the
absolute value of it.

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Because at some points f of x might be
larger than the polynomial.

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Sometimes the polynomial might be larger
than f of x.

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What we care is the absolute distance
between them.

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And so what I want to do in this video is

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try to bound, try to bound our error at
some b.

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Try to bound our error.

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So say it's less than or equal to some
constant value.

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Try to bound it at b for some b is greater
than a.

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We're just going to assume that b is
greater than a.

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And we saw some tantalizing, we, we got to
a bit of a tantalizing

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result that seems like we might be able to
bound it in the last video.

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We saw that the n plus 1th derivative of
our error

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function is equal to the n plus 1th
derivative of our function.

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Or their absolute values would also be
equal to.

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So if we could somehow bound the n plus
1th derivative

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of our function over some interval, an
interval that matters to us.

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An interval that maybe has b in it.

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Then, we can, at least bound the n plus
1th derivative our error function.

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And then, maybe we can do a little bit of

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integration to bound the error itself at
some value b.

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So, let's see if we can do that.

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Well, let's just assume, let's just assume
that we're in a reality where

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we do know something about the n plus1
derivative of f of x.

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Let's say we do know that this.

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We do it in a color I that haven't used
yet.

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Well, I'll do it in white.

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So let's say that that thing over there
looks something like that.

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So that is f the n plus 1th derivative.

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The n plus 1th derivative.

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And I only care about it over this
interval right over here.

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Who cares what it does later, I just gotta
bound it over the interval

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cuz at the end of the day I just wanna
balance b right over there.

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So let's say that the absolute value of
this.

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Let's say that we know.

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Let me write it over here, let's say that
we know.

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We know that the absolute value of the n
plus 1th derivative, the n plus 1th.

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And, I apologize I actually switch between
the capital N and

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the lower-case n and I did that in the
last video.

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I shouldn't have, but now that you know

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that I did that hopefully it doesn't
confuse it.

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N plus 1th, so let's say we know that the
n plus 1th

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derivative of f of x, the absolute value
of it, let's say it's bounded.

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Let's say it's less than or equal to some
m

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over the interval, cuz we only care about
the interval.

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It might not be bounded in general, but
all we

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care is it takes some maximum value over
this interval.

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So over, over, over the interval x, I
could write it this way,

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over the interval x is a member between a
and b, so this includes both of them.

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It's a closed interval, x could be a, x

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could be b, or x could be anything in
between.

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And we can say this generally that,

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that this derivative will have some
maximum value.

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So this is its, the absolute value,
maximum value, max value, m for max.

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We know that it will have a maximum value,
if this thing is continuous.

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So once again we're going to assume that
it is continuous,

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that it has some maximum value over this
interval right over here.

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Well this thing, this thing right over
here, we know is the

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same thing as the n plus 1 derivative of
the error function.

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So then we know, so then that, that
implies, that implies that,

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that implies that the, that's a new color,
let me do that in blue, or that green.

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That implies that the, the, the end plus
one derivative of the error function.

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The absolute value of it because these are
the

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same thing is also, is also bounded by m.

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So that's a little bit of an interesting
result but it gets us no where near there.

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It might look similar but this is the n
plus 1 derivative of the error function.

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And, and we'll have to think about how we
can get an m in the future.

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We're assuming that we some how know it
and maybe

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we'll do some example problems where we
figure that out.

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But this is the m plus 1th derivative.

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We bounded it's absolute value but we

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really want to bound the actual error
function.

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The 0 is the derivative, you could say,
the actual function itself.

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What we could try to integrate both sides
of this and see

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if we can eventually get to e, to get to e
of x.

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To get our, to our error function or our
remainder function so let's do that.

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Let's take the integral, let's take the
integral of both sides of this.

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Now the integral on this left hand side,
it's a little interesting.

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We take the integral of the absolute
value.

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It would be easier if we were taking the
absolute value of the integral.

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And lucky for us, the way it's set up.

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So let me just write a little aside here.

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We know generally that if I take, and it's
something for you to think about.

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If I take, so if I have two options, if I
have two

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options, this option versus and I don't
know, they look the same right now.

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I know they look the same right now.

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So over here, I'm gonna have the integral
of the absolute value

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and over here I'm going to have the
absolute value of the interval.

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Which of these is going to be, which of
these can be larger?

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Well, you just have to think about the
scenarios.

