Calculus Chapter 11 Summary

Definitions

(a) A set S of real numbers is called bounded from above if there is a real number k such that k ≥ s for all s in S.

The element k is called an upper bound of S.

(b) A set S of real numbers is called boundedif there is a real number m such that

|s| ≤ m for all s in S.

(c) b is called the least upper bound (or supremum) of S, denoted by b = lub S or sup S, if (i) b ≥ s for all s in S, i.e. b is an upper bound of S.

(ii) if k is another upper bound of S then k ≥ b.

Equivalently,

b = lub S ⇐⇒ ∀  > 0 ∃s ∈ S such that b −  < s ≤ b =⇒ |b − s| < .

(d) A sequence {x_n} ⊂ R is called a Cauchy sequence if for any  > 0 there exists K ∈ N such that if n, m ≥ K then |x_n− x_m| < .

Facts: It is easy to show that the followings are true.

(i) If {x_n} ⊂ R is a Cauchy sequence then {x_n} is bounded, i.e. there exists L ∈ R such that |x_n| ≤ L for all n ∈ N.

(ii) If {x_n} isa convergent sequence then {x_n} is a Cauchy sequence.

Remarks

(a) Completeness Axiom of R: Every nonempty, bounded from above set of real numbers has a least upper bound (or supremum) in the set of real numbers.

(b) Least Upper Bound Property: Let S be a nonempty, bounded from above subset of R and let b = lub S. By taking n = 1

n, for n = 1, 2, . . . , there exists x_n∈ S such that b − 1

n = b − _n < x_n ≤ b for all n ∈ N.

This implies that

for each n ∈ N there exists xn∈ S such that lim

n→∞x_n= lub S.

(c) Theorem R is complete iff every Cauchy sequence {xn} ⊂ R converges to a point in R.

Proof.

=⇒ Given  > 0 since {x_n} is Cauchy,

there exists K ∈ N such that if n, m ≥ K then |xn− x_m| <  2.

Case 1: {x_n | n ∈ N} = {r1, r₂, . . . , r_m} is a finite subset of R. Then there is an element appears infinitely many times in the sequence {x_n}.

Let r_j be an element that appears infinitely many times in the sequence {x_n} and let {x_nk | k ∈ N} be the subsequence of {xn} such that

x_nk = r_j for all k ∈ N =⇒|x_nk− r_j| = 0 for all k ∈ N. This implies that if n ≥ K,since n_K ≥ K, we have

|x_n− r_j|= |x_n− x_nK + x_nK − r_j| ≤ |x_n− x_nK| + |x_nK − r_j|<

2+ 0 <.

Hence

n→∞lim x_n= r_j. Case 2: {x_n | n ∈ N} is an infinite subset of R.

Since a Cauchy sequence is bounded, there exists L > 0 such that x_n∈ [−L, L] for all n ∈ N.

Let I1 = [−L, L] and let xn1 = x1 ∈ I1.

Divide I₁ into 2 equal length subintervals and let I₂ be a closed subinterval such that I₂∩ {x_n | n > n₁} = an infinite subset of R

and let

xn2 ∈ I2∩ {xn | n > n1}.

Continuing this process, we obtain a nested sequence of closed intervals {I_k} and a subse- quence {xn_k} such that

x_nk ∈ I_k∩ {x_n | n > n_k−1} = an infinite subset of R for all k ∈ N.

Since

I₁ ⊃ I₂ ⊃ · · · ⊃ I_k⊃ · · · ; lim

k→∞|I_k| = lim

k→∞

L 2^k−2 = 0, the sequence {x_nk} converges and since

∃! x ∈

∞

\

k=1

I_k6= ∅, lim

k→∞x_nk = x.

With the given , since lim

k→∞x_nk = x,

there exists K₁ ≥ K such that |x − x_nK1| <  2. This implies that if n ≥ K,since nK1 ≥ K1 ≥ K, we have

|x_n− x|= |x_n− x_nK1 + x_nK1 − x| ≤ |x_n− x_nK1| + |x_nK1 − x|< 2 + 

2 =.

Hence

n→∞lim x_n = x.

⇐= Let S be a nonempty, bounded from above subset of R. Choose a sufficiently large M > 0 such that

M is an upper bound of S and [−M, M ] ∩ S 6= ∅.

