1
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Mean value.

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00:00:04,137 --> 00:00:08,260
[SOUND].

3
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Remember the mean value theorem?
Well here's what the theorem says.

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Suppose I got some function f.

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It's defined in the closed interval a to
b.

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And its values are real numbers.

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It's a continuous function, so its
continuous

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in the closed interval between a and b.

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And on the open interval between a and b,
it's differentiable.

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Now, with these assumptions, here's what
happens.

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Then there's some point c between a and b.

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00:00:30,310 --> 00:00:33,380
Doesn't tell you how to find that point,
but there is one, so that

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this happens.

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The derivative of the function at the
point c is equal to

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f of b minus f of a divided by b minus a.

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Let me rearrange this a bit.

17
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So I'm just going to change the names of
some of these variables.

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So I'll write f prime not of c but of z,
is equal to f not of b, but of x.

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And I'm going to keep the name

20
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of a the same.
So I'll call that f of a.

21
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Okay, and I'll again divide by x minus

22
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a.
Now I'm going to multiply both sides of

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this by the quantity x minus a, and what
do I get?

24
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Well I get f prime of z,

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times the quantity x minus a

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and that's equal to this numerator here, f
of x minus f of a.

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So I'll put an x in there and an a in
there.

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And I'm going to add f of a to both sides.

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And I'm going to start with f of x on this
side.

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So,

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I'm just going to write this,

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f of x is equal to what, f of a,

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f of a, plus this quantity

34
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here, f' prime of z, times x

35
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minus a.
Why would

36
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I want to write it that way?

37
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Well I could write this in yet another
way.

38
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I could write it like this.

39
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This is the same thing here, but just
written a little bit differently.

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F of x equals f of a plus a remainder
term,

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r subzero of x, where r subzero of x is
this.

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The derivative of f at some point z
between x

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and a divided by 1 factorial, which is
just 1.

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Times x minus a to the

45
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first power, but when I read it like this,

46
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it looks exactly like an instance of
Taylor's Theorem.

47
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Let's remember what Taylor's Theorem says.

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Suppose you've got some function f.
Takes real inputs, produces real outputs.

49
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And I'm just going to assume it's smooth.

50
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I don't want to worry too much

51
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about the exact differentiability
conditions that I need.

52
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Alright.

53
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And f of x is a sum, little n goes from
zero

54
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to big n of the little nth derivitate of f
at the point

55
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a, divided by n factorial, times x minus a
to the nth

56
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power, plus this remainder term, big R sub
big N of x.

57
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Then this happens.

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Big R sub big N of x is given by this, the
big

59
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N plus first dirivative of f at some point
z between x and a.

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Divided by the big N plus 1 factorial,
times

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x minus a to the big N plus 1st power.

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And think about what happens

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here when you plug in big N equals 0.

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And in that case, what you get is exactly
what we had before.

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It's f of x equals, this is just the N
equals 0 term here, plus a remainder term.

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And that remainder term then, when big n
is zero, is exactly the derivative of f

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at some point z divided by one factorial
times x minus a to the first power.

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It looks exactly like the mean value
theorem.

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So Taylor theorem is a whole lot like a
supped up version of mean value theorem.

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00:04:03,280 --> 00:04:06,760
The Mean Value Theorem tells you something
about your first derivative,

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and Taylor's Theorem is telling you
information about your higher derivatives.

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And this isn't just a theoretical story.

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I mean, you can actually sort of see this
in the real world.

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Let's imagine the following scenario.

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Well, let's let f of t be your position

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at time t seconds after the beginning of
the experiment.

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And when the experiment with f of 0, where
are

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you, f of zero is 0, so you're at the
origin.

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And you're not moving.

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Your velocity at time 0 is 0, so the

81
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derivative of your position at time 0 is
0.

82
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And I'd prefer that we survive the
experience, so

83
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I'd like to control the acceleration that
that we experience.

84
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So the second derivative

85
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of f at any time t, will be no bigger than
250 meters per second squared.

86
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This is, you know, 25G or so and I think
humans don't do so well

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[LAUGH]

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above 25G.

89
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So this is a reasonable thing if you want
to you know, make this a healthy trip.

90
00:05:03,750 --> 00:05:08,310
And now the question is how big can f of
60 seconds be?

91
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How far can you get starting at zero, not

92
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moving, how far away can you be after 60
seconds?

93
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I want to bound our accelerations so that
we survive the trip.

94
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This problem doesn't have to be approached
using

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Taylor series.

96
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But we can do it that way, so let's start
down that path.

97
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So I'll write this down.

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F of t is f of 0 plus f

99
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prime of 0 times t minus 0, right those
are the first two terms in the Taylor

100
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series expansion for f, Plus R sub 1 of t,
this is the remainder term.

101
00:05:46,980 --> 00:05:48,860
And what do I know about the remainder
term?

102
00:05:48,860 --> 00:05:49,070
Well,

103
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R sub 1 of t is equal to the second
derivative, some point z, divided by

104
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2 factorial, times t minus 0 squared.
I can bound that remainder term.

105
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Right, because I don't want to be injured
during this trip.

106
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I don't want to be accelerating more than
about

107
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25g's, I can now control something about R
sub

108
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1, right I know something about how big R
sub 1 is.

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The absolute value of R sub 1 of t is less
than

110
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or equal to how big can the second
derivative be at any point.

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250 divided by 2 factorial times t
squared.

112
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Now I can say something about how large f
of t is.

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So f of t is less than or equal to f of

114
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0 plus f prime of zero, times t plus this
quantity, 250 divided by 2

115
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factorial is 125, times t squared.
Now I know what f of zero is.

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I'm assuming that I'm starting at a point
I'm calling the origin.

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And I know something about the derivative
of f

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at 0, I'm assuming that I start my journey
motionless.

119
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So this quantity here is equal

120
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to zero plus zero times t plus just 125t
squared.

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I gave us one minute of travel time.
So f of 60 seconds

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is no bigger than 125 times 60

123
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squared.
Which is equal to 450,000.

124
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And this is meters.
So f of 60 seconds

125
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is no bigger than 450 km.

126
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So, if we're currently stationary, I'm not
moving currently, one minute from now, I

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could be 450 km away and probably live to,
to tell the tale, maybe.

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But the other issue of course

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[LAUGH]

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is that after that minute I'm traveling at
just some insanely high speed.

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But eh, you know, at least I'm 450
kilometers away.

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[SOUND]