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So, if f of x is always positive over the
interval that

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you're taking the integration, then
they're going to be the same thing.

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They're, you're gonna get positive values.

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Take the absolute of a value of a positive
value.

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It doesn't make a difference.

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What matters is if f of x is negative.

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If f of x, if f of x is negative the

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entire time, so if this our x-axis, that
is our y-axis.

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If f of x is, well we saw if it's positive
the entire

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time, you're taking the absolute value of
a positive, absolute value of positive.

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It's not going to matter.

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These two things are going to be equal.

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If f of x is negative the whole time, then
you're going

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to get, then this integral going to
evaluate to a negative value.

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But then, you would take the absolute
value of it.

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And then over here, you're just going to,
this is, the integral going to

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value to a positive value and it's still
going to be the same thing.

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The interesting case is when f of x is
both

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positive and negative, so you can imagine
a situation like this.

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If f of x looks something like that, then

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this right over here, the integral, you'd
have positive.

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This would be positive and then this would
be negative right over here.

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And so they would cancel each other out.

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So this would be a smaller value than

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if you took the integral of the absolute
value.

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So the integral, the absolute value of f
would look something like this.

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So all of the areas are going to be, if
you view the

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integral, if you view this it is

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definitely going to be a definite
integral.

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All of the areas, all of the areas would
be positive.

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So when you it, you are going to get a

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bigger value when you take the integral of
an absolute value.

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Then you will, especially when f of x

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goes both positive and negative over the
interval.

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Then you would if you took the integral
first and then the absolute value.

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Cuz once again, if you took the integral
first, for something like

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this, you'd get a low value cause this
stuff would cancel out.

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Would cancel out with this stuff right
over here then you'd

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take the absolute value of just a lower, a
lower magnitude number.

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And so in general, the integral, the

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integral, sorry the absolute value of the
integral

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is going to be less than or equal to the
integral of the absolute value.

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So we can say, so this right here is the
integral of

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the absolute value which is going to be
greater than or equal.

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What we have written over here is just
this.

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That's going to be greater than or equal
to, and I

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think you'll see why I'm why I'm doing
this in a second.

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Greater than or equal to the absolute
value, the absolute

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value of the integral of, of the n plus
1th derivative.

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The n plus 1th derivative of, x, dx.

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And the reason why this is useful, is that
we can still

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keep the inequality that, this is less
than, or equal to this.

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But now, this is a pretty straight forward
integral to evaluate.

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The indo, the anti-derivative of the n
plus

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1th derivative, is going to be the nth
derivative.

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So this business, right over here.

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Is just going to the absolute value of the
nth derivative.

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The absolute value of the nth derivative
of our error function.

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Did I say expected value?

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I shouldn't.

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See, it even confuses me.

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This is the error function.

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I should've used r, r for remainder.

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But this all error.

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The, noth, nothing about probability or
expected value in this video.

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This is.

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E for error.

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So anyway, this is going to be the nth
derivative of our

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error function, which is going to be less
than or equal to this.

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Which is less than or equal to the
anti-derivative of M.

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Well, that's a constant.

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So that's going to be mx, mx.

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And since we're just taking indefinite
integrals.

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We can't forget the idea that we have a
constant over here.

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And in general, when you're trying to
create an upper

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bound you want as low of an upper bound as
possible.

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So we wanna minimize, we wanna minimize
what this constant is.

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And lucky for us, we do have, we do know
what this,

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what this function, what value this
function takes on at a point.

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We know that the nth derivative of our
error function at a is equal to 0.

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00:09:08,430 --> 00:09:09,940
I think we wrote it over here.

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The nth derivative at a is equal to 0.

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And that's because the nth derivative of
the function and the

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approximation at a are going to be the
same exact thing.

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And so, if we evaluate both sides of this
at a, I'll

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do that over here on the side, we know
that the absolute value.

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We know the absolute value of the nth
derivative at a, we know

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that this thing is going to be equal to
the absolute value of 0.

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Which is 0.

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Which needs to be less than or equal to
when you evaluate this

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thing at a, which is less than or equal to
m a plus c.

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And so you can, if you look at this part

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of the inequality, you subtract m a from
both sides.

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You get negative m a is less than or equal
to c.

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So our constant here, based on that little
condition

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that we were able to get in the last
video.

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Our constant is going to be greater than
or equal to negative ma.