Let I₁ = [−M, M ]. Divide I₁ into 2 equal length subintervals and let I₂ be the closed subinterval such that

the right endpoint of I₂ is an upper bound of S and I₂∩ S 6= ∅.

Continuing this process, we obtain a sequence of closed intervals {I_n} such that the right endpoint of I_n is an upper bound of S, I_n∩ S 6= ∅ ∀ n ≥ 1.

and

I₁ ⊃ I₂ ⊃ · · · ⊃ I_n⊃ · · · ; lim

n→∞|I_n| = lim

n→∞

M 2ⁿ⁻² = 0, For each n ∈ N, since In∩ S 6= ∅, let x_n be a point in I_n∩ S.

Since lim

n→∞|I_n| = lim

n→∞

M

2ⁿ⁻² = 0, {x_n} is a Cauchy sequence and hence it converges to a point, say x, in R.

To show that x = lub S, we need to show that

(i) x ≥ s for all s in S, i.e. x is an upper bound of S.

(ii) if u is another upper bound of S then u ≥ x.

Proof of (i): Suppose that x is not an upper bound of S, i.e. suppose that there exists s ∈ S such that s > x. Let  = s − x. Since lim

n→∞x_n = x, x_n ∈ I_n and |I_n| = M

2ⁿ⁻², there exists K ∈ N such that

if n ≥ K then |In| = M 2ⁿ⁻² < 

2 and |xn− x| <  2. This implies that







x_n− 

2 ≤ y ≤ x_n+ 

2 for all y ∈ I_n∩ S ⇐⇒ I_n ⊂ (x_n− 

2, x_n+  2), x − 

2 < x_n< x + 

2 ⇐⇒ x_n∈ (x − 

2, x + 

2) for all n ≥ K.

Combining these, we get

x −  < y < x +  = s for all y ∈ I_n∩ S ⇐⇒ I_n ⊂ (x − , x + ) = (x − , s)

x_n+ ^₂ x_n

x_n− ^₂

x x + ^₂

x − ^₂ x +  = s

x −  y

which implies that the right endpoint of I_n is smaller than s ∈ S. This contradicts to the definition of I_n. Hence, x is an upper bound of S.

Proof of (ii): Let u be another upper bound of S. Suppose that u < x, then  = x − u 2 > 0.

Since lim

n→∞x_n= x, there exists K ∈ N such that if n ≥ K then |xn− x| < .

Thus 





 = x − u

2 =⇒u = x − 2,

x −  < x_n< x +  ⇐⇒ x_n∈ (x − , x + ) for all n ≥ K.

This implies that

u = x − 2 <x −  <x_n< x +  for all x_n∈ S ∩ I_n, n ≥ K=⇒ u < x_n∈ S

x +  x

xn

xn− ₂^ xn+ ^₂ x − 

u = x − 2

which contradicts to the assumption that u being an upper bound of S.

Hence u ≥ x.

(d) Every bounded, monotonic sequence {x_n} ⊂ R is convergent.

Proof: Without loss of generality, we may assume that {x_n} is an increasing sequence and it is bounded from above. By the completeness of R,

x = lub {xn | n ∈ N} exists.

Thus

for each  > 0 there exists K ∈ N such that x −  < xK ≤ x.

Now {x_n} is an increasing sequence, we also have

x_K ≤ x_n for each n ≥ K.

Hence

x −  < x_K ≤ x_n≤ x =⇒ |x_n− x| <  for all n ≥ K.

This proves that

n→∞lim x_n = x.

Definition The series

∞

X

k=1

ak is called convergent if lim

n→∞Sn = lim

n→∞

n

X

k=1

ak exists and the limit is called the sumof the series. If the sequence {S_n} of n^th partial sum is divergent, then the series is calleddivergent.

Remarks

(a) Test for Divergence If lim

n→∞a_n does not exist or if lim

n→∞a_n 6= 0, then the series

∞

X

n=1

a_n is divergent.

Proof: Since

∞

X

n=1

a_n is convergent ⇐⇒ {S_n =

n

X

k=1

a_k} is a Cauchy sequence

=⇒ lim

m, n→∞|Sm− Sn| = 0

=⇒ lim

n→∞|a_n| = lim

n→∞|S_n− S_n−1| = 0.

