Math 335 - Homework Assignment 1- Solutions

Turn in the problems with *.

. Ÿx  1  m1
1. Rewrite the power series ! m0 as a sum of 2 power series, one consists of all terms with even
2 m m!
indices and the other consists of all terms with odd indices. Rewrite it also as a sum of 3 power series which
consist of terms with indices as m  3n, 3n  1, and 3n  2, respectively.
. . .
Ÿx  1  m1 Ÿx  1  2n1 Ÿx  1  2n2
! 2 m m!
let m2n and m2n1

n0,1,...
! 2 2n Ÿ2n !
 ! 2 2n1 Ÿ2n  1 !
m0 n0 n0
. . . .
Ÿx  1  m1 Ÿx  1  3n1 Ÿx  1  3n2 Ÿx  1  3n3
! 2 m m!
let m2n and m2n1

n0,1,...
! 2 3n Ÿ3n !
 !  !
2 3n1 Ÿ3n  1 ! n0 2 3n2 Ÿ3n  2 !
m0 n0 n0

2. *Page 245: 2, 3, 4 - for each power series,

a. give the partial sum with the first 4 non-zero terms;
b. find the radius and interval of convergence;
c. (extra points) find the limits of the power series if possible.

. Ÿ100  n
#2. ! n0 Ÿx  7  n
n!
a.
2 3
S 3  1  100Ÿx  7   100 Ÿx  7  2  100 Ÿx  7  3
2 6
2
 1  100Ÿx  7   5000Ÿx  7   500000 Ÿx  7  3
3
b.
cn Ÿ100  n Ÿn  1 ! n1  .
R  lim
nv. c n1
 lim  lim
nv. n! 100 n1 nv. 100

So, the radius of convergence is R  ., and the interval of convergence I  "., . .

c.
. . .
Ÿ100  n Ÿ100  n Ÿ100Ÿx  7   n
! n!
Ÿx  7  n  ! n!
Ÿx  7  n  ! n!
 e 100Ÿx7 
n0 n0 n0
100Ÿx7 
fŸx   e is the limit function of this power series.

. Ÿ"1  k
#3. ! k1 Ÿx " 5  k
10 k
a.
S 4  " 1 Ÿx " 5   1 Ÿx " 5  2 " 1 Ÿx " 5  3  1 Ÿx " 5  4
10 100 1000 10000
b.

1
 R  lim cck1
k1
k
 lim 1 k 10  10
kv. kv. 10 1
. . .
Ÿ"1  k Ÿ"1  k
When x " 5  10, ! 10 k
Ÿx " 5  k
 ! 10 k
Ÿ10  k  !Ÿ"1  k diverges
k1 k1 k1
. k . k .
Ÿ"1  Ÿ"1 
When x " 5  "10, ! 10 k
Ÿx " 5  k  ! 10 k
Ÿ"10  k  ! 1 diverges
k1 k1 k1

|x " 5|  10 « " 10  x " 5  10, " 10  5  x  10  5 « " 5  x  15

So, the radius of convergence is R  10, and the interval of convergence I  "5, 15 .
c. Recognize the series is of the form
.
Ÿ"1 
! r k  r  r 2  r 3 . . .  r n . . . where r 
10
Ÿx " 5 .
k1
.
The sum for ! k1 r k is:
.
! r k  r  r 2  r 3 . . .  r n . . .  rŸ1  r  r 2 . . .    r 1 1" r  r
1"r
k1
. . .
Ÿ"1  k "Ÿx " 5  k
"Ÿx " 5  k
! Ÿx " 5  k  !  "x"5 !
10 k 10 10 10
k1 k1 k0

 "x"5 1 " x"5  "x"5  5"x

10 1 x"5 10  x " 5 x5 x5
10
fŸx   5 " x is the limit function of this power series.
x5

.
4. ! k0 k!Ÿx " 1  k
a.
S 3  1  Ÿx " 1   2Ÿx " 1  2  6Ÿx " 1  3
b.
R  lim cck1
k
 lim k!  lim 1  0
kv. kv. Ÿk  1 ! kv. k  1

So, the radius of convergence is R  0, and the interval of convergence I  £1¤.

