Mean value.
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Remember the mean value theorem? Well here's what the theorem says.
Suppose I got some function f.
It's defined in the closed interval a to b.
And its values are real numbers.
It's a continuous function, so its continuous
in the closed interval between a and b.
And on the open interval between a and b, it's differentiable.
Now, with these assumptions, here's what happens.
Then there's some point c between a and b.
Doesn't tell you how to find that point, but there is one, so that
this happens.
The derivative of the function at the point c is equal to
f of b minus f of a divided by b minus a.
Let me rearrange this a bit.
So I'm just going to change the names of some of these variables.
So I'll write f prime not of c but of z, is equal to f not of b, but of x.
And I'm going to keep the name
of a the same. So I'll call that f of a.
Okay, and I'll again divide by x minus
a. Now I'm going to multiply both sides of
this by the quantity x minus a, and what do I get?
Well I get f prime of z,
times the quantity x minus a
and that's equal to this numerator here, f of x minus f of a.
So I'll put an x in there and an a in there.
And I'm going to add f of a to both sides.
And I'm going to start with f of x on this side.
So,
I'm just going to write this,
f of x is equal to what, f of a,
f of a, plus this quantity
here, f' prime of z, times x
minus a. Why would
I want to write it that way?
Well I could write this in yet another way.
I could write it like this.
This is the same thing here, but just written a little bit differently.
F of x equals f of a plus a remainder term,
r subzero of x, where r subzero of x is this.
The derivative of f at some point z between x
and a divided by 1 factorial, which is just 1.
Times x minus a to the
first power, but when I read it like this,
it looks exactly like an instance of Taylor's Theorem.
Let's remember what Taylor's Theorem says.
Suppose you've got some function f. Takes real inputs, produces real outputs.
And I'm just going to assume it's smooth.
I don't want to worry too much
about the exact differentiability conditions that I need.
Alright.
And f of x is a sum, little n goes from zero
to big n of the little nth derivitate of f at the point
a, divided by n factorial, times x minus a to the nth
power, plus this remainder term, big R sub big N of x.
Then this happens.
Big R sub big N of x is given by this, the big
N plus first dirivative of f at some point z between x and a.
Divided by the big N plus 1 factorial, times
x minus a to the big N plus 1st power.
And think about what happens
here when you plug in big N equals 0.
And in that case, what you get is exactly what we had before.
It's f of x equals, this is just the N equals 0 term here, plus a remainder term.
And that remainder term then, when big n is zero, is exactly the derivative of f
at some point z divided by one factorial times x minus a to the first power.
It looks exactly like the mean value theorem.
So Taylor theorem is a whole lot like a supped up version of mean value theorem.
The Mean Value Theorem tells you something about your first derivative,
and Taylor's Theorem is telling you information about your higher derivatives.
And this isn't just a theoretical story.
I mean, you can actually sort of see this in the real world.
Let's imagine the following scenario.
Well, let's let f of t be your position
at time t seconds after the beginning of the experiment.
And when the experiment with f of 0, where are
you, f of zero is 0, so you're at the origin.
And you're not moving.
Your velocity at time 0 is 0, so the
derivative of your position at time 0 is 0.
And I'd prefer that we survive the experience, so
I'd like to control the acceleration that that we experience.
So the second derivative
of f at any time t, will be no bigger than 250 meters per second squared.
This is, you know, 25G or so and I think humans don't do so well
[LAUGH]
above 25G.
So this is a reasonable thing if you want to you know, make this a healthy trip.
And now the question is how big can f of 60 seconds be?
How far can you get starting at zero, not
moving, how far away can you be after 60 seconds?
I want to bound our accelerations so that we survive the trip.
This problem doesn't have to be approached using
Taylor series.
But we can do it that way, so let's start down that path.
So I'll write this down.
F of t is f of 0 plus f
prime of 0 times t minus 0, right those are the first two terms in the Taylor
series expansion for f, Plus R sub 1 of t, this is the remainder term.
And what do I know about the remainder term?
Well,
R sub 1 of t is equal to the second derivative, some point z, divided by
2 factorial, times t minus 0 squared. I can bound that remainder term.
Right, because I don't want to be injured during this trip.
I don't want to be accelerating more than about
25g's, I can now control something about R sub
1, right I know something about how big R sub 1 is.
The absolute value of R sub 1 of t is less than
or equal to how big can the second derivative be at any point.
250 divided by 2 factorial times t squared.
Now I can say something about how large f of t is.
So f of t is less than or equal to f of
0 plus f prime of zero, times t plus this quantity, 250 divided by 2
factorial is 125, times t squared. Now I know what f of zero is.
I'm assuming that I'm starting at a point I'm calling the origin.
And I know something about the derivative of f
at 0, I'm assuming that I start my journey motionless.
So this quantity here is equal
to zero plus zero times t plus just 125t squared.
I gave us one minute of travel time. So f of 60 seconds
is no bigger than 125 times 60
squared. Which is equal to 450,000.
And this is meters. So f of 60 seconds
is no bigger than 450 km.
So, if we're currently stationary, I'm not moving currently, one minute from now, I
could be 450 km away and probably live to, to tell the tale, maybe.
But the other issue of course
[LAUGH]
is that after that minute I'm traveling at just some insanely high speed.
But eh, you know, at least I'm 450 kilometers away.
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