This proves that if lim

n→∞a_ndoes not exist or if lim

n→∞a_n 6= 0, then the series

∞

X

n=1

a_nis divergent.

(b) The Integral Test Suppose f is a continuous, positive, decreasing function on [1, ∞) and let a_n = f (n). Then the series

∞

X

n=1

a_n := lim

k→∞

k

X

n=1

a_n is convergent if and only if the improper integral

Z ∞ 1

f (x) dx := lim

M →∞

Z M 1

f (x) dx is convergent, i.e.

Z ∞ 1

f (x) dx and

∞

X

n=1

a_n converge or diverge simultaneously.

Proof: For each M ∈ N, consider the partition {1, 2, , 3, . . . , n, n + 1, . . . , M } of [1, M], i.e. x_n = n + 1 and ∆x_n = 1 for each 1 ≤ n ≤ M − 1. Since f is a continuous, positive, decreasing function on [1, ∞),

M −1

X

n=1

f (n + 1) =

M −1

X

n=1

inf

x∈[n,n+1]f (x) ∆x_n ≤ Z M

1

f (x) dx ≤

M −1

X

n=1

sup

x∈[n,n+1]

f (x) ∆x_n =

M −1

X

n=1

f (n).

Tis implies that Z ∞

1

f (x) dx and

∞

X

n=1

a_n converge or diverge simultaneously.

(c) The Ratio Test Suppose |a_n| > 0 for all n ∈ N and assume that

n→∞lim

|a_n+1|

|an| = L ∈ [0, ∞].

Then the series

∞

X

n=1

a_n is

(absolute convergent if L < 1,

divergent if L > 1 or L = ∞.

However, the ratio test is inconclusive if L = 1.

Proof: Suppose that L 6= 1 then  = |L − 1|

2 > 0. Since lim

n→∞

|a_n+1|

|a_n| = L ∈ (0, ∞), there exists K ∈ N such that if n ≥ K then

|a_n+1|

|a_n| − L    

< .

Thus

L −  < |a_n+1|

|a_n| < L +  ∀ n ≥ K

=⇒ L − )|a_n| < |a_n+1| < L + 
|an| ∀ n ≥ K

=⇒ L − )²|a_n−1| < |a_n+1| < L + 
2

|a_n−1| ∀ n ≥ K

· · · ·

=⇒ L − )^n−K+1|a_K| < |a_n+1| < L + 
n−K+1

|a_K| ∀ n ≥ K

=⇒







|a_n+1| < L + 
n−K+1

|a_K| ∀ n ≥ K if L < 1

L − 
n−K+1

|a_K| < |a_n+1| ∀ n ≥ K if L > 1

⇐⇒











|an+1| < L + 1 − L 2

n−K+1

|aK| ∀ n ≥ K if L < 1

L − L − 1 2

n−K+1

|a_K| < |a_n+1| ∀ n ≥ K if L > 1

⇐⇒











|a_n+1| < 1 + L 2

n−K+1

|a_K| ∀ n ≥ K if L < 1 L + 1

2

n−K+1

|aK| < |an+1| ∀ n ≥ K if L > 1.

This imples that

∞

X

n=K

|an+1|























<

∞

X

n=K

1 + L 2

n−K+1

|a_K| =

|a_K| 1 + L 2

1 −1 + L

2

since 0 < 1 + L

2 < 1 if L < 1,

>

∞

X

n=K

L + 1 2

n−K+1

|a_K| = ∞ since L + 1

2 > 1 if L > 1.

Hence the series

∞

X

n=1

a_n is

(absolute convergent if L < 1,

divergent if L > 1 or L = ∞.

(d) The Root Test Suppose that

n→∞lim |an|^1/n = L ∈ [0, ∞].

Then the series

∞

X

n=1

a_n is

(absolute convergent if L < 1,

divergent if L > 1 or L = ∞.

However, the root test is inconclusive if L = 1.

Proof: Suppose that L 6= 1 then  = |L − 1|

2 > 0. Since lim

n→∞|a_n|^1/n = L ∈ (0, ∞), there exists K ∈ N such that if n ≥ K then 

|a_n|^1/n− L < .