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00:10:00,820 --> 00:10:03,880
So if we want to minimize the constant, if
we wanna get this as low

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00:10:03,880 --> 00:10:08,090
of a bound as possible, we would wanna
pick c is equal to negative Ma.

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00:10:08,090 --> 00:10:10,250
That is the lowest possible c that will

201
00:10:10,250 --> 00:10:13,170
meet these constraints that we know are
true.

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00:10:13,170 --> 00:10:16,969
So, we will actually pick c to be negative
Ma.

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00:10:16,969 --> 00:10:19,364
And then we can rewrite this whole thing
as the

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00:10:19,364 --> 00:10:22,590
absolute value of the nth derivative of
the error function.

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The nth derivative of the error function.

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Not the expected value.

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00:10:25,970 --> 00:10:28,010
I have a strange suspicion I might have
said expected value.

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But, this is the error function.

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00:10:29,790 --> 00:10:30,440
The nth der.

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The absolute value of the nth derivative
of the error function

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00:10:33,230 --> 00:10:38,600
is less than or equal to M times x minus
a.

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00:10:38,600 --> 00:10:40,820
And once again all of the constraints
hold.

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00:10:40,820 --> 00:10:43,880
This is for, this is for x as part of the
interval.

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00:10:43,880 --> 00:10:48,910
The closed interval between, the closed
interval between a and b.

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00:10:48,910 --> 00:10:50,220
But looks like we're making progress.

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00:10:50,220 --> 00:10:52,910
We at least went from the m plus 1
derivative to the n derivative.

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00:10:52,910 --> 00:10:55,170
Lets see if we can keep going.

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So same general idea.

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This if we know this then we know that

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00:11:00,090 --> 00:11:00,740
we can take the integral of both sides of
this.

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00:11:00,740 --> 00:11:02,850
So we can take the integral of both sides
of this

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00:11:06,280 --> 00:11:08,360
the anti derivative of both sides.

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00:11:08,360 --> 00:11:10,740
And we know from what we figured out up
here

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that something's that's even smaller than
this right over here.

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00:11:14,780 --> 00:11:19,820
Is, is the absolute value of the integral
of the expected value.

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00:11:19,820 --> 00:11:21,070
Now [LAUGH] see, I said it.

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Of the error function, not the expected
value.

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Of the error function.

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00:11:23,900 --> 00:11:27,170
The nth derivative of the error function
of x.

230
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The nth derivative of the error function
of x dx.

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00:11:29,940 --> 00:11:33,510
So we know that this is less than or equal
to based on the exact same logic there.

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And this is useful because this is just
going to be, this is just

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00:11:37,450 --> 00:11:42,640
going to be the nth minus 1 derivative of
our error function of x.

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00:11:42,640 --> 00:11:45,160
And of course we have the absolute value
outside of it.

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And now this is going to be less than or
equal to.

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00:11:46,650 --> 00:11:48,390
It's less than or equal to this, which is
less than or equal

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00:11:48,390 --> 00:11:50,940
to this, which is less than or equal to
this right over here.

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00:11:50,940 --> 00:11:53,340
The anti-derivative of this right over
here is going

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00:11:53,340 --> 00:11:58,060
to be M times x minus a squared over 2.

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00:11:58,060 --> 00:12:01,410
You could do U substitution if you want or
you could just say hey look.

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00:12:01,410 --> 00:12:03,820
I have a little expression here, it's
derivative is 1.

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00:12:03,820 --> 00:12:06,480
So it's implicitly there so I can just
treat it as kind of a U.

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00:12:06,480 --> 00:12:09,320
So raise it to an exponent and then divide
that exponent.

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00:12:09,320 --> 00:12:11,460
But once again I'm taking indefinite
integrals.

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So I'm going to say a plus C over here.

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00:12:14,350 --> 00:12:16,600
But let's use that same exact logic.

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00:12:16,600 --> 00:12:19,130
If we evaluate this at A, you're going to
have it.

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00:12:19,130 --> 00:12:22,250
If you evaluate this while, let's evaluate
both sides of this at A.

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the left side, evaluated at A, we know, is
going to be zero.

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00:12:25,990 --> 00:12:29,250
We figured that out, all, up here in the
last video.

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So you get, I'm gonna do it on the right
over here.

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You get zero, when you valued the left
side of a.

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00:12:34,130 --> 00:12:36,820
The right side of a, if you, the right
side of the

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00:12:36,820 --> 00:12:39,850
value of a you get m times a menus a
square over 2.