.
c. When x  1, ! k0 k!Ÿx " 1  k  1

m
3. *Consider the power series ! m1 2m x 2m .
.

a. Give the partial sum with the first 4 non-zero terms.

b. Find the radius and interval of convergence.
c. (extra points) Find the limits of the power series if possible.

a. Give the partial sum with the first 4 non-zero terms.

.
! 2 m x 2m X S 8  2x 2  2x 4  8 x 6  4x 8
m 3
m1

b. Find the radius and interval of convergence.

2
 lim b m1  mv.
m1
L  mv. lim 2 x 2m2 m  mv.
lim 2m x 2  2|x| 2  1
bm m 1 2 m x 2m m1
|x|  1 , R  1
2 2
When x  o 1 ,
2
. 2m . .
m
! 2m o 1  ! 2m 1  ! 1
m 2 m 2 m diverges since it is a harmonic series.
m1 m1 m1

So the interval of convergence is: I  1 , 1 " .

2 2
c. (extra points) Find the limits of the power series.
Since
. .
lnŸ1  x   !Ÿ"1  1 x m1 
m
!Ÿ"1  m1 m1 x m , for " 1  x t 1
m1
m0 m1
. . .
Ÿ"1  m m
lnŸ1 " x   !Ÿ"1  m1 m1 Ÿ"x  m  !Ÿ"1  m1 m x  " ! m x , " 1  "x t 1
1 m
m1 m1 m1
.
! 1 m
m x  " lnŸ1 " x , " 1 t x  1
m1

So,
. . m
2 m x 2m  Ÿ2x 2  
! m ! m  " lnŸ1 " 2x 2   for " 1 t x  1
m1 m1

4. Page 245: 5 - Find also S 1 , S 2 , S 3 and sketch the graphs of sinŸx  cosŸx , S 1 , S 2 , and S 3 for x in ¡"1, 1¢.
It is know that
3 5 7 2 4 6
sinŸx   x " x  x " x . . . and cosŸx   1 " x  x " x . . .
3! 5! 7! 2! 4! 6!
3 5 7 2 4 6
sinŸx  cosŸx   x " x  x " x . . . 1 " x  x " x . . .
3! 5! 7! 2! 4! 6!
 x " 1 " 1 3
x  1  1 1  1 x5  " 1 " 1 1 " 1 1 " 1 x7. . .
3! 2! 5! 3! 2! 4! 7! 5! 2! 3! 4! 6!
2
 x" x  3 2 5
x " 4 7
x . . .
3 15 315
The first 4 terms of sinŸx  cosŸx  are
x, " 2 x 3 , 2 x5, " 4 x7.
3 15 315
S 1  x, S 2  x, S 3  x " 2 x 3
3

3
 y 1

0.5

0
-1 -0.5 0 0.5 1

x
-0.5

-1

– y  sinŸx  cosŸx , - - y  S 1 , S 2 .-.- y  S 3

5. *Use the known power series for functions e x , sin x, cos x, 1 , and lnŸ1  x  to find the power series
1"x
representation of each function including the interval of convergence I.
a. fŸx   x cosŸ2x 2  
. 2m . .
Ÿ2x 2   2m 4m1 m 4m1
fŸx   x cosŸ2x 2    x !Ÿ"1  m  !Ÿ"1  m 2 Ÿ2m !
x  !Ÿ"1  m 4Ÿ2m !
x
m0
Ÿ2m ! m0 m0
I "., . .

b. gŸx   e "x/3
m
. " x . m
gŸx   e "x/3
 ! 3
m!
 !Ÿ"1  m 3 mxm!
m0 m0
I "., .
c. (extra points) hŸx   1
4x
. .
m
hŸx   1  1  1 ! "x  1 !Ÿ"1  m xm
4x 4 1" " x 4 4 4 4m
m0 m0
4
" 1  " x  1 « 4  x  "4
4
I "4, 4