Fooling Newton’s Method
You might think that if the Newton sequence of a function converges to a number, that the
number must be a zero of the function. Let’s look at the Newton iteration and see what might
go wrong:
f  xn 
xn1  xn 
f   xn 
Most textbooks give examples where the Newton sequence gets stuck(oscillates), hits a
horizontal tangent and fails, or simply converges to a different zero than the one intended, but I
don’t see textbooks give examples of Newton sequences converging to nonzeros.
Normally the Newton sequence  xn  converges to a number L and the function and its
derivative are continuous, so we can let n   in the Newton formula to conclude that
f  L
LL
f  L 

f  L
 0.
f  L
Assuming that f   L   0 , we conclude that the Newton sequence converges to a zero of f.
If we can get a Newton sequence
 f   x  diverges to  , then
n
 xn 
to converge to a number L with the property that
L might not be a zero of f.
1
x3
x2
x1
Consider
the
function
1

1  2 x sin  x  ; x  0
, and let’s start the
f  x  
1
;
x

0


1
sequence with x1 
.
2
x3 x 2
x1
a) Find a formula for the Newton sequence, and verify that it converges to a nonzero of f.
b) Find a formula for f   xn  and determine its behavior as n   .
A Stirling-like Inequality
n
n
Stirling’s asymptotic approximation n!  2n   comes from the inequality
e
n
n
1 
n
n 
2n    n!  2n   1 
.
e
 e   12n  1 
Let’s use some elementary calculus to derive a weaker inequality:

n
I. ln  n! 
ln  i 
i 1
1
n
2
From the two graphs, you can deduce the
following double inequality:
n 1
n

ln x dx  ln  n! 
1

ln x dx .
1
2
Integrate the left and right sides,
exponentiate, and complete the inequality:

e

n


  n!  e  





n 1
.
3
n n+1

II. Find the interval of convergence of the power series

n 1
nn xn
.
n!
Use part I. for the
endpoints. (Most textbooks just ask for the radius of convergence!)

III. a) If k is a positive integer, find the radius of convergence of the power series

n 0
b) If k  1 check the endpoints.
c) If k  2 , use the result of I. to check the endpoints.
 n! x n .
 kn !
k
Evaluating Proper/Improper Integrals with little or no Integration.


I. For the improper integral

ln x
dx , let’s look at the two cases:
1  x2
0

For


1
ln x
dx and
1  x2
1
1

ln x
dx .
1  x2
0
1
ln x
x2
x2 1
ln x
,


 3 , so it’s convergent by comparison.
dx
1  x2 1  x2 x2 x 2
1  x2
1
1
For


1
ln x
ln x
ln x

 ln x , but
dx ,
2
2
1 x
1  x2
1 x
0


ln x
dx is
1  x2
{Hint:
1
z

.}
z
z
ln x dx is convergent. So
0
0
absolutely convergent by comparison.
Use the substitution u 
1
to find its value.
x

II. Evaluate

0
x ln x
1
dx using the substitution u  .
2
x
 x  1  x  x  1

For

1
x ln x
x
1
x ln x

 2
dx ,
2
2
2
 x  1  x  x  1  x  1  x  x  1  x  1  x  x  1 x
1
For

0
x ln x
dx ,
2
x

1
x

x

1
 

x ln x
x ln x

 ln x
2
2
x

1
x

x

1
x

1
x

x

1
 
  


1
III. If you use the substitution u 
x
in the integral

x2  1
dx , you arrive at
x2
0

0


x2  1
dx  0 ? Explain.
x2
0


 1 1
 2 du 
1
u
2
u
x2  1
dx 
x2

1
u2



1
u2
0

u2 1
 1 du  
du .
u2

Is it okay to conclude that
0
0
IV. a) Use the substitution u 



 x along with the identities sin   x   cos x and
2
2




cos   x   sin x to evaluate the definite integral
2

2

sin x
dx .
cos x  sin x
0

 sin x 
dx
n
n
cos x    sin x 

0
2
b) Evaluate the definite integral

n
for n a positive integer.

V. Evaluate

0
x sin x
dx using the substitution u    x and the identities sin   x   sin x
1  cos 2 x
and cos   x    cos x .
.

