WEN-CHINGLIEN
Department of Mathematics National Cheng Kung University
WEN-CHINGLIEN Advanced Calculus (I)
A power series is a series of the form
S(x ) :=
∞
X
k =0
ak(x − x0)k
WEN-CHINGLIEN Advanced Calculus (I)
An extended real number R is said to be the radius of convergence of a power series S(x ) :=
∞
P
k =0
ak(x − x0)k if and only if S(x ) converges absolutely for |x − x0| < R and S(x ) diverges for |x − x0| > R.
WEN-CHINGLIEN Advanced Calculus (I)
An extended real number R is said to be the radius of convergence of a power series S(x ) :=
∞
P
k =0
ak(x − x0)k if and only if S(x ) converges absolutely for |x − x0| < R and S(x ) diverges for |x − x0| > R.
WEN-CHINGLIEN Advanced Calculus (I)
Let S(x ) = P
k =0
ak(x − x0)k be a power series centered at x0, If R = 1
lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞, then R is the radius of convergence of S.
In fact,
(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval
[a, b] ⊆ (x0− R, x0+R),
(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
Let S(x ) = P
k =0
ak(x − x0)k be a power series centered at x0, If R = 1
lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞, then R is the radius of convergence of S.
In fact,
(i)S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval
[a, b] ⊆ (x0− R, x0+R),
(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
Let S(x ) = P
k =0
ak(x − x0)k be a power series centered at x0, If R = 1
lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞, then R is the radius of convergence of S.
In fact,
(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval
[a, b] ⊆ (x0− R, x0+R),
(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
Let S(x ) = P
k =0
ak(x − x0)k be a power series centered at x0, If R = 1
lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞, then R is the radius of convergence of S.
In fact,
(i)S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii)S(x) converges uniformly on any closed interval
[a, b] ⊆ (x0− R, x0+R),
(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
Let S(x ) = P
k =0
ak(x − x0)k be a power series centered at x0, If R = 1
lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞, then R is the radius of convergence of S.
In fact,
(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval
[a, b] ⊆ (x0− R, x0+R),
(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
Let S(x ) = P
k =0
ak(x − x0)k be a power series centered at x0, If R = 1
lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞, then R is the radius of convergence of S.
In fact,
(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii)S(x) converges uniformly on any closed interval
[a, b] ⊆ (x0− R, x0+R),
(iii)and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
Let S(x ) = P
k =0
ak(x − x0)k be a power series centered at x0, If R = 1
lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞, then R is the radius of convergence of S.
In fact,
(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval
[a, b] ⊆ (x0− R, x0+R),
(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
Let S(x ) = P
k =0
ak(x − x0)k be a power series centered at x0, If R = 1
lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞, then R is the radius of convergence of S.
In fact,
(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval
[a, b] ⊆ (x0− R, x0+R),
(iii)and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
Let S(x ) = P
k =0
ak(x − x0)k be a power series centered at x0, If R = 1
lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞, then R is the radius of convergence of S.
In fact,
(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval
[a, b] ⊆ (x0− R, x0+R),
(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
Fix x ∈R, x 6= x0, and set ρ := 1 lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞. To apply the Root Test to S(x ), consider
r (x ) := lim sup
n→∞
|ak(x − x0)k|1/k = |x − x0| · lim sup
k →∞
|ak|1/k.
WEN-CHINGLIEN Advanced Calculus (I)
Fix x ∈R, x 6= x0,and set ρ := 1 lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞. To apply the Root Test to S(x ),consider
r (x ) := lim sup
n→∞
|ak(x − x0)k|1/k = |x − x0| · lim sup
k →∞
|ak|1/k.
WEN-CHINGLIEN Advanced Calculus (I)
Fix x ∈R, x 6= x0, and set ρ := 1 lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞. To apply the Root Test to S(x ), consider
r (x ) := lim sup
n→∞
|ak(x − x0)k|1/k = |x − x0| · lim sup
k →∞
|ak|1/k.
WEN-CHINGLIEN Advanced Calculus (I)
Fix x ∈R, x 6= x0, and set ρ := 1 lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞. To apply the Root Test to S(x ),consider
r (x ) := lim sup
n→∞
|ak(x − x0)k|1/k = |x − x0| · lim sup
k →∞
|ak|1/k.
