MAT 5610, Analysis I Part III

Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series

Wm C Bauldry

[email protected]

Spring, 2011

MAT 5610: 0 0 Today Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series

MAT 5610: 0 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Outerlude

MAT 5610: 19 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Seriesly Fun

Deﬁnition • A series is a sequence of partial sums with the nth term given by Pn Sn = ak for some ﬁxed p Z k=p ∈ Pn • A sequence bn n p can be written as a series Sn = k p ak by { } ≥ = setting an = bn bn 1 and bp 1 = 0. − − − Examples

X∞ 1 π2 X∞ ( 1)k+1 1. = 4. − = ln(2) k2 6 k k=1 k=1

X∞ 1 k 2. = e X∞ ( 1) +1 sin(k) k! 5. − = k=0 k ∞ ∞ k=1 X 1 X 1 3. =1= sin(1) k(k + 1) k(k − 1) arctan k=1 k=2 1 + cos(1)

MAT 5610: 20 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Harmonious Series

RECALL: Deﬁnition P A series ∞ ak converges to A R iff for every ε > 0 there is an k=p ∈ n∗ N s.t. whenever n n∗, then Sn A < ε. ∈ ≥ | − | Proposition P • If ak converges, then ak 0. P → • If ak 0, then ak diverges. 6→ Example (Nicole Oresme (c.1360)) The harmonic series P 1/k diverges.

Take segments that are greater than 1/2 to show S n 1 + n/2. 2 ≥

MAT 5610: 21 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series A Critical Series

Proposition The geometric series P a rk converges iff the constant r < 1, then | | X∞ a a rk = . 1 r k=0 −

Proof. 1. If r 1, then a rk 0. Whence P a rk diverges. | | ≥ 6→ n+1 2. Let r < 1 and n > 1. Then Sn rSn = a a r . Hence | | − − a(1 rn+1) Sn = − 1 r − n 3. Since r < 1, then r 0. Thence Sn converges. | | → What about P sin(k)k? MAT 5610: 22 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series A Geometric Sample

Example Write 0.1011 as a rational number.

4 42 43 1. 0.1011 = 1011 10− + 1011 10− + 1011 10− + · · · ··· X∞ 4k 2. = 1011 10− · k=1 ! X∞ 4k 3. = 1011 1 + 10− − k=0 1 4. = 1011 1 + − 1 10 4 − − 10000 1 337 5. = 1011 1 + = 1011 = − 9999 9999 3333

MAT 5610: 23 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series A Maple Sample

Maple Summing

X∞ ( 1)k+1 sin(k) sin(1) How does Maple ﬁnd that − = arctan ? k 1 + cos(1) k=1

1. First, enter > infolevel[sum]:=5: to raise the level of “user information” printed.

2. Now enter the summation X∞ ( 1)k+1 > − sin(k) k k=1 and wait for the results.1

1PlanB. MAT 5610: 24 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Cauchy and Friends

Theorem (Cauchy Criterion for Series) P ak converges iff for each ε > 0 there is an n∗ N s.t. whenever ∈ n m n∗, then ≥ ≥ Sn Sm 1 = am + am+1 + am+2 + + an < ε | − − | | ··· | Proof. X (Series are sequences; cf. Cauchy’s Criterion for Sequences.)

Corollary P P 1. If ak = bk for k k∗, and bk converges, then ak converges. P ≥ P 2. If ak converges, then ak converges. P | | P 3. If ak diverges, then ak diverges | |

MAT 5610: 25 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, I

Theorem (The Integral Test) If f :[1, ) R is continuous, decreasing, and nonnegative, then ∞ → Z X∞ ∞ f(k) converges iff f(x) dx converges k=1 1 Proof.

Let ak = f(k). Since f is decr and cont, ak f(x) ak for k 1. +1 ≤ ≤ ≥ Whence Z k+1 ak+1 f(x) dx ak ≤ k ≤ Therefore Z k+1 Sk+1 a1 f(x) dx Sk − ≤ 1 ≤ The series bounds the integral, the integral bounds the series. Since all terms are nonneg, Bolzano-Weierstrass Thm gives the result.

MAT 5610: 26 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Integral Test Figure

∞ ∞ X Z ∞ Z ∞ X ak ≤ f(x) dx f(x) dx ≤ ak k=2 1 1 k=1 Z Z ∞ X∞ ∞ X∞ So f(x) dx ak f(x) dx ak ≤ ≤ ≤ 2 k=2 1 k=1 MAT 5610: 27 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, II

Theorem (The Comparison Test) P P Suppose ak and bk are two series with bk 0 for all k. ≥ + P 1. If ak Mbk eventually (for some M R ) and bk converges, | |P ≤ P ∈ then ak and ak both converge. | | + P 2. If mbk ak eventually (for some m R ) and bk diverges, P≤ | | 1 ∈ then ak diverges. | | Proof. P 1. Wolog ak Mbk for all k. Since bk converges, so does P | | ≤ Mbk, say to S. Then N N X X X∞ 0 ak Mbk Mbk = S ≤ | | ≤ ≤ k=1 k=1 k=1 P Whence ak is a bounded, increasing series, thus convergent. | | 1 P k+1 ak may or may not diverge; e.g., ak = (−1) /k and bk = 1/k. MAT 5610: 28 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series A Useful Series Set