Thus

L −  < |a_n|^1/n < L +  ∀ n ≥ K

=⇒ L − )ⁿ < |an| < L + 
n

∀ n ≥ K

=⇒







|a_n| < L + 
n

∀ n ≥ K if L < 1

L − 
n

< |a_n+1| ∀ n ≥ K if L > 1

⇐⇒











|a_n| < L + 1 − L 2

n

∀ n ≥ K if L < 1

L −L − 1 2

n

< |a_n| ∀ n ≥ K if L > 1

⇐⇒











|a_n| < 1 + L 2

n

∀ n ≥ K if L < 1 L + 1

2

n

< |a_n| ∀ n ≥ K if L > 1.

This imples that

∞

X

n=K

|a_n|























<

∞

X

n=K

1 + L 2

n

=

1 + L 2

K

1 −1 + L 2

since 0 < 1 + L

2 < 1 if L < 1,

>

∞

X

n=K

L + 1 2

n

= ∞ since L + 1

2 > 1 if L > 1.

Hence the series

∞

X

n=1

a_n is

(absolute convergent if L < 1,

divergent if L > 1 or L = ∞.

(e) The Rearrangement Theorem Let

∞

X

k=1

a_k be an absolutely convergent series in R. Then

any rearragement of

∞

X

k=1

ak converges absolutely to the same value.

Proof: Let s =

∞

X

k=1

a_k, let

∞

X

j=1

b_j be an rearrangement of

∞

X

k=1

a_k, and let L =

∞

X

k=1

|a_k|.

Since |a_k| ≥ 0 for all k ∈ N, L is an upper bound for each partial sum of

∞

X

k=1

|a_k|, i.e.

n

X

k=1

|a_k| ≤ L ∀ n ∈ N.

This implies that

r

X

j=1

|b_j| = |b₁| + · · · + |b_r| ≤ L ∀ r ∈ N,

i.e. the sequence {

r

X

j=1

|b_j| | r ∈ N} is bounded from above by L.

Also since |b_j| ≥ 0 for all j ∈ N, the sequence {

r

X

j=1

|b_j| | r ∈ N} is increasing in r.

Therefore, the series

∞

X

j=1

bj is absolutely convergent to some element t ∈ R.

Claim: s = t.

Proof of the Claim: For each n ∈ N, let sn =

n

X

k=1

a_k be the n^th partial sum of

∞

X

k=1

a_k.

For each  > 0, since

∞

X

k=1

a_k is convergent absolutely to s,

there exists M ∈ N such that if m > n ≥ M, then |s − sn| <  and

m

X

k=n+1

|a_k| < .

Also since

∞

X

j=1

b_j is absolutely convergent to t, there exists a partial sum t_r =

r

X

j=1

b_j such that

|t − t_r| <  and such that each a₁, a₂, · · · , a_n appears in t_r.

Having done this, choose m > n so large that every bj appearing in tr also appears in sm. Therefore

|s − t| ≤ |s − s_m| + |s_m− t_r| + |t_r− t| <  +

m

X

k=n+1

|a_k| +  < 3.

Since  > 0 is arbitrary, this proves that s = t.

(f) Remarks on Rearrangement:

(i) Let a_k ∈ R for all k ∈ N. If

∞

X

k=1

a_k is conditionally convergent, i.e. it is convergent but not absolutely convergent, then the series of potive terms is divergent (to ∞) and the series of negative terms is divergent (to −∞).

(ii) If

∞

X

k=1

a_k is conditionally convergent and if c is an arbitrary real number, then there

exists a rearragement

∞

X

j=1

b_j of

∞

X

k=1

a_k such that c =

∞

X

j=1

b_j. Proof:

1. Construction of a rearrangement of

∞

X

k=1

a_k:

Step 1: take positive terms until we obtain a partial sum greater than c,

Step 2: take negative terms from the remaining series until we obtain a partial sum of terms less than c,

Step 3: take positive terms from the remaining series until we obtain a partial sum of terms greater than c,

Step 4: take negative terms from the remaining series until we obtain a partial sum of terms less than c,

· · · ·

Step 2r − 1: take positive terms from the remaining series until we obtain a partial sum of terms greater than c,

Step 2r: take negative terms from the remaining series until we obtain a partial sum of terms less than c,

· · · etc.

2. Convergence of the rearrangement: For each  > 0, since

∞

X

k=1

ak converges =⇒ ∃ K ∈ N such that if n ≥ K then |

∞

X

k=n+1

ak| < .