WEN-CHINGLIEN Advanced Calculus (I)
Fix x ∈R, x 6= x0, and set ρ := 1 lim sup
k →∞
|ak|1/k, with the convention that 1
∞ =0 and 1
0 = ∞. To apply the Root Test to S(x ), consider
r (x ) := lim sup
n→∞
|ak(x − x0)k|1/k = |x − x0| · lim sup
k →∞
|ak|1/k.
WEN-CHINGLIEN Advanced Calculus (I)
ρ =0. By our convention,ρ =0 implies that r (x ) = ∞ > 1, so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1,so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ =0. By our convention,ρ =0 implies that r (x ) = ∞ > 1, so by the Root Test,S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1,so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1, so by the Root Test,S(x) does not converge for any x 6= x0. Hence,the radius of convergence of S is R = 0 = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1, so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1, so by the Root Test, S(x) does not converge for any x 6= x0. Hence,the radius of convergence of S is R = 0 = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1, so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence,the radius of convergence of S is R = ∞ = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence,the radius of convergence of S is R = ∞ = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1 if and only if |x − x0| < ρ,it follows from the Root Test that S(x ) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1 if and only if |x − x0| < ρ,it follows from the Root Test that S(x ) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ,we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ,we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
ρ ∈ (0, ∞). Then r (x ) = |x − x0|
ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.
WEN-CHINGLIEN Advanced Calculus (I)
|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that
∞
P
k =0
Mk
converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2
WEN-CHINGLIEN Advanced Calculus (I)
|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that
∞
P
k =0
Mk
converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2
WEN-CHINGLIEN Advanced Calculus (I)
|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that
∞
P
k =0
Mk
converges.Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2
WEN-CHINGLIEN Advanced Calculus (I)
|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that
∞
P
k =0
Mk
converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N,it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2
WEN-CHINGLIEN Advanced Calculus (I)
|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that
∞
P
k =0
Mk
converges.Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2
WEN-CHINGLIEN Advanced Calculus (I)
|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that
∞
P
k =0
Mk
converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N,it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2
WEN-CHINGLIEN Advanced Calculus (I)
|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that
∞
P
k =0
Mk
converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2
WEN-CHINGLIEN Advanced Calculus (I)
If the limit
R = lim
k →∞
ak
ak +1
exists as an extended real number, then R is the radius of convergence of the power series S(x ) =
∞
P
k =0
ak(x − x0)k.
WEN-CHINGLIEN Advanced Calculus (I)
If the limit
R = lim
k →∞
ak
ak +1
exists as an extended real number, then R is the radius of convergence of the power series S(x ) =
∞
P
k =0
ak(x − x0)k.
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.
By Theorem 7.21, for a given power series S =
∞
P
k =0
ak(x − x0)k, there are only three possibilities:
(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),
(ii) R = 0, in which case the interval of convergence of S is {x0}, and
(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.
By Theorem 7.21, for a given power series S =
∞
P
k =0
ak(x − x0)k, there are only three possibilities:
(i)R = ∞, in which case the interval of convergence of S is (−∞, ∞),
(ii) R = 0, in which case the interval of convergence of S is {x0}, and
(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.
By Theorem 7.21, for a given power series S =
∞
P
k =0
ak(x − x0)k, there are only three possibilities:
(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),
(ii) R = 0, in which case the interval of convergence of S is {x0}, and
(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.
By Theorem 7.21, for a given power series S =
∞
P
k =0
ak(x − x0)k, there are only three possibilities:
(i)R = ∞, in which case the interval of convergence of S is (−∞, ∞),
(ii)R = 0, in which case the interval of convergence of S is {x0}, and
(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.
By Theorem 7.21, for a given power series S =
∞
P
k =0
ak(x − x0)k, there are only three possibilities:
(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),
(ii) R = 0, in which case the interval of convergence of S is {x0}, and
(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.
By Theorem 7.21, for a given power series S =
∞
P
k =0
ak(x − x0)k, there are only three possibilities:
(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),
(ii)R = 0, in which case the interval of convergence of S is {x0}, and
(iii)0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.
By Theorem 7.21, for a given power series S =
∞
P
k =0
ak(x − x0)k, there are only three possibilities:
(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),
(ii) R = 0, in which case the interval of convergence of S is {x0}, and
(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.