Proposition (p-Series1) ∞ X 1 The real series ζ(p) = converges iff p > 1. kp k=1

Proof. X (Apply the integral test.) Examples ∞ ∞ X X 4 4 1. 1/k = ∞ 4. 1/k = π /90 k=1 k=1 ∞ ∞ X 2 2 X 5 2. 1/k = π /6 5. 1/k = ζ(5) k=1 k=1 ∞ ∞ X 3 X 6 6 3. 1/k = ζ(3) 6. 1/k = π /945 k=1 k=1

1See the Riemann Zeta function. Also $ MAT 5610: 29 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, III

Theorem (The Limit Comparison Test) P P Suppose ak and bk are two positive-term series. an P P 1. If lim is ﬁnite and bn converges, then an converges. n→∞ bn bn P P 2. If lim is ﬁnite and bn diverges, then an diverges. n→∞ an Proof.

Set L = lim an/bn. + 1. If L = 0, then ∃M ∈ R s.t. an ≤ Mbn. Now compare. 1 3 2. If L > 0, then 2 L bn ≤ an ≤ 2 L bn eventually. Again compare.

Corollary P P an Suppose ak & bk are positive-term series, and lim = L with n→∞ b P P n 0 < L < ∞. Then ak & bk converge or diverge together.

MAT 5610: 30 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Fun Time

Exercises ∞ ∞ √ ∞ X (−1)kk X ( 5 − 1)m X (−1)nn 1. 5. 9. (k + 1) ek m2 + 1 n2 + 1 k=1 m=1 i=1

∞ ∞ ∞ X 102n X k! X (−1)mm3 2. 6. 10. (2n − 1)! kk (m2 + 1)4/3 n=1 k=1 m=1

∞ j ∞ n−1 1 1·4 1·4·7 X 3 X (−1) 11. 3 + 3·6 + 3·6·9 +··· 3. 7. j3 n2 + 1 2 2·5 2·5·8 j=1 n=1 12. 9 + 9·12 + 9·12·15 + ∞ ∞ j ∞ X (−1)i 23i X (−1) X k + 2 4. 8. 13. √ 32i j ln(j) i=1 j=2 k=1 (k+1) k+3

∞ X k a − (a − b)r Challenge: (a + kb)r = ; |r| < 1 (1 − r)2 k=1

(cacdcadcac2c2cddcd) MAT 5610: 31 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, IV

Theorem (Ratio Comparison Test) P P Let an & bn be positive term series. If an+1/an ≤ bn+1/bn eventually, P P and bn converges, then an converges.

Theorem (D’Alembert’s Ratio Test) P Let an be a series with positive terms. If there is a constant α ∈ [0, 1) s.t. P an+1/an ≤ α eventually, then an converges. If an+1/an ≥ 1 eventually, P then an diverges.

Corollary (Cauchy’s Ratio Test) P an+1 Let an be a series with positive terms. Set r = limn→∞ . Then: an • If r < 1, the series converges. • If r > 1, the series diverges. • If r = 1, the test fails.

MAT 5610: 32 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Proofs

Proof (Ratio Comparison Test).

1. an/bn is decreasing eventually, thus bounded. Thus an Mbn. P P ≤ 2. Since bn converges, then an converges by comparison.

Proof (D’Alembert’s Ratio Test). 1. α < 1 gives P αn converges. n+1 n P 2. Then an /an α /α = α implies that an converges. +1 ≤ Proof (Cauchy’s Ratio Test). X

MAT 5610: 33 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, V

Theorem (Root Test) √ Let P a be a series with positive terms. If ∃α ∈ [0, 1) s.t. n a ≤ α n √ n P n P eventually, then an converges. If an > 1 eventually, then an diverges.

Theorem (Cauchy’s Root Test) √ P n Let an be a series with positive terms. Set ρ = lim an. Then: n→∞ • If ρ < 1, the series converges. • If ρ > 1, the series diverges. • If ρ = 1, the test fails.

Proof. √ n n 1. an ≤ ρ < 1 implies an ≤ ρ < 1. P n P 2. Since ρ < 1, then ρ converges. By comparison, an converges.

MAT 5610: 34 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, VI

Theorem (Raabe’s Test1) h i P an+1 an+1 Let an be a series with lim = 1. Set ρ = lim n 1 . n an n an →∞ →∞ − • If ρ < 1, the series diverges. • If ρ > 1, the series converges absolutely. • If ρ = 1, the test fails. Example

1 X∞ 1 3 (2n 3) Consider the series S = 1 · ··· − − 2 − 2n n! k=2 an+1 2n 1 1. = − an 2n + 2 an+1 3n 3 2. n 1 = − an 2n + 2 → 2

1Raabe’s Test (1832) is a special case of Kummer’s Test (1835) MAT 5610: 35 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Challenge Prize Problem

Ready. Set. Go. . . What conditions are needed on α, β, and γ (in addition to γ = 0, 1, 2,... ) so that the hypergeometric function 6 − − k X∞ (α)k(β)k z 2F1(α, β; γ; z) = (γ)k · k! k=0

X∞ α(α + 1) (α + k 1) β(β + 1) (β + k 1) zk = ··· − · ··· − γ(γ + 1) (γ + k 1) · k! k=0 ··· − converges at z = 1. (See Gauss’s Hypergeometric Theorem.)