VI. Show that if f is continuous then

xf  sin x  dx 
0




 x   f  sin x  dx  0
2

0
symmetry.
using the substitution

2


f  sin x  dx by showing that
0
u x



, sin  x    cos x , and
2
2

Limit Problems
x sin x
?
x  x 2  1
I. What happens if you try L’Hopital’s Rule on lim
x sin x
x
x sin x
x
by considering the inequality 2
which is valid for
 2
 2
2
x  x  1
x 1 x 1 x 1
x  0.
Find lim
x  sin x
.{See problem I.}
x 
x
II. lim
 xc
III. Find the value of c so that lim 
  9.
x  x  c


x
IV. Find a simple formula for lim
xb
xb  b x
, for b  0 .
x x  bb
1
x 2 sin  
 x  . L’Hopital’s Rule won’t work, so try something else.
V. Find lim
x0
tan x
VI. Find the following limits:
 ex  1 
ln 
x 

a) lim
x 0
x
 ex  1 
ln 
x 

b) lim
x 
x
n
VII. Find lim
 n  1 n  2   n  n 
n
n
by observing the following:
 n  n  1 n  2   n  n   1
  ln  n  1  ln  n  2    ln(n  n)   ln n
ln 
n

 n
1
 ln  n 1  1n    ln  n 1  n2     ln  n 1  nn    ln n
n

1
1
 ln 1  1n   ln 1  n2    ln 1  nn    ln n  ln n  ln n   ln n
n
n
n terms

1
 ln 1  1n   ln 1  n2    ln 1  nn  
n
The last expression is a Riemann sum of some definite integral.


VIII. The alternating series
 1
n 1
converges by the Alternating Series Test, but what
n
n 1
does it converge to?
Let’s look at the even partial sums:
S2 n  1 
 1 1
1   
 2 3

1 1 1
  
2 3 4
1   1 1 1
  1    
2n   2 3 4


1
2n
1   1 1
  1   
2n   2 3
1
 
n
S2 n
Solving the previous equation for S 2n , we get
1   1 1
 1 1
S2 n  1    
  1   
2n   2 3
 2 3
1
1
1


 
n 1 n  2
2n

So

n 1
 1
n
n 1
1
 1
 lim S2 n  lim 


n
n n  1
n2

1
 1


Find lim 
n n  1
n2


1
 
n

1 
.
2n 
1 
 , and you’ll know the sum of the series.
2n 
1
1 
 1
Method 1: Calculate lim 
by rewriting it as

 
n n  1
n2
n  n 

1 1
1
1 
lim  1 


and identifying it as a definite integral.
n
2
n n 1 
1

1

n
n
 n
Method 2:
nn
n 1
nn
n 1
From the pictures you get
2 n 1

n 1
2n
1
1
1
dx 


x
n 1 n  2
1


nn

1
dx
x
n
IX. Telescopers

a)

n n   n  1 n 1
1
1

n 1

b)

n 1

c)

n 1
n 1  n
n2  n
1


tan 1  2

 n  n 1
{Hint: n 2  n  n  n  1 .}
 x y 
{Hint: tan 1 x  tan 1 y  tan 1 
 , choose x and y carefully.}
 1  xy 
Assorted Series

I.

n2

II.
1
ln n 

n 3
{Hint: For n  ee ,  ln n 
2
ln n
1
ln  ln n  
ln n
ln n
  e2 
ln n
.}
{Try something like I.}
n
 1
n 1
n
1  
 1
 1  n
III. a) Show that 1    1   
.
n
 n
 n
b) Show that if an  is a sequence of positive numbers, then if ln  an  is decreasing, then
an  is decreasing.
In other words, show that if ln  an1   ln  an  , then an1  an .
x
c) For x  0 , show that ln 1  x   x .
{Hint: ln 1  x  

1
1
dt , and
 1.}
1 t
1 t
0
  1 n 
 1   
n 
d) Show that an  ln  
is a decreasing


n




 1
f  x   x ln 1    ln x has a negative derivative.
x

sequence
by
showing
{Hint:Use part c).}
that


 1 n1  1 n 
e) Determine whether the alternating series
 1 1    1    is absolutely
 n  
 n 
n 1
convergent, conditionally convergent, or divergent using the previous results.

IV. a) Starting with e 
x

n 0
n
xn
, you get that xe x 
n!


n 0
x n1
. Now integrate from x  0 to
n!
1
1
x  1 and get


xe dx 
x


0
n!
n 0
0
x n1dx
. Evaluate the integrals on both sides of the equation
and find the sum of a series.
b)
You


can
verity


the
sum
 n  1 
 n  2 !


you
found
 n  2  1 
 n  2 !


in
part
a)
by
noticing
that
 1
1 


 . So find the sum
n

1
!
n

2
!