Fix n ≥ K, we may take a sufficiently large r such that each a₁, a₂, · · · , a_n appears in the partial sum of terms obtained from Step r. This implies that

r→∞lim

∞

X

j=r+1

b_j = 0 ⇐⇒

∞

X

j=1

b_j = lim

m→∞

m

X

j=1

b_j exists.

Since any subsequece {

mk

X

j=1

bj | k ∈ N} converges to the same value, and since the partial sums are greater than c in the odd number steps and the partial sums are less than c in the even number steps, one conclude that the constructed rearrangement converges to c.

Power Series

(a) Definition Let R be the radius of convergence of the power seies

∞

X

n=0

a_nxⁿ
, let Sm =

m

X

n=0

anxⁿ
for each m = 0, 1, 2, . . . , and let E = [a, b] ⊂ (−R, R). Then Sm is said to be uniformly convergent on E = [a, b] if for each  > 0, there exists K = K() ∈ N such that

if m ≥ K then |

∞

X

n=m+1

a_nxⁿ
| = |

∞

X

n=0

a_nxⁿ
−

m

X

n=0

a_nxⁿ
| <  ∀ x ∈ E = [a, b].

(b) Cauchy Criterion S_m =

m

X

n=0

a_nxⁿ
converges uniformly to f (x) =

∞

X

n=0

a_nxⁿ
on E = [a, b] if and only if for each  > 0, there exists K = K() ∈ N such that

if m, n ≥ K then |S_m(x) − S_n(x)| <  ∀x ∈ E = [a, b].

(c) Let R be the radius of convergence of

∞

X

n=1

a_nxⁿ
and let K = [a, b] be a bounded, closed subset of the interval of convergence (−R, R). Then the power series converges uniformly on K.

(d) Continuity Theorem Let E = [a, b] ⊂ (−R, R) and let S_m =

m

X

n=0

a_nxⁿ
for each m =

0, 1, 2, . . . . If S_m converges uniformly to f (x) =

∞

X

n=0

a_nxⁿ
on E then f is continuous on E = [a, b].

Proof: For each  > 0, since S_m converges uniformly to f (x) =

∞

X

n=0

a_nxⁿ
on E = [a, b], there exists K ∈ N such that

if m ≥ K then |f (x) − S_m(x)| = |

∞

X

n=0

a_nxⁿ
−

m

X

n=0

a_nxⁿ
| <  ∀ x ∈ E = [a, b].

On the other hand, since S_K is continuous on E = [a, b], S_K is uniformly continuous on E = [a, b]. With the given  > 0,there exists δ_K > 0 such that

if x, y ∈ E and |x − y| < δ_K then |S_K(x) − S_K(y)|| < .

This implies that if x, y ∈ E and |x − y| < δ_K then

|f (x) − f (y)| ≤ |f (x) − S_K(x)| + |S_K(x) − S_K(y)| + |S_K(y) − f (y)| < 3.

Since  > 0 is arbitrary, this proves that f is (uniformly) continuous on E = [a, b].

(e) Differentiation Theorem: A power series can be differentiated term-by-term within the interval of convergence. In fact, if

f (x) =

∞

X

n=1

anxⁿ
, then f⁰(x) =

∞

X

n=1

nanxⁿ⁻¹
.

Both series have the same radius of convergence.

Proof: Since lim

n→∞ n
1/n

= 1, the sequence |na_n|^1/n
is bounded if and only if the sequence

|a_n|^1/n
is bounded. Moreover, it is easily seen that

n→∞lim |nan|^1/n
= lim

n→∞ |an|^1/n
.

Therefore, the radius of convergence of the two series is the same, so S_m⁰ =

m

X

n=1

na_nxⁿ⁻¹
is uniformly convergent on each bounded closed subset E = [a, b] ⊂ (−R, R).

Thus,for each  > 0, there exists K = K() ∈ N such that

if m, n ≥ K then |S_m⁰ (x) − S_n⁰(x)| <  ∀x ∈ E = [a, b], .

and, by the Mean Value Theorem, there exists z, depending on m, n ≥ K, such that {S_m(x) − S_n(x)} − {S_m(c) − S_n(c)} = (x − c){S_m⁰ (z) − S_n⁰(z)}

which implies that    

S_m(x) − S_m(c)

x − c − S_n(x) − S_n(c) x − c

= |S_m⁰ (z) − S_n⁰(z)| < .