By Theorem 7.21, for a given power series S =
∞
P
k =0
ak(x − x0)k, there are only three possibilities:
(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),
(ii) R = 0, in which case the interval of convergence of S is {x0}, and
(iii)0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.
By Theorem 7.21, for a given power series S =
∞
P
k =0
ak(x − x0)k, there are only three possibilities:
(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),
(ii) R = 0, in which case the interval of convergence of S is {x0}, and
(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence may contain none, one, or both its endpoints.
WEN-CHINGLIEN Advanced Calculus (I)
The interval of convergence may contain none, one, or both its endpoints.
WEN-CHINGLIEN Advanced Calculus (I)
If f (x ) = P
k =0
ak(x − x0)k is a power series with positive radius of convergence R, then f is continuous on (x0− R, x0+R).
WEN-CHINGLIEN Advanced Calculus (I)
If f (x ) = P
k =0
ak(x − x0)k is a power series with positive radius of convergence R, then f is continuous on (x0− R, x0+R).
WEN-CHINGLIEN Advanced Calculus (I)
Suppose that [a, b] is nondegenerate. If f (x ) :=
∞
P
k =0
ak(x − x0)k converges on [a,b], then f(x) is continuous and converges uniformly on [a,b].
WEN-CHINGLIEN Advanced Calculus (I)
Suppose that [a, b] is nondegenerate. If f (x ) :=
∞
P
k =0
ak(x − x0)k converges on [a,b], then f(x) is continuous and converges uniformly on [a,b].
WEN-CHINGLIEN Advanced Calculus (I)
If a power series S(x ) =
∞
P
k =0
ak(x − x0)k converges at some x1 >x0, then S(x) converges uniformly on [x0,x1] and absolutely on [x0,x1). It might not converge
absolutely at x = x1.
WEN-CHINGLIEN Advanced Calculus (I)
If a power series S(x ) =
∞
P
k =0
ak(x − x0)k converges at some x1 >x0, then S(x) converges uniformly on [x0,x1] and absolutely on [x0,x1). It might not converge
absolutely at x = x1.
WEN-CHINGLIEN Advanced Calculus (I)
If an ∈ R for n ∈ N, then x := lim sup
n→∞
(n|an|1/(n−1)) = y := lim sup
n→∞
|an|1/n
WEN-CHINGLIEN Advanced Calculus (I)
If an ∈ R for n ∈ N, then x := lim sup
n→∞
(n|an|1/(n−1)) = y := lim sup
n→∞
|an|1/n
WEN-CHINGLIEN Advanced Calculus (I)
Let > 0. Since n1/(n−1) → 1 as n → ∞,choose N ∈N so that n ≥ N implies 1 − < n1/(n−1) <1 + , i.e.,
(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).
WEN-CHINGLIEN Advanced Calculus (I)
Let > 0.Since n1/(n−1) → 1 as n → ∞, choose N ∈ N so that n ≥ N implies 1 − < n1/(n−1) <1 + , i.e.,
(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).
WEN-CHINGLIEN Advanced Calculus (I)
Let > 0. Since n1/(n−1) → 1 as n → ∞,choose N ∈N so that n ≥ N implies 1 − < n1/(n−1) <1 + ,i.e.,
(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).
WEN-CHINGLIEN Advanced Calculus (I)
Let > 0. Since n1/(n−1) → 1 as n → ∞, choose N ∈ N so that n ≥ N implies 1 − < n1/(n−1) <1 + , i.e.,
(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).
WEN-CHINGLIEN Advanced Calculus (I)
Let > 0. Since n1/(n−1) → 1 as n → ∞, choose N ∈ N so that n ≥ N implies 1 − < n1/(n−1) <1 + ,i.e.,
(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).
WEN-CHINGLIEN Advanced Calculus (I)
Let > 0. Since n1/(n−1) → 1 as n → ∞, choose N ∈ N so that n ≥ N implies 1 − < n1/(n−1) <1 + , i.e.,
(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank If nk n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank k n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank If nk n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank k n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n,it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank k n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank k n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n,it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank k n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank k n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly,the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank k n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank k n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly,the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
zk := |ank k n ≥ N, then
|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it
follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2
WEN-CHINGLIEN Advanced Calculus (I)
If f (x ) = P
k =0
ak(x − x0)k is a power series with positive radius of convergence R, then f0(x ) =
∞
P
k =1
kak(x − x0)k −1 for x ∈ (x0− R, x0+R).