MAT 5610: 36 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Converge? Absolutely!

Deﬁnition P k P k+1 • A series of the form ( 1) ak or ( 1) ak with ak 0 is an alternating series. − − ≥ P P • If ak converges, then ak converges absolutely. P | | P P • If ak diverges but ak converges, then ak converges conditionally| | .

Theorem (Leibniz’s Alternating Series Test)

Let an 0 for all n. If 1. an is eventually decreasing, and 2. an 0, P≥ k+1 → then ( 1) ak converges. (N.B. S Sn < an .) − | − | +1 Examples

k+1 k+1 P 1 P ( 1) P 1 π2 P ( 1) π2 1. = ; − = ln(2) 2. 2 = ; − 2 = k ∞ k k 6 k 12

MAT 5610: 37 0 Fun Time Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Rearranging the Terms

Theorem P P P • If ak converges to S absolutely and bk is a rearrangement of ak, P then bk converges to S absolutely. P • If ak converges to S conditionally and α ∈ [−∞, +∞], then there is a P rearrangement bk that converges to α. P • If ak converges to S conditionally and α, β ∈ R, then there is a P rearrangement bk that oscillates between α and β.

Example ∞ X (−1)k+1 = ln(2), conditionally convergent. Choose α ∈ . k R k=1

1. Add positive terms until 1 + 1/3 + ··· + 1/(2n1 − 1) ﬁrst exceeds α

2. Add negative terms until 1 + ··· + 1/(2n1 − 1) − 1/2 − · · · − 1/(2n2) ﬁrst is less than α. 3. Rinse and repeat.

MAT 5610: 38 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Outerlude

MAT 5610: 39 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Functioning Sequences & Series

Deﬁnition (Pointwise Convergence)

A sequence of fcns fn converges pointwise to f on D R iff for { } ⊆ each xo D, the sequence fn(x ) f(x ). In symbols: ∈ 0 → 0 h i fn f ( ε>0)( x0 D)( n∗ N)( n>n∗) fn(x0) f(x0) < ε → ⇐⇒ ∀ ∀ ∈ ∃ ∈ ∀ | − |

Examples ( n 0 x < 1 n 1. fn(x) = x f(x) = | | on D = ( 1, 1]. (( 1) div.) → 1 x = 1 − − 1 2. gn(x) = cos(nx) 0 on D = R n →

3. hn(x) = sin(nx) 0 on D = kπ k Z and div otherwise. → { | ∈ } 2 4. jn(x) = arctan(nx) signum(x) on D = R π →

MAT 5610: 40 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Problems

Examples n 2 Continuity: Both x and π arctan(nx) are continuous fcns that have discontinuous limits

lim lim fn(x) = lim lim fn(x) n x a x a n →∞ → 6 → →∞

1 Differentiability: fn(x) = n sin(nx) 0, but fn0 (x) = cos(nx) diverges for all x = 0 → 6 d d lim fn(x) = lim fn(x) n n dx →∞ 6 →∞ dx

Integrability: nxn 0 on [0, 1), but R 1 nxn dx 1 → 0 → Z b Z b lim fn(x) dx = lim fn(x) dx n n a →∞ 6 →∞ a

MAT 5610: 41 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series

Interlude: L2

Deﬁnition (L2-norm)

For f, deﬁne the mean-square or L2-norm on [a, b] to be Z b 1/2 2 kfk2 = |f(x)| dx a

Deﬁnition (Mean-Square Convergence)

A seq {fn} converges to f in the mean on [a, b] iff lim kfn − fk2 = 0. n→∞

Examples q √ n n R 1 2n 1. Let fn(x)= x on D =[0, 1]. Since kx k2 = 0 x dx =1/ 2n+1→0, thence xn → 0 in the mean on [0, 1]. 2. Theorem. Riesz-Fischer (1907). The Fourier series of an L2 fcn f converges to f in the mean.

MAT 5610: 42 1 Fourier Squares Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Uniform Convergence

Deﬁnition (Uniform Convergence)

A sequence of fcns fn converges uniformly to f on D R iff for { } ⊆ every ε > 0 there is an n∗ = n∗(ε) N such that for any x D, if ∈ ∈ n > n∗, then fn(x) f(x) < ε. | − | Examples xn 1. fn(x) = 0 uniformly on [0, 1] 1 + nxn → 1 2. gn(x) = cos(nx) 0 uniformly on R √n → 1 1 3. hn(x) = 0 uniformly on R (but doesn’t) n(x2 + 1) → nx2 + 1 ( n 0 x < 1 4. jn(x) = x f(x) = | | pw, not uniformly, on [0, 1]. → 1 x = 1

MAT 5610: 43 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Two Graphs

Uniform Convergence Nonuniform Convergence

MAT 5610: 44 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series How To...

