This preview shows page 1. Sign up to view the full content.
Unformatted text preview: then the number m! e is not an integer and deduce that the number
e is irrational. Solution: This problem also appeared in Chapter 10 where it was provided with a solution. 375 Some Exercises on Binomial Series
1. Given that 1, prove that Ý 2 .
n n0 Solution: We know from earlier work that the series
n converges and it therefore follows from Abel’s theorem that
Ý 2 lim 1 x
x1 x1 2. Given that 0, prove that xn
n lim
n0 Ý 1
n n n0 Ý .
n n0 0. 3. Given that and are any real numbers and x  1, apply Cauchy’s theorem on products of series to the
Maclaurin expansions of 1 x and 1 x to deduce that the equation
n n j0
j j
n holds for every positive integer n.
Since the nth term of the Cauchy product of the series
n
j0 n xn j j
x
j j
n n n j0 n0 Ý xn
n n
nx n0 1x x n is n
x
j j and since this Cauchy product must converge on the interval
Ý
n x n and 1, 1 to 1x 1x the equation
Ý
n0 holds for all x
deduce that n n j0 Ý
j j xn
n0 n
x
n 1, 1 . From the theorem on uniqueness of coefficients of a power series we
n
j0 n j
j
.
n 4. The motivation and notation used in this exercise is the theory of Cesàro summability of series that
can be found in the book Divergent Series by G. H. Hardy.
In this exercise we define
1 2 n
A
n
n!
whenever 1 and n is a positive integer. We also define A 1.
0
a. Prove that if 1 and n is a nonnegative integer we have
1 1 nA.
n
n 376 This identity is obvious.
b. Use the preceding exercise to prove that if and are greater than 1 and n is a nonnegative integer
then
n 1 A jAj An
n . j0 We observe that
n n A jAj
n
j0 1 nj 1 j0
n 1 j0 1 n
n 2 1
j 1
nj n 1
nj j 1
j 1 An c. Prove that if and n are nonnegative integers then
n1 n2 n
A
n
!
where the numerator of the right side is understood to be 1 when 0.
This identity is obvious.
d. Prove that if , and n are nonnegative integers we have
n
j0 j1 j
! j1 j
! n1 n1
! This identity follows at once from parts b and c.
5. a. Prove that if
whenever 0 x
interval 0, 1 . fx 1 x
1 then there exists a sequence of polynomials that converges uniformly to f on the Solution: Since the series
1/2
n
is absolutely convergent, the uniform convergence of the Maclaurin series of f on the interval 0, 1
follows from the comparison test for uniform convergence.
b. Prove that if f is the function defined in part a and if
gx f1 x2 whenever 1 x 1 then there exists a sequence of polynomials that converges uniformly to g on the
interval 1, 1 .
Choose a sequence p n of polynomials that converges uniformly on the interval 0, 1 to the
function f. For each positive integer n and for each x
1, 1 we now define
2
qn x pn 1 x
and we observe that q n converges uniformly on
function q n is also a polynomial. 1, 1 to the function g. Of course, each c. Prove that if g x x  for all x
1, 1 th...
View
Full
Document
This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details