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18.01 Single Variable Calculus
Fall 2006
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Lecture 38
18.01 Fall 2006
Lecture 38: Final Review
Review: Differentiating and Integrating Series.
If f (x) =
∞
�
an xn , then
n=0
�
f (x) =
∞
�
nan x
n−1
�
f (x)dx = C +
and
n=1
∞
�
an xn+1
n+1
n=0
Example 1: Normal (or Gaussian) Distribution.
�
� x
� x�
2
(−t2 )2
(−t2 )3
e−t dt =
1 − t2 +
+
+ · · · dt
2!
3!
0
0
�
� x�
t4
t6
t8
2
=
1 − t + − + − ... dt
2! 3! 4!
0
=x−
x
�
x3
1 x5
1 x7
+
−
+ ...
3
2! 5
3! 7
2
e−t dt isn’t an elementary function, we can still compute it. Elementary functions
Even though
0
are still a little bit better, though. For example:
sin x = x −
x3
x5
π
π (π/2)3
(π/2)5
+
− · · · =⇒ sin = −
+
− ···
3!
5!
2
2
3!
5!
But to compute sin(π/2) numerically is a waste of time. We know that the sum if something very
simple, namely,
π
sin = 1
2
It’s not obvious from the series expansion that sin x deals with angles. Series are sometimes com­
plicated and unintuitive.
π
π
Nevertheless, we can read this formula backwards to find a formula for . Start with sin = 1.
2
2
Then,
� 1
�1
dx
π
π
�
√
= sin−1 x� = sin−1 1 − sin−1 0 = − 0 =
2
2
2
0
1−x
0
We want to find the series expansion for (1 − x2 )−1/2 , but let’s tackle a simpler case first:
�
��
�
�
��
��
�
1
1
1
1
1
�
�
−
− −1
− −2
−
− −1
1
2
2
2
2
2
(1 + u)−1/2 = 1 + −
u+
u2 +
u3 + · · ·
1·2
1·2·3
2
1
1 · 3 2 1 · 3 · 5
3
=1− u+
u −
u + · · ·
2·4
2
2·4·6
Notice the pattern: odd numbers go on the top, even numbers go on the bottom, and the signs
alternate.
1
Lecture 38
18.01 Fall 2006
Now, let u = −x2 .
1
1·3 4 1·3·5 6
x +
x + ···
(1 − x2 )−1/2 = 1 + x2 +
2
2·4
2·4·6
�
�
�
1 x3
1 · 3 x5
1 · 3 · 5 x7
2 −1/2
(1 − x )
dx = C + x +
+
+
+ ···
2 3
2·4 5
2·4·6 7
� � �
�� � �
�� �
� 1
π
1 1
1·3
1
1·3·5
1
=
(1 − x2 )−1/2 dx = 1 +
+
+
+ ···
2
2 3
2·4
5
2·4·6
7
0
Here’s a hard (optional) extra credit problem: why does this series converge?
L’Hôpital’s rule to find out how quickly the terms decrease.
Hint: use
The Final Exam
Here’s another attempt to clarify the concept of weighted averages.
Weighted Average
A weighted average of some function, f , is defined as:
�b
a
Average(f ) =
�
w(x)f (x) dx
�b
w(x) dx
a
b
w(x) dx is the total, and w(x) is the weighting function.
Here,
a
Example: taken from a past problem set.
You get $t if a certain particle decays in t seconds. How much should you pay to play? You were
given that the likelihood that the particle has not decayed (the weighting function) is:
w(x) = e−kt
Remember,
�
∞
e−kt dt =
0
1
k
The payoff is
f (t) = t
The expected (or average) payoff is
�∞
� ∞ −kt
f (t)w(t) dt
te
dt
0�
= �0∞ −kt
∞
w(t) dt
e
dt
0
0
∞
�
te−kt dt =
=k
0
∞
�
(kt)e−kt dt
0
Do the change of variable:
u = kt and du = k dt
2
Lecture 38
18.01 Fall 2006
∞
�
ue−u
Average =
0
du
k
∞
�
ue−u du = 1.
On a previous problem set, you evaluated this using integration by parts:
0
�
∞
ue−u
Average =
0
du
1
=
k
k
On the problem set, we calculated the half-life (H) for Polonium120 was (131)(24)(60)2 seconds. We
also found that
ln 2
k=
H
Therefore, the expected payoff is
1
H
=
k
ln 2
where H is the half-life of the particle in seconds.
Now, you’re all probably wondering: who on earth bets on particle decays?
In truth, no one does. There is, however, a very similar problem that is useful in the real world.
There is something called an annuity, which is basically a retirement pension. You can buy an
annuity, and then get paid a certain amount every month once you retire. Once you die, the annuity
payments stop.
You (and the people paying you) naturally care about how much money you can expect to get
over the course of your retirement. In this case, f (t) = t represents how much money you end up
with, and w(t) = e−kt represents how likely your are to be alive after t years.
What if you want a 2-life annuity? Then, you need multiple integrals, which you will learn about
in multivariable calculus (18.02).
Our first goal in this class was to be able to differentiate anything. In multivariable calculus, you
will learn about another chain rule. That chain rule will unify the (single-variable) chain rule, the
product rule, the quotient rule, and implicit differentiation.
You might say the multivariable chain rule is
One
One
One
And
thing to rule them all
thing to find them
thing to bring them all
in a matrix bind them.
(with apologies to JRR Tolkien).
3