Proposition x f (x) = converges uniformly to f(x) = 0 on [0, ∞). n nx + 1

Proof.

1. If x = 0, then fn(0) ≡ 0. Hence |fn(0) − 0| < ε for any ε > 0.

x x 1 2. If x > 0, then |fn(x) − 0| = < = < ε for n sufﬁc large. nx + 1 nx n

Proposition 2 −1 gn(x)→0 pw, not unif on [0, 1], where gn(x) = 4n x for 0 ≤ x < (2n) , and −1 −1 gn(x) = 4n(1−nx) for (2n) ≤x≤n , and 0 otherwise.

Proof.

Consider gn(1/(2n)) = 2n 6< ε for n sufﬁciently large.

MAT 5610: 45 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Outerlude

MAT 5610: 46 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Props

Query We know that a function continuous on a compact set is uniformly continuous. Does a sequence converging on a compact set converge uniformly?

Theorem (Cauchy Criterion for Sequences)

A sequence of fcns fn converges uniformly on D R iff for each { } ⊆ ε > 0 there is an n∗ s.t. for all x D, if n, m n∗, then ∈ ≥ fn(x) fm(x) < ε. | − | Proof.

1. Build an f by considering fn(x) for each x D ∈ 2. shows fn f uniformly 4 →

MAT 5610: 47 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Interlude: L ∞

Deﬁnition (L∞-norm)

For f, deﬁne the sup- or L∞-norm on [a, b] to be

kfk∞ = sup |f(x)| x∈[a,b]

x 2 =1 x =1 x =1 ∞ 1

Theorem

A sequence fn converges to f in the sup norm on [a, b]; i.e., kfn − fk∞ → 0 iff fn converges to f uniformly on [a, b].

Proof.

Observe that |fn(x) − f(x)| < ε for all x ∈ [a, b] iff kfn − fk∞ < ε.

MAT 5610: 48 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Interlude: L Examples ∞ Example x Consider fn(x) = .(fn 0 pw on [ 1, 1]) 1 + n2x2 → −

L : fn 0 in L [ 1, 1]. ∞ → ∞ − 2 2 0 1 − n x 0 • fn(x) = =⇒ fn(x) = 0 at x = ±1/n. Thus (1 + n2x2)2 −1 1 ≤ f (x) ≤ 2n n 2n

• 1 So kfn − 0k∞ = 2n → 0 on [−1, 1] (fn → 0 unif on [−1, 1])

L2: fn 0 in L2[ 1, 1]. → − s q arctan(n) 1 • R 1 |f (x)|2dx = − → 0. −1 n n3 n2(n2 + 1)

• Thus kfn − 0k2 → 0 on [−1, 1]

MAT 5610: 49 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Interlude II: Extra for Experts

Theorem (Dini’s Theorem (1878))

Suppose fn(x) is a sequence of functions converging to f on { } D R. If ⊆ 1. each fn is continuous on D, 2. f is continuous on D,

3. fn f monotonically, and → 4. D is compact (on R: compact = closed and bounded), then fn converges to f uniformly on D.

1. fn(x) = 1 on 0 < x < 1/n and 0 otherwise converges to 0 on D = [0, 1] (fn is not continuous) 2. xn U(x 1) on [0, 1] (f = U is not continuous) → − 3. “Marching triangles” 0 on [0, 1] (convergence is not monotonic) → 4. xn 0 on [0, 1) ([0, 1) is not compact) → MAT 5610: 50 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Props to UniCon

Theorem

If a sequence of continuous functions fn converges uniformly on D, the limit function f is continuous on D.{ }

Proof. The proof hinges on the inequalities

f(x) f(x ) f(x) fn (x) + fn (x) fn (x ) + fn (x ) f(x ) | − 0 | ≤ | − 1 | | 1 − 1 0 | | 1 0 − 0 | | {z ∗ } | {z } | {z ∗ } n1>n x x0 <δ n1>n | − | ε ε ε < + + = ε 3 3 3

for some n > n∗ and all x B(x , δ). 1 ∈ 0

Converse? No — the “marching triangles”.

