Physics 105
Solutions to Problem Set 2
By Stephen Glassman, Fall 2007
2.8
As suggested, we begin by writing Newton’s second law as
mdv/F
(
v
) =
dt
and
integrate both sides from time
t
= 0 when
v
=
v
0
to some arbitrary time
t
=
T
when
v
=
v
(
T
).
v
(
T
)
v
0
mdv

cv
3
/
2
=
T
0
dt
2
m
cv
1
/
2
v
(
T
)
v
0
=
T
2
m
cv
(
T
)
1
/
2

2
m
cv
1
/
2
0
=
T
On the last line, we readily see that
v
(
T
) = 0 =
⇒
T
=
∞
.
Therefore, our mass
never comes to rest. Finally, we solve for
v
(
T
) and obtain:
v
(
T
) =
cT
2
m
+
1
v
1
/
2
0

2
2.12
Consider a function
v
(
x
(
t
)). Then by the chain rule,
dv
dt
=
dv
dx
dx
dt
=
dv
dx
v
. Also by
the chain rule,
d
dx
v
2
= 2
v
dv
dx
. Therefore,
F
= ˙
p
=
m
˙
v
=
mv
dv
dx
=
m
1
2
dv
2
dx
. We separate
variables to obtain
2
m
Fdx
=
dv
2
. If we integrate both sides from some initial state
v
0
,
x
0
to some final state
v
,
x
:
x
x
0
2
m
F
(
x
)
dx
=
v
v
0
dv
2
=
⇒
2
m
x
x
0
F
(
x
)
dx
=
v
2

v
2
0
For
F
(
x
) =
F
0
, we can write this expression as
F
0
Δ
x
=
1
2
mv
2

1
2
mv
2
0
= Δ
KE
.
1
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2.18
Unfortunately, Taylor
†
used the variable
x
for the base point in his expression
for the Taylor series. I’m going to take it back and express Taylor’s theorem as:
f
(
a
+
δ
) =
f
(
a
) +
f
(
a
)
δ
+
1
2!
f
(
a
)
δ
2
+
· · ·
where
f
(
x
) is some sufficiently reasonable function, so long as the right hand side
converges.
It should be noted that an expansion for a particular
f
(
x
) around a
particular base point
a
will generally only converge for some interval in
δ
. I will use
the notation
f
(
n
)
(
x
) to denote the
n
th
derivative of
f
(
x
). Of particular interest is the
special case
a
= 0 for which the Taylor series simplifies to:
f
(
δ
) =
∞
n
=0
f
(
n
)
(0)
n
!
δ
n
(a)
Let
f
(
x
) = ln(1 +
x
).
Claim:
f
(
n
)
(
x
) = (

1)
n

1
(
n

1)!(1 +
x
)

n
∀
n
≥
1
Proof:
We proceed using mathematical induction. For
n
= 1, the expression eval
uates to
f
(
x
) = (1 +
x
)

1
which is precisely the derivative of ln(1 +
x
).
Now we
assume the claim for
n
≤
k
and show it for
n
=
k
+ 1.
f
(
k
)
(
x
)
=
(

1)
k

1
(
k

1)!(1 +
x
)

k
=
⇒
f
(
k
+1)
(
x
)
=
(

k
)(

1)
k

1
(
k

1)!(1 +
x
)

k

1
=
(

1)
k
k
!(1 +
x
)

(
k
+1)
which is precisely the claim for
n
=
k
+ 1.
Then
f
(
n
)
(0) = (

1)
n

1
(
n

1)! and:
f
(
δ
)
=
∞
n
=0
f
(
n
)
(0)
n
!
δ
n
=
f
(0) +
∞
n
=1
f
(
n
)
(0)
n
!
δ
n
=
0 +
∞
n
=1
(

1)
n

1
(
n

1)!
n
!
δ
n
=

∞
n
=1
(

1)
n
n
δ
n
†
by which I mean John Taylor, the author of the textbook.
This is not to be confused with
subsequent uses of “Taylor” all of which refer to Brook Taylor of Taylor series fame. Ug!
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 Fall '07
 EdgarKnobloch
 Physics, Derivative, Taylor Series, Taylor's theorem, John Taylor, DV DV, dv dv dv

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