n 0
n 0
n 0
n 0 
of this telescopic series and verify the previous result.
1

 n  2  n!
The Goat/Cow Grazing in the Grass/Seaweed Problem
I. Suppose that after a string is wound clockwise around a circle of radius a, its free end is at the
point A  a,0  . Now the string is unwound, always stretched tight so the unwound portion
TP is tangent to the circle at T . The set of points traced out by the free end of the string is
called the involute of the circle.
T  a cos t , a sin t 
at sin t
t
t
t
at cos t
P  ?,? 
at
Find the parametric equations of the involute of the circle.
x
y
T
t
P
A  a,0 
II. Suppose the circle in the previous problem represents the cross-section of a cylindrical water
tank of radius a, and the string is a rope of length  a . The rope is anchored at the point B
opposite point A. If the other end of the rope is tied to a cow, let’s examine the region that
can be grazed by the cow. Here is a diagram showing the rope in various positions:
Q
P
B
A
S
R
The boundary of the grazing region can be broken down into three pieces: APQ is a portion
of the involute of the circle, QR is a semicircle, and RSA is the reflection across the x-axis of
a portion of the involute.
Find the length of the boundary of the grazing region.


L 





x  t 
2


 y  t  dt 


2
III. Find the area of the grazing region.
b
 ydx 

a
Area   b

 xdy 

a




y  t  x  t  dt ; curve and x-axis


 
x t y  t  dt ; curve and y-axis

IV. Now suppose that a sea cow(manatee) is tied to a point on the surface of a sphere of radius a
by a rope of length  a . Try to find the surface area and the volume of the grazing region of
the sea cow.

 b
 y 2 dx   y  t 2 x  t  dt ; discs

 a

Volume   b


2 xydy  2 x  t  y  t  y  t  dt ; shells


 a





Surface Area  2


y  t  x  t   y  t  dt
2
2
V. Now suppose that the rope in the previous problem has length 2 a and is anchored at the
point A before being wound completely around the tank.
R
S
A
P
Q
APQ is a portion of the involute, QR is a semicircle, and RSA is a reflection of a portion of the
involute.
Attempt all the previous calculations: Length, area, surface area, and volume.
Iteration and More Grazing
I. Analyze the following recursively defined sequences using a cobweb diagram:
a) a1  6, an1  2 3an  2
b) a1  18, an1  2 3an  2
lim an 
n
c) a1  15, an1  6  an
lim an 
n
lim an 
n
d) a1  6, an1  15 an2
lim an 
n
e) a1  0, an1
 a  2
 n
2
1
4
lim an 
f) a1  4, an1
 a  2
 n
2
1
4
lim an 
n
n

g) Determine the convergence or divergence of the series
a  5  an 
3
; an1  n
2
2
{Hint: Determine lim an by cob-web analysis.}
recursively by the following: a1 
n 

n 1
an whose terms are defined
II. A farmer has a fenced circular pasture of radius a and wants to tie a cow to the fence with a
rope of length b so as to allow the cow to graze half the pasture. How long should the rope
be to accomplish this?
2a
b
a
The length of the rope, b, must be longer than a and shorter than
find the area of the grazing region, we can use polar coordinates:
2a , i.e. a  b  2a . To
sin 1  2ba 
1
1


 sin  2ba 
 sin  2ba 
2
2
1
1

2
2
2
2
2
4a sin  d  
b d  
4a sin  d 
b 2d .
The grazing area  2  
2
2


0
0
sin 1  2ba 
sin 1  2ba 






We want this to equal half the pasture area which is
sin 1
 2ba 

 a2
2
, so we get the equation

2
4a sin  d 
2
0
2

sin 1
b d 
2
 2ba 
 a2
2
. If we multiply both sides by
2
and perform the
a2

b2 
b2  b2
 b  b
integrations, we arrive at the equation  4  2 2  sin 1   
4 2  2  .
a 
a
a
 2a  a

a) Verify the previous equation.
b
 x
, we get the simplified equation  4  2 x 2  sin 1    x 4  x 2   x 2   , and
a
2
we’re looking for the solution x, with 1  x  2 . Here’s a plot of the leftside and rightside
of the equation:
 4  2 x2  sin 1  2x   x 4  x2   x2
If we let x 

solution
If we rearrange the equation, we can produce a sequence that will converge to the solution:
 x
 x 2    x 4  x 2   4  2 x 2  sin 1  
2
x2 
x
Let x1  1, and xn1 
 x
 
  x 4  x 2   4  2 x 2  sin 1  
2

 x
 .
  x 4  x 2   4  2 x 2  sin 1  
2

 xn 

 2.
  xn 4  xn 2   4  2 xn 2  sin 1 

b) Complete the cobweb diagram for the recursive sequence.
Here are the first 14 terms of the sequence generated by Excel:
x1
1
x2 1.10363
x3 1.13795
x4 1.15068
x5 1.15558
x6 1.15749
x7 1.15824
x8 1.15854
x9 1.15865
x10 1.15870
x11 1.15872
x12 1.15872
x13 1.15873
x14 1.15873
So the rope length, b, should be approximately 1.15873a .