MAT 5610: 51 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series More Props to UniCon

Theorem

If a sequence of continuous functions fn converges uniformly to f on D = [a, b], then { } " # Z b Z b h i lim fn(x) dx = lim fn(x) dx. n n →∞ a a →∞ R b R b I.e., if Fn = fn and F = f, then Fn F. a a → Proof. The proof hinges on the inequalities

Z b Z b Z b

fn(x) dx f(x) dx = [fn(x) f(x)] dx a − a a − Z b Z b fn(x) f(x) dx < ε dx = (ε) ≤ a | − | a O

MAT 5610: 52 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Yet Even More Props to UniCon

Theorem 1 Suppose fn is a sequence of continuously differentiable functions { } converging pointwise to f on [a, b]. If the sequence of derivatives fn0 converges uniformly on [a, b], then f is continuously differentiable{ and}

• fn f uniformly on [a, b], and → • f 0 f 0 uniformly on [a, b]. n → Proof. The proof is based on three ideas: R x 1. f 0 (t) dt = fn(x) fn(a) a n − R x R xh i 2. f(x) f(a)= lim [fn(x) fn(a)] = lim fn0 (t) = lim fn0 (t) n n a a n − →∞ − →∞ →∞ d R x 3. f 0(x) = dx a f 0(t) dt

1‘Continuously’ can be relaxed. ‘Pointwise’ can be relaxed to a single point. MAT 5610: 53 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series A Sample Case

Exercises

npn o 1 1 1. sin(x) converges uniformly on [, 2 π] for > 0, but pw on (0, 2 π]. a Z 2 2. Determine lim e−nx dx where a > 0. n→∞ −a

−nx2 3. Set Sn(x) = nxe . What does Sn converge to? Show Z a Z a lim Sn(x) dx 6= lim Sn(x) dx n→∞ 0 0 n→∞

x 4. Let f (x) = for |x| ≤ 1. n 1 − nx2

4.1 What is f(x) = lim fn(x)? n→∞ 4.2 Is the convergence uniform?

0 0 4.3 Is f (x) = lim fn(x)? Explain. n→∞

MAT 5610: 54 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Series Deﬁnition

n P Function Series: The sequence of partial sums Sn(x) = fk(x) k=1

Series Conv PW: The series converges pointwise on D if {Sn} converges pointwise on D

Series Conv Unif: The series converges uniformly on D if {Sn} converges uniformly on D

Series Conv Abs: The series converges absolutely on D if {Sn} converges absolutely on D Series Representation: A function f has a series representation on D iff ∃Sn → f on D

Theorem P If a series k fk(x) of continuous functions converges uniformly to S(x), then S is continuous.

MAT 5610: 55 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Samples

Exercise X∞ x2 1. S(x) = (1 + x2)k k=1 • S(0) = 0 1 • 0 < < 1 for x 6= 0 (1 + x2) ∞ • Set a = x2 and r = 1/(1 + x2), then use P ark = a/(1 − r). k=0 Whence for x 6= 0, ∞ 2 2 X x x 2 S(x) = = − x = 1 (1 + x2)k 1 − 1 k=1 1+x2 ( 1 x 6= 0 • Thus S(x) = . 0 x = 0

• Since S is not continuous, but each fn is, the convergence is pointwise and absolutely.

MAT 5610: 56 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Modes

Pointwise convergence

Cesàro Absolute convergence convergence Uniform convergence

n P Cesaro` Convergence: Let Sn(x) = fn(x) be a series. Then k=0 Cesaro` 1 Sn S(x) iff (S + S + + Sn) S −→ n + 1 0 1 ··· → P∞ n Cesaro` 1 E.g., ( 1) 2 n=0 − −→ MAT 5610: 57 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Two Big Theorems

Theorem (Cauchy Criterion) P The series S(x) = fn(x) converges uniformly on D iff for any ε > 0 there is an n∗ N s.t. whenever m, n n∗, then for all x D, ∈ ≥ ∈ n X Sn(x) Sm(x) = fn(x) < ε | − | k=m+1

Theorem (Weierstrass M-Test) P Suppose that fn(x) Mn for all x D and for each n. If Mn | P | ≤ ∈ converges, then fn(x) converges uniformly and absolutely on D.

Example

X∞ sin(k4x) What can be said about ( 1)k+1 ? − k2 k=1

MAT 5610: 58 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series The Example

1.2 1.2

0.8 0.8

0.4 0.4

-5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5 -0.4 -0.4

-0.8 -0.8

-1.2 -1.2

n = 25

16 16

8 8

-5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5

-8 -8

-16 -16

d X∞ sin(k4x) X∞ d sin(k4x) ( 1)k+1 = ( 1)k+1 dx − k2 6 − dx k2 k=1 k=1

MAT 5610: 59 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Integrating Series

Theorem n P If Sn(x) = fn(x) is integrable on [a, b] for each n N and Sn S uniformly, then ∈ → " n # n " # Z b X X Z b fn(x) dx = fn(x) dx a k=0 k=0 a Example

1 X∞ = ( 1)kt2k on t < 1 and uniformly on t ρ < 1 1 + t2 − | | | | ≤ k=0 x x Z 1 X∞ Z Whence dt = ( 1)k t2kdt 1 + t2 − 0 k=0 0 X∞ x2k+1 Wherefore arctan(x) = ( 1)k for x ρ < 1 − 2k + 1 | | ≤ k=0

MAT 5610: 60 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Arctan? < 1?

