38.34
38.35
Show that
Substitute jc
2 for x in the series of Problem 38.34.
for
Substitute -x for x in Problem 37.113.
Show that
Hence, the radius of convergence is k
k . (Problem 38.25
Use the ratio test.
38.33 Let k be a fixed positive integer. Find the radius of convergence of
For |x|
(by Problem 37.111). Thus, the radius of convergence of Y,(a n + b n )x" isr,.
38.32
If E a,,x" has a radius of convergence r l and if E b n x" has a radius of convergence /•-, >r,, what is the
radius of convergence of the sum E (a n + b n )x"!
Hence, the radius of convergence is 1.
38.31
Find the radius of convergence of the binomial series
Use the ratio test.
Therefore, by Problem 38.29, the radius of convergence is l/e
POWER SERIES
329
38.27
Prove that, if the radius of convergence of E a n x" is R, then the radius of convergence of E a,,x
2 " is Vfl.
Assume \u\
2 " converges.
Now, assume \u\>VR. Then u
2 >R, and, therefore, E a a (u
2 )" diverges. Thus, E a a u
2 " diverges.
38.28 Find the radius of convergence of
By Problem 38.25, the radius of convergence of
is 4. Hence, by Problem 38.27, the radius of
convergence of
is
38.29
If
show that the radius of convergence of E a n x" is 1/L.
Assume |x|
and, therefore, \a n \ < r".
Hence, for n > k, \a n x"\ < r"\x\" = \rx\". Thus, eventually, E \a n x"\ is term by term less than the convergent
geometric series L \rx\ and is convergent by the comparison test. Now, on the other, hand, assume |jt| > 1IL.
Then L> 1/1x1. Choose r so that L>r>l/|x|. Then |rx|>l. Since lim
= L, there exists
an integer k such that, if n > k, then
and, therefore, |a n |>r". Hence, for n^k, \a n x"\>
r \x\ = |rx| > 1. Thus, we cannot have hm a n x =0, and, therefore, L a n x cannot converge, (the
theorem also holds when L = 0: then the series converges for all x.)
38.30
Find the radius of convergence of
is a special case of this result.)
38.35
Show that
Substitute jc
2 for x in the series of Problem 38.34.
for
Substitute -x for x in Problem 37.113.
Show that
Hence, the radius of convergence is k
k . (Problem 38.25
Use the ratio test.
38.33 Let k be a fixed positive integer. Find the radius of convergence of
For |x|
(by Problem 37.111). Thus, the radius of convergence of Y,(a n + b n )x" isr,.
38.32
If E a,,x" has a radius of convergence r l and if E b n x" has a radius of convergence /•-, >r,, what is the
radius of convergence of the sum E (a n + b n )x"!
Hence, the radius of convergence is 1.
38.31
Find the radius of convergence of the binomial series
Use the ratio test.
Therefore, by Problem 38.29, the radius of convergence is l/e
POWER SERIES
329
38.27
Prove that, if the radius of convergence of E a n x" is R, then the radius of convergence of E a,,x
2 " is Vfl.
Assume \u\
2 " converges.
Now, assume \u\>VR. Then u
2 >R, and, therefore, E a a (u
2 )" diverges. Thus, E a a u
2 " diverges.
38.28 Find the radius of convergence of
By Problem 38.25, the radius of convergence of
is 4. Hence, by Problem 38.27, the radius of
convergence of
is
38.29
If
show that the radius of convergence of E a n x" is 1/L.
Assume |x|
and, therefore, \a n \ < r".
Hence, for n > k, \a n x"\ < r"\x\" = \rx\". Thus, eventually, E \a n x"\ is term by term less than the convergent
geometric series L \rx\ and is convergent by the comparison test. Now, on the other, hand, assume |jt| > 1IL.
Then L> 1/1x1. Choose r so that L>r>l/|x|. Then |rx|>l. Since lim
= L, there exists
an integer k such that, if n > k, then
and, therefore, |a n |>r". Hence, for n^k, \a n x"\>
r \x\ = |rx| > 1. Thus, we cannot have hm a n x =0, and, therefore, L a n x cannot converge, (the
theorem also holds when L = 0: then the series converges for all x.)
38.30
Find the radius of convergence of
is a special case of this result.)