WEN-CHINGLIEN Advanced Calculus (I)
If f (x ) = P
k =0
ak(x − x0)k is a power series with positive radius of convergence R, then f0(x ) =
∞
P
k =1
kak(x − x0)k −1 for x ∈ (x0− R, x0+R).
WEN-CHINGLIEN Advanced Calculus (I)
If f (x ) =
∞
P
k =0
ak(x − x0)k has a positive radius of convergence R, then f ∈C∞(x0− R, x0+R) and
f(k )(x ) =
∞
X
n=k
n!
(n − k )!an(x − x0)n−k for x ∈ (x0− R, x0+R) and k ∈ N.
WEN-CHINGLIEN Advanced Calculus (I)
If f (x ) =
∞
P
k =0
ak(x − x0)k has a positive radius of convergence R, then f ∈C∞(x0− R, x0+R) and
f(k )(x ) =
∞
X
n=k
n!
(n − k )!an(x − x0)n−k for x ∈ (x0− R, x0+R) and k ∈ N.
WEN-CHINGLIEN Advanced Calculus (I)
Let f (x ) = P
k =0
ak(x − x0)k be a power series and let a, b ∈R with a < b.
(i)If f(x) converges on [a,b], then f is integrable on [a,b]
and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
(ii) If f(x) converges on [a,b) and if
∞
P
k =0
ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
WEN-CHINGLIEN Advanced Calculus (I)
Let f (x ) = P
k =0
ak(x − x0)k be a power series and let a, b ∈R with a < b.
(i) If f(x) converges on [a,b], then f is integrable on [a,b]
and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
(ii) If f(x) converges on [a,b) and if
∞
P
k =0
ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
WEN-CHINGLIEN Advanced Calculus (I)
Let f (x ) = P
k =0
ak(x − x0)k be a power series and let a, b ∈R with a < b.
(i)If f(x) converges on [a,b], then f is integrable on [a,b]
and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
(ii)If f(x) converges on [a,b) and if
∞
P
k =0
ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
WEN-CHINGLIEN Advanced Calculus (I)
Let f (x ) = P
k =0
ak(x − x0)k be a power series and let a, b ∈R with a < b.
(i) If f(x) converges on [a,b], then f is integrable on [a,b]
and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
(ii) If f(x) converges on [a,b) and if
∞
P
k =0
ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
WEN-CHINGLIEN Advanced Calculus (I)
Let f (x ) = P
k =0
ak(x − x0)k be a power series and let a, b ∈R with a < b.
(i) If f(x) converges on [a,b], then f is integrable on [a,b]
and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
(ii)If f(x) converges on [a,b) and if
∞
P
k =0
ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
WEN-CHINGLIEN Advanced Calculus (I)
Let f (x ) = P
k =0
ak(x − x0)k be a power series and let a, b ∈R with a < b.
(i) If f(x) converges on [a,b], then f is integrable on [a,b]
and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
(ii) If f(x) converges on [a,b) and if
∞
P
k =0
ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and
Z b a
f (x )dx =
∞
X
k =0
ak
Z b a
(x − x0)kdx .
WEN-CHINGLIEN Advanced Calculus (I)
(i)
By Abel’s Theorem,f(x) converges uniformly on [a,b].
Hence, by Theorem 7.14ii, f(x) is term-by-term integrable on [a,b].
WEN-CHINGLIEN Advanced Calculus (I)
(i)
By Abel’s Theorem, f(x) converges uniformly on [a,b].
Hence, by Theorem 7.14ii, f(x) is term-by-term integrable on [a,b].
WEN-CHINGLIEN Advanced Calculus (I)
(i)
By Abel’s Theorem,f(x) converges uniformly on [a,b].
Hence, by Theorem 7.14ii,f(x) is term-by-term integrable on [a,b].
WEN-CHINGLIEN Advanced Calculus (I)
(i)
By Abel’s Theorem, f(x) converges uniformly on [a,b].
Hence, by Theorem 7.14ii, f(x) is term-by-term integrable on [a,b].
WEN-CHINGLIEN Advanced Calculus (I)