X∞ x2k+1 arctan(x) = ( 1)k for x ρ < 1 − 2k + 1 | | ≤ k=0 | {z } Why?

z f(z) = |arctan(z)|

y x

f(z) = arctan(z) has poles at z = i ± MAT 5610: 61 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Differentiating Series

Theorem n P Suppose Sn(x) = fn(x) is a series of continuously differentiable functions converging pointwise to S(x) on [a, b]. If the series of n P derivatives fn0 (x) converges uniformly on [a, b], then • S(x) is continuously differentiable, P • fn(x) S(x) uniformly on [a, b], and P → • f 0 (x) S0(x) uniformly on [a, b]. n → Example

X n 1 X n−1 1 x = =⇒ nx = =⇒ 1 − x (1 − x)2 X n−2 2 X n−3 3! n(n − 1)x = =⇒ n(n − 1)(n − 2)x = (1 − x)3 (1 − x)4

MAT 5610: 62 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Fun Day

Problem I. Where does II. Let a > 0. Investigate

∞ n X n(x 1) n n − d X∞ x ? X∞ d x 2n(3n 1) = n=1 − dx n2an dx n2an n=1 n=1 converge?

nx2 III. Set fn(x) = nxe− and IV. Bessel’s function is P on . n 2n+p f(x) = n fn(x) [0, 1] X∞ ( 1) (x/2) Investigate Jp(x) = − . n!(n + p)! n Z 1 Z 1 =0 ? X∞ f(x) dx = fn(x) dx Show that y = Jp(x) solves 0 n=0 0 2 2 2 x y00 + xy0 + (x p )y = 0. −

MAT 5610: 63 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Fun Day, II

Problem V. Is VI. Is X∞ 1 ( 1)n X∞ n 1 ( 1)(n+1)/2 − − ( 1)n − − · 2 − · n + 1 n=1 n=1 Cesaro-convergent?` Cesaro-convergent?`

1 1 VII. Calculate when VIII. Calculate when f(x) g(x)

X∞ X∞ xn f(x) = xn. g(x) = . n! n=0 n=0

MAT 5610: 64 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Very Powerful Series

Deﬁnition (Power Series)

X∞ k • A power series is a series of the form ak(x a) . − k=0 • The radius of convergence is the largest value R [0, ] s.t. the power series converges for all x a < R. ∈ ∞ | − | Theorem P k Suppose k ak(x a) converges for x∗ s.t. x∗ a = r0. Then P k − | − | ak(x a) converges absolutely for all x s.t. x a < r . k − | − | 0 Proof.

Based on the inequalities (x a)/r0 < 1 and | − | k k k k x a x a ak(x a) akr0 − M − − ≤ · r0 ≤ · r0

MAT 5610: 65 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series More Power

Corollary P k P k If ak(x a) diverges for x∗, then ak(x a) diverges for all x k − k − s.t. x a x∗ a . | − | ≥ | − | Theorem P k P k Let R be the radius of conv of ak(x a) . Then ak(x a) k − k − 1. converges absolutely on x a < R, | − | 2. converges uniformly on x a ρ < R, and | − | ≤ 3. diverges on x a > R. | − | P k P k • If k ak(R) conv, then k ak(x a) conv uniformly on R < ρ x a R. − − − ≤ | − | ≤ P k P k • If k ak( R) conv, then k ak(x a) conv uniformly on R x − a ρ < R. − − ≤ | − | ≤

MAT 5610: 66 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series More Power

Examples (“Try not. Do, or do not. There is no try.”)

2k+1 "r # 2k 1 X∞ k x 1 + x X∞ x − 1. sin(x) = ( 1) 4. ln = − (2k + 1)! 1 x 2k 1 k=0 − k=1 − X∞ X∞ ln(k + 1) 2. f(x) = k! xk 5. g(x) = (x 5)k √ − k=0 k=1 k + 1 k x X∞ k+1 x xe X∞ k kx 3. ln(1 + x) = ( 1) 6. = k x e− − k (ex x)2 k=1 − k=1

1 1 1 3 1 3 5 1 3 5 7 7. = 1 x + · x2 · · x3 + · · · x4 ... √1 + x − 2 2 4 − 2 4 6 2 4 6 8 − · · · · · · x 3 1 1 1 1 8. ln − = + + + + ... x 4 x 3 2(x 3)2 3(x 3)3 4(x 3)4 − − − − − MAT 5610: 67 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series More Power Solutions, I Examples 1 1 1 3 1 3 5 1 3 5 7 7. = 1 x + · x2 · · x3 + · · · x4 ... √1 + x − 2 2 4 − 2 4 6 2 4 6 8 − · · · · · · 1 · 3 · 5 1 · 2 · 3 · 4 · 5 · 6 6! 6! a. = = = (n = 3) 2 · 4 · 6 (2 · 4 · 6)2 26 · (3!)2 43 · (3!)2 1 · 3 · 5 · 7 8! 8! b. = = (n = 4) 2 · 4 · 6 · 8 28 · (4!)2 44 · (4!)2 ! (−1)n(2n)! (−1)n 2n c. a = = · n 4n(n!)2 4n n

n+1 n 2 (2(n + 1))! x 4 (n!) 2n + 1 d. rn(x) = · = |x| −→ |x| 4n+1((n + 1)!)2 (2n)! xn 2n + 2 n→∞ Whence the RoC is 1; i.e., the series converges on x < 1. | | Testing the endpoints shows the series converges at x = 1 and diverges at x = 1. (The recursive form of an is easier.) −