Taking the limit with respect to m, we have 

f (x) − f (c)

x − c − S_n(x) − S_n(c) x − c

= lim

m→∞

S_m(x) − S_m(c)

x − c − S_n(x) − S_n(c) x − c

≤ .

Also, for each x 6= c ∈ [a, b], since S_K =

K

X

n=1

a_nxⁿ
is differentiable on E = [a, b] ⊂ (−R, R), there exists δ_K() > 0 such that

if 0 < |x − c| < δ_K() then    

S_K(x) − S_K(c)

x − c − S_K⁰ (c)    

< .

Therefore, it follows that if 0 < |x − c| < δ_K() and n ≥ K, then 

f (x) − f (c)

x − c − S_K⁰ (c)    

=    

f (x) − f (c)

x − c − S_n(x) − S_n(c)

x − c +S_n(x) − S_n(c)

x − c − S_K(x) − S_K(c) x − c + S_K(x) − S_K(c)

x − c − S_K⁰ (c)    

≤    

f (x) − f (c)

x − c − S_n(x) − S_n(c) x − c

+    

S_n(x) − S_n(c)

x − c − S_K(x) − S_K(c) x − c

    +

S_K(x) − S_K(c)

x − c − S_K⁰ (c)    

< 3.

This shows that f⁰(c) exists and equals lim

K→∞S_K⁰ (c) =

∞

X

n=1

na_ncⁿ⁻¹
for all c ∈ E = [a, b].

(f) Integration Theorem Let E = [a, b] ⊂ (−R, R) and let S_m =

m

X

n=0

a_nxⁿ
for each m =

0, 1, 2, . . . . If S_m converges uniformly to f (x) =

∞

X

n=0

a_nxⁿ
on E then

Z b a

f (x) dx =

∞

X

n=0

Z b a

a_nxⁿ
dx,

i.e.

Z b a

m→∞lim S_m(x) dx = lim

m→∞

Z b a

S_m(x) dx.

Proof: For each  > 0, since S_m converges uniformly to f (x) =

∞

X

n=0

a_nxⁿ
on E = [a, b], there exists K ∈ N such that

if m ≥ K then |f (x) − S_m(x)| < 

b − a ∀ x ∈ E = [a, b].

For any partition P = {a = x₀ < x₁ < · · · < x_n= b} of [a, b], since

|

n

X

i=1

f (x^∗_i) − S_m(x^∗_i)
∆x_i| ≤

n

X

i=1

|f (x^∗_i) − S_m(x^∗_i)| ∆x_i <  b − a

n

X

i=1

∆x_i = ,

by letting max

1≤i≤n∆x_i → 0, we get

| Z b

a

∞

X

n=0

a_nxⁿ
dx − Z b

a

S_m(x) dx| = | Z b

a

f (x) dx − Z b

a

S_m(x) dx| ≤  ∀ m ≥ K.

Since  > 0 is arbitrary, this proves that Z b

a

∞

X

n=0

a_nxⁿ
dx = Z b

a

f (x) dx = lim

m→∞

Z b a

S_m(x) dx =

∞

X

n=0

Z b a

a_nxⁿ
dx.

Taylor’s Theorem

(a) Suppose that n ∈ N, that f and its derivatives f⁰, f⁰⁰, . . . , f⁽ⁿ⁻¹⁾ are defined and continuous on [a, b], and that f⁽ⁿ⁾ exists in (a, b). If α, β ∈ [a, b], then there exists a number γ between α and β such that

f (β) = f (α) + f⁰(α)

1! (β − α) + f⁰⁰(α)

2! (β − α)²+ · · · + f⁽ⁿ⁻¹⁾(α)

(n − 1)! (β − α)ⁿ⁻¹+ f⁽ⁿ⁾(γ)

n! (β − α)ⁿ

=

n−1

X

k=0

f^(k)(α)

k! (β − α)^k+f⁽ⁿ⁾(γ)

n! (β − α)ⁿ

=

n−1

X

k=0

f^(k)(α)

k! (β − α)^k+R_n(β).