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00:12:39,850 --> 00:12:45,220
So you are gonna get 0 plus c, so you are
gonna get, 0 is less or equal to c.

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00:12:45,220 --> 00:12:47,620
Once again we want to minimize our constant,

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00:12:47,620 --> 00:12:49,800
we wanna minimize our upper boundary up
here.

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00:12:49,800 --> 00:12:52,930
So we wanna pick the lowest possible c
that we talk constrains.

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00:12:52,930 --> 00:12:57,440
So the lowest possible c that meets our
constraint is zero.

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00:12:57,440 --> 00:13:01,070
And so the general idea here is that we
can keep doing this, we can

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00:13:01,070 --> 00:13:07,270
keep doing exactly what we're doing all
the way, all the way, all the way until.

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00:13:07,270 --> 00:13:10,440
And so we keep integrating it at the exact
same, same way that I've

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00:13:10,440 --> 00:13:14,040
done it all the way that we get and using
this exact same property here.

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00:13:14,040 --> 00:13:19,180
All the way until we get, the bound on the
error function of x.

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00:13:19,180 --> 00:13:21,550
So you could view this as the 0th
derivative.

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00:13:21,550 --> 00:13:22,740
You know, we're going all the way to the

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00:13:22,740 --> 00:13:25,360
0th derivative, which is really just the
error function.

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00:13:25,360 --> 00:13:27,620
The bound on the error function of x is
going to

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00:13:27,620 --> 00:13:29,660
be less than or equal to, and what's it
going to be?

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00:13:29,660 --> 00:13:31,940
And you can already see the pattern here.

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00:13:31,940 --> 00:13:36,270
Is that it's going to be m times x, minus
a.

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00:13:36,270 --> 00:13:39,490
And the exponent, the one way to think
about it, this exponent

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00:13:39,490 --> 00:13:42,950
plus this derivative is going to be equal
to n plus 1.

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00:13:42,950 --> 00:13:46,980
Now this derivative is zero so this
exponent is going to be n plus 1.

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00:13:46,980 --> 00:13:50,210
And whatever the exponent is, you're going
to have,a nd maybe I should

276
00:13:50,210 --> 00:13:54,280
have done it, you're going to have n plus
one factorial over here.

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00:13:54,280 --> 00:13:56,950
And if say wait why, where does this n
plus 1 factorial come from?

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00:13:56,950 --> 00:13:58,370
I just had a two here.

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00:13:58,370 --> 00:14:01,120
Well think about what happens when we
integrate this again.

280
00:14:01,120 --> 00:14:04,700
You're going to raise this to the third
power and then divide by three.

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00:14:04,700 --> 00:14:07,050
So your denominator is going to have two
times three.

282
00:14:07,050 --> 00:14:08,540
Then when you integrate it again, you're
going to raise

283
00:14:08,540 --> 00:14:10,800
it to the fourth power and then divide by
four.

284
00:14:10,800 --> 00:14:12,960
So then your denominator is going to be
two times three times four.

285
00:14:12,960 --> 00:14:14,140
Four factorial.

286
00:14:14,140 --> 00:14:15,530
So whatever power you're raising to, the

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00:14:15,530 --> 00:14:18,500
denominator is going to be that power
factorial.

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00:14:18,500 --> 00:14:21,240
But what's really interesting now is if we
are

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00:14:21,240 --> 00:14:24,360
able to figure out that maximum value of
our function.

290
00:14:24,360 --> 00:14:28,510
If we're able to figure out that maximum
value of our function right there.

291
00:14:28,510 --> 00:14:31,800
We now have a way of bounding our error
function

292
00:14:31,800 --> 00:14:36,500
over that interval, over that interval
between a and b.

293
00:14:36,500 --> 00:14:39,530
So for example, the error function at b.

294
00:14:39,530 --> 00:14:42,040
We can now bound it if we know what an m
is.

295
00:14:42,040 --> 00:14:49,190
We can say the error function at b is
going to be less than or equal to m times

296
00:14:49,190 --> 00:14:57,190
b minus a to the n plus 1th power over n
plus 1 factorial.

297
00:14:57,190 --> 00:15:00,030
So that gets us a really powerful, I guess
you

298
00:15:00,030 --> 00:15:03,720
could call it, result, kinda the, the math
behind it.

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00:15:03,720 --> 00:15:06,849
And now we can show some examples where
this could actually be applied.