MAT 5610: 68 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series More Power Solutions, II

Examples x 3 1 1 1 1 8. ln − = + + + + ... x 4 x 3 2(x 3)2 3(x 3)3 4(x 3)4 − − − − − a. Let the series be S. Substitute y = 1/(x − 3). Then 1 1 1 S = y + y2 + y3 + y4 + ... 2 3 4

b. Thus an = 1/n. n+1 y /(n + 1) n c. So rn = = |y| −→ |y| yn/n n + 1 n→∞ 1 d. S converges for |y| < 1. That is, for < 1. Or |x − 3| > 1. |x − 3| Testing the endpoints shows S(4) diverges and S(2) converges.

MAT 5610: 69 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Cauchy–Hadamard Theorem

Theorem (Cauchy (1821)–Hadamard (1888) Theorem)

X∞ k The radius of convergence R of the power series ak(x a) is − k=0 given by 1 pn = lim sup an R n | | →∞

Proof. Wolog a = 0. Since

pn n pn x lim sup anx = x lim sup an = | | | | | | | | R the result follows from Cauchy’s root test.

MAT 5610: 70 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Differentiate and Integrate

Theorem

X∞ k If the power series ak(x a) has RoC R (0, ], the power − ∈ ∞ k=0 series

X∞ k 1 X∞ ak k+1 k ak(x a) − and (x a) − k + 1 − k=1 k=0 have the same radius of convergence R.

Proof. L T T R A A E!

P 1 n NB: Differentiating may lose convergence at the endpoints; e.g., n x

MAT 5610: 71 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Unicity

Theorem (Unicity of Power Series Representation) P n P n Suppose f(x) = an(x a) and f(x) = bn(x a) on some n − n − interval x a < R where R > 0. Then an = bn for all n N. | − | ∈ Proof.

Setting x = a shows that a0 = b0. Differentiate both series, then set x = a again. Now we have a1 = b1.(Do it again!) Induction shows (n) an = bn for all n. Actually, we can show an = f (a)/n!.

Example sin(x) = x 1 x3 + 1 x5 1 x7 + 1 x9 + ... − 6 120 − 5040 362880 sin(x) = (x π) + 1 (x π)3 1 (x π)5 + 1 (x π)7 ... − − 6 − − 120 − 5040 − − k+1 k ( 1) 2k 1 ( 1) 2k P∞ P∞ a0 = b0 = k=1 (2−k 1)! π − , a1 = b1 = k=0 (2−k)! π ,... −

MAT 5610: 72 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Unicity: Caveat Emptor

T11(x) 2 f(x) = tan(sin(x))

-5 -4 -3 -2 -1 0 1 2 3 4 5

-1 T (x 1) 11 − -2

-3

MAT 5610: 73 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Unicity: Caveat Emptor, II

f(x) = tan(sin(x))

3 5 7 9 11 T , (x) = x + 0.1667x 0.0250x 0.0212x 0.0030x + 0.0011x 11 0 − − − converging on ( 1.87, 1.87) −

2 3 T , (x) = 1.119 + 1.217(x 1) 0.2122(x 1) 0.7861(x 1) 11 1 − − − − − 0.6239(x 1)4 + 0.0625(x 1)5 + 0.4961(x 1)6 − − − − + 0.3716(x 1)7 0.0054(x 1)8 0.3161(x 1)9 − − − − − 0.2243(x 1)10 + 0.0042(x 1)11 − − − = 0.0910 + 2.045x 5.126x2 + 15.50x3 31.53x4 − − − + 44.06x5 44.75x6 + 30.28x7 13.97x8 + 4.241x9 − − 0.6887x10 + 0.0435x11 − converging on ( 0.17, 2.17)

− MAT 5610: 74 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Taylor Series

Theorem (Taylor Series1) P∞ k Suppose f(x) = k=0 ak(x − a) converges with RoC R > 0. Then • f ∈ C∞(|x − a| < R) (n) • f (a) = ann! for n = 1, 2,...

Example (Bad example! No Biscuit! — Cauchy (1823))

2 Consider f(x) = e−1/x with f(0) = 0. Since f (n)(0) = 0 for all n, we have the Maclaurin series for f is 0.

Example (Good example! Biscuit!) Consider f(x) = sin(x2). ∞ 2k−1 ∞ 2(2k−1) X k+1 x X k+1 x Since sin(x)= (−1) , then f(x)= (−1) . (2k − 1)! (2k − 1)! k=1 k=1

1Taylor (1712). Earlier: Gregory (1671), Leibniz (1670s), Newton (1691), J Bernoulli (1694), de Moivre (1708). Remainder formulas: Cauchy (1822), Lagrange (1772) MAT 5610: 75 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Taylor Series Convergence

Taylor Convergence IF 1. f (n)(a) has a “reasonable” form, and 2. one of the remainders

(n+1) f (c) n+1 Lagrange’s form Rn(x) = (x a) (n+1)! − (n+1) f (c) n Cauchy’s form Rn(x) = (x c) (x a) n! − −

1 R x (n+1) n Integral form Rn(x) = f (t)(x t) dt n! a − etc., is “reasonably” easy to ﬁnd the limit of as n , → ∞ THEN we can determine a Taylor series and its radius of convergence.