Proof: Let P be the real number defined by the relation (β − α)ⁿ

n! P = f (β) −

n−1

X

k=0

f^(k)(α)(β − α)^k

k! ,

and consider the function g : [a, b] → R defined by

g(x) = f (β) −

n−1

X

k=0

f^(k)(x)(β − x)^k

k! − P

n!(β − x)ⁿ ∀ x ∈ [a, b].

Then g is differentiable on [a, b] with

g⁰(x) = −

n−1

X

k=0

d dx

f^(k)(x)(β − x)^k

k!
− d

dx P

(n − 1)!(β − x)ⁿ⁻¹

= −

n−1

X

k=0

f^(k+1)(x)(β − x)^k

k! +

n−1

X

k=1

f^(k)(x)(β − x)^k−1

(k − 1)! + P

(n − 1)!(β − x)ⁿ⁻¹

= −

n−1

X

k=0

f^(k+1)(x)(β − x)^k

k! +

n−2

X

k=0

f^(k+1)(x)(β − x)^k

k! + P

(n − 1)!(β − x)ⁿ⁻¹

= −f⁽ⁿ⁾(x)(β − x)ⁿ⁻¹

(n − 1)! + P

(n − 1)!(β − x)ⁿ⁻¹

= P − f⁽ⁿ⁾(x)

(n − 1)! (β − x)ⁿ⁻¹.

Clearly, g is continuous on [a, b] and is differentiable on (a, b). Also it is evident that g(β) = 0 and it follows from the definition of P that g(α) = 0. By the Rolle’s Theorem, there exists a point γ between α and β such that g⁰(γ) = 0.

Since g⁰(γ) = 0, then P = f⁽ⁿ⁾(γ) and this proves that

f (β) =

n−1

X

k=0

f^(k)(α)

k! (β − α)^k+ f⁽ⁿ⁾(γ)

n! (β − α)ⁿ=

n−1

X

k=0

f^(k)(α)

k! (β − α)^k+ R_n(β).

(b) Integral Form for the Remainder Suppose that f and its derivatives f⁰, f⁰⁰, . . . , f⁽ⁿ⁾ are continuous on [a, b] to R. Then

f (b) = f (a) + f⁰(a)

1! (b − a) + f⁰⁰(a)

2! (b − a)²+ · · · + f⁽ⁿ⁻¹⁾(a)

(n − 1)! (b − a)ⁿ⁻¹+ R_n(b), where the remainder is given by

R_n(b) = 1 (n − 1)!

Z b a

(b − t)ⁿ⁻¹f⁽ⁿ⁾(t) dt.

Proof: Integrate R_n by parts to obtain R_n(b) = 1

(n − 1)!



(b − t)ⁿ⁻¹f⁽ⁿ⁻¹⁾(t)|^t=b_t=a+ (n − 1) Z b

a

(b − t)ⁿ⁻²f⁽ⁿ⁻¹⁾(t) dt



= −f⁽ⁿ⁻¹⁾(a)

(n − 1)! (b − a)ⁿ⁻¹+ 1 (n − 2)!

Z b a

(b − t)ⁿ⁻²f⁽ⁿ⁻¹⁾(t) dt

= −f⁽ⁿ⁻¹⁾(a)

(n − 1)! (b − a)ⁿ⁻¹−f⁽ⁿ⁻²⁾(a)

(n − 2)!(b − a)ⁿ⁻²+ 1 (n − 3)!

Z b a

(b − t)ⁿ⁻³f⁽ⁿ⁻²⁾(t) dt

· · · ·

= −f⁽ⁿ⁻¹⁾(a)

(n − 1)! (b − a)ⁿ⁻¹−f⁽ⁿ⁻²⁾(a)

(n − 2)!(b − a)ⁿ⁻²− · · · − f⁰(a)

1! (b − a) + Z b

a

f⁰(t) dt

= −

n−1

X

k=0

f^(k)(a)

k! (b − a)^k+ f (b).

This proves the stated formula

f (b) =

n−1

X

k=0

f^(k)(a)

k! (b − a)^k+ R_n(b).

Examples:

(a) Let f (x) = sin x for x ∈ (−∞, ∞).

For each n = 0, 1, 2, . . . , since

f⁽⁴ⁿ⁾(x) = sin x, f⁽⁴ⁿ⁺¹⁾(x) = cos x, f⁽⁴ⁿ⁺²⁾(x) = − sin x, f⁽⁴ⁿ⁺³⁾(x) = − cos x, thus, for all n and for all t,

|f⁽ⁿ⁾(t)| ≤ 1.

It follows that

|R_n(x)| ≤ |x|ⁿ

n! → 0 for all − ∞ < x < ∞ as n → ∞,

i.e. Maclaurin series (Taylor series at a = 0) of sin x converges for −∞ < x < ∞.

Also since

f⁽²ⁿ⁾(0) = 0 and f⁽²ⁿ⁺¹⁾(0) = (−1)ⁿ for n = 0, 1, 2, . . . , we have

sin x =

∞

X

n=0

(−1)ⁿ

(2n + 1)!x²ⁿ⁺¹ = x −x³ 3! + x⁵

5! − · · · for all − ∞ < x < ∞.

(b) Let f (x) = ln(1 + x) for x ∈ (−1, ∞).

Since

f⁰(x) = 1

1 + x, f⁰⁰(x) = − 1

(1 + x)², f⁰⁰⁰(x) = 2

(1 + x)³, f⁽⁴⁾(x) = − 3!

(1 + x)⁴, · · · we have

f⁽ⁿ⁾(x) = (−1)ⁿ⁻¹(n − 1)!

(1 + x)ⁿ for n ≥ 1, and

R_n(x) = 1 (n − 1)!

Z x 0

(x − t)ⁿ⁻¹f⁽ⁿ⁾(t) dt

= 1

(n − 1)!

Z x 0

(x − t)ⁿ⁻¹ (−1)ⁿ⁻¹(n − 1)!

(1 + t)ⁿdt

= (−1)ⁿ⁻¹ Z x

0

(x − t)ⁿ⁻¹ (1 + t)ⁿ dt.

For 0 ≤ x ≤ 1 and for 0 ≤ t ≤ x, we have

1 + t = |t − (−1)| ≥ 1 =⇒ 1

(1 + t)ⁿ ≤ 1 ∀ t ∈ [0, x]

and hence

|R_n(x)| = Z x

0

(x − t)ⁿ⁻¹ (1 + t)ⁿ dt ≤

Z x 0

(x − t)ⁿ⁻¹ dt = xⁿ

n → 0 as n → ∞.

For −1 < x < 0 and for x ≤ t ≤ 0, we have

1 + t = |t − (−1)| ≥ |x − t| = t − x =⇒ |x − t|ⁿ⁻¹

(1 + t)ⁿ⁻¹ ≤ 1 ∀ t ∈ [x, 0]

and hence

|Rn(x)| =    

Z x 0

(x − t)ⁿ⁻¹ (1 + t)ⁿ dt

= Z 0

x

(t − x)ⁿ⁻¹ (1 + t)ⁿ⁻¹

1 1 + t dt.

By the Mean Value Theorem for Integrals [i.e. If f is continuous on [a, b], then there is at least one number c in (a, b) for which

Z b a

f (x)dx = f (c)(b − a)] and since (t − x)ⁿ⁻¹ (1 + t)ⁿ⁻¹

1 1 + t is continuous on [x, 0], there exists x_n ∈ (x, 0) for each n ∈ N such that

Z 0 x

(t − x)ⁿ⁻¹ (1 + t)ⁿ⁻¹

1

1 + t dt = (x_n− x)ⁿ⁻¹ (1 + xn)ⁿ⁻¹

1 1 + xn

(0 − x).

Since −x = |x| and 0 < 1 + x < 1 + x_n, we conclude that

|R_n(x)| < x_n− x 1 + xn

n−1  |x|

1 + x

 . Since |x| < 1 and x_n < 0, we have

x_n< |x|x_n, 0 < x_n+ |x| < |x|x_n+ |x| = |x| 1 + x_n
, and thus

x_n+ |x|

1 + x_n < |x| =⇒ x_n− x 1 + x_n < |x|.

It follows that

|R_n(x)| < |x|ⁿ⁻¹ |x|

1 + x
→ 0 as n → ∞ since |x| < 1,

i.e.Maclaurin series (Taylor series at a = 0) of ln(1 + x) converges for −1 < x ≤ 1.

Hence

ln(1 + x) =

∞

X

n=1

(−1)ⁿ⁻¹

n xⁿ= x − x² 2 +x³

3 − · · · for − 1 < x ≤ 1.