MAT 5610: 76 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Weierstrass’s Two Stone Caps

Theorem (Weierstrass (1872)) Let X∞ f(x) = bn cos(anx). n=1 Then 1. f converges uniformly for b < 1, and 3 2. f is nowhere differentiable when ab > 1 + 2 π.

Theorem (Weierstrass (1885))

Let f ([a, b]). For every ε > 0 there is a polynomial pε(x) s.t. ∈ C f(x) pε(x) < ε | − | for all x [a, b]. This is also written as f pε < ε. ∈ k − k∞

A continuous function f may not have a derivative anywhere, but is still within ε of an inﬁnitely differentiable function everywhere. MAT 5610: 77 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Interlude: Algebra for Analysis

Deﬁnition (The Field of Rational Functions on R) The ﬁeld of rational functions on R is p(x) R(x) = p, q R[x] and q 0 . q(x) ∈ 6≡ with the usual addition and multiplication.

Deﬁnition (Derivation) An operator D on R(x) is a derivation if and only if for any c R and ∈ any α and β R(x), we have ∈ 1. D(x) = 1, 2. D(c α) = c D(α) for any constant c, 3. D(α + β) = D(α) + D(β), 4. D(α β) = D(α) β + α D(β) (the Leibniz rule). · · ·

MAT 5610: 78 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Properties of D

Theorem Let c R and n N. Then ∈ ∈ 1. D(1) = 0 2. D(c) = 0 n n 1 3. D(x ) = n x − n n 1 4. D(x− ) = n x− − − Proof. 1. Apply the Leibniz rule to 1 = 1 1 · 2. Use linearity on c = c 1 · 3. Apply the Leibniz rule to xk+1 = xk x and induct n n · 4. Apply the Leibniz rule to x x− = 1 ·

MAT 5610: 79 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Chains

Deﬁnition (The Chain Rule) For α and β ∈ R(x), deﬁne D(α ◦ β) = (D(α) ◦ β) · D(β)

Theorem Let p, q ∈ N and α, β ∈ R(x) with β 6≡ 0. Then 1. D(x1/q) = (1/q) x1/q−1 2. D(xp/q) = (p/q) xp/q−1 3. D (α/β) = [D(α) · β − α · D(β)]/β2

Proof. 1. Apply the chain rule to (x1/q)q = x 2. Modiﬁed ditto 3. Look at (α/β) · β = α

MAT 5610: 80 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Two Extensions to R(x)

Definition • Define the function L by the relation: for any nonzero α R(x), ∈ D(α) D(L α) = ◦ α • Define the function E by the relation: for any nonzero α R(x), ∈ D(E α) = E(α) D(α) ◦ · Exercise 1. Determine D(L α) when α(x) = x. ◦ 2. Determine D(E β) when β(x) = x. ◦ Definition R For any γ R(x)[L, E], define γ = Γ iff D(Γ) = γ. ∈

MAT 5610: 81 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series J-Lio’s Result

Theorem (Liouville’s Principle) If a function f has an elementary integral, then the integral has the form n Z X f = F0 + ckL(Fk) k=1 where the ck R and the Fk R(x). The function L is the logarithm deﬁned above.∈ ∈

Problem Z Z Z • Does xx + xx ln(x) =? xx + xx ln(x)? No!

Read Risch’s two papers: • “The Problem of Integration in Finite Terms,” Robert H. Risch, Trans AMS, Vol. 139 (May, 1969), pp. 167–189. • “Algebraic Properties of the Elementary Functions of Analysis,” Robert H. Risch, Am J Math, Vol. 101, No. 4 (Aug., 1979), pp. 743–759. MAT 5610: 82 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Bad Function! No Biscuit!

Theorem 2 There is no elementary antiderivative for f(x) = ex .

Proof.

Z 2 2 2 Suppose F = ex . We know D(ex ) = ex 2x. Applying Liouville’s principle

x2 0 0 x2 x2 x2 implies F = α(x) e with α ∈ R(x). Thus F = α e + α 2x e = e . So α0 + 2x α = 1. Comparing degrees shows α can’t be a poly, and so must be a rational function. Let α = p(x)/q(x). Then q must have a root (possibly m complex), so write α = h(x)/(x − z0) where h(z0) 6= 0 is a rational function. Substitute into the DE α0 + 2x α = 1: h0(x) h(x) h(x) m − m m+1 + 2x m = 1 (x − z0) (x − z0) (x − z0)

Taking the limit of both sides as x → z0 yields the contradiction ∞ = 1. Whence F cannot exist in R(x, L, E).

MAT 5610: 83