Lecture 24

We can now see that k-th term is (−1)k 1/k, and that there are 100 terms, so we would write the sum in sigma notation as X100 k=1 (−1)k 1 k Key Point To write a sum in sigma notation, try to ﬁnd a formula involving a variable k where the ﬁrst term can be obtained by setting k = 1, the second term by k = 2, and so on Exercises 3

ORIGINAL HAND PUMPS - Sigma Engineering

The Sigma ‘K’ Pump incorporates many original technical features which make it the best of its type • Reinforced inlet, discharge and lugs – prevent breakage even if subjected to rough treatment • The tapered internal body – this unique technological advantage assures automatic settlement of the wings in

41 Sigma Notation and Riemann Sums

k=3 1 k using sigma notation J Practice 4 Evaluate the sum of the rectangular areas in the margin ﬁgure, then write the sum using sigma notation

Tutorial 7: VASP Calculations With Model Solvation

SIGMA = 0 05 ISYM = 0 Write flags LWAVE = F LCHARG = F Solvation LSOL = TRUE • Default solvent is water • Specify solvent parameters EB_K, SIGMA_K, NC_K, TAU for other solvents e g acetonitrile KPOINTS (Γ-only) 0 Gamma 111 000 ’ POSCAR H 2O in 15 Å box

Proteinase K (P2308) - Datasheet - Sigma-Aldrich

Proteinase K from Engyodontium album (formerly Tritirachium album) for Molecular Biology Catalog Number P2308 Storage Temperature –20 C CAS RN 39450-01-6 E C 3 4 21 64 Synonyms: Peptidase K, Endoproteinase K, Endopeptidase K Temperature prof Product Description Proteinase K is a stable serine protease with broad substrate specificity

Proteinase K, recombinant, PCR Grade - Sigma-Aldrich

sigma-aldrich com For life science research only Not for use in diagnostic procedures Proteinase K, recombinant, PCR Grade From Pichia pastoris Lyophilizate Cat No 03 115 836 001 25 mg y Version 07

Statistical Process Control & Process Capability

k Cp k 8 Rules of Control Charts 1 One (1) point > 3 Standard Deviations from the Center Line (Outside the UCL or LCL {3 Sigma Limit}) 2 Nine (9) points in a row on same side of center line 3 Six (6) points in a row, all increasing or all decreasing 4 Fourteen (14) points in a row, alternating up and down 5

Lecture 24 - UH

k→∞ a k 6= 0 p-series: X 1 kp, if p ≤ 1 Convergence Tests (1) Basic Test for Convergence KeepinMindthat, if a k 9 0, then the series P a k diverges; therefore there is no reason to apply any special convergence test Examples 1 P xk with x ≥ 1 (e g, P (−1)k) diverge since xk 9 0 [1ex] X k k +1 diverges since k k+1 → 1 6= 0 [1ex

Ductile Iron Pipes and Fittings - Sigma Commercial Products

FREEPHONE 08081 449 888 Sigma Pipeline Solutions your ﬁtting choice for so much more quality 3 Sigma Pipeline Solutions’ technical team stay ahead of the game - developing new and improved products and ensuring full compliance with any relevant existing or proposed legislation or quality standards Since its inception, Sigma Pipeline

[PDF] resultat tpe 2016

[PDF] inventer une ruse de renart

[PDF] que signifie le mot roman au moyen age

[PDF] pierre de saint cloud

[PDF] auteurs du roman de renart

[PDF] roman de renart texte

[PDF] notation a b c d e sur 20

[PDF] les formes poétiques pdf

[PDF] notation anglaise scolaire

[PDF] les registres poétiques

[PDF] systeme de notation quebecois

[PDF] genre poétique caractéristiques

[PDF] type de poeme wikipedia

[PDF] poème forme libre exemple

[PDF] notation lv2 bac

Lecture 24Section 11.4 Absolute and Conditional

Convergence; Alternating Series

Jiwen He

1 Convergence Tests

Basic Series that Converge or Diverge

Basic Series that Converge

Geometric series:

?xk,if|x|<1 p-series:?1k p,ifp >1

Basic Series that Diverge

Any series

?a kfor which limk→∞ak?= 0 p-series:?1k

Convergence Tests (1)

Basic Test for Convergence

Keep in Mind that, ifak?0, then the series?akdiverges; therefore there is no reason to apply any special convergence test.Examples1.? xkwith|x| ≥1 (e.g,?(-1)k)divergesincexk?0. [1ex] ?kk+ 1divergessincekk+1→1?= 0. [1ex]?? 1-1k k divergessince a k=?1-1k k→e-1?= 0.

Convergence Tests (2)

Comparison Tests

Rational termsare most easily handled bybasic comparisonorlimit comparison withp-series?1/kp

Basic Comparison Test1

12k3+ 1converges by comparison with?1k

3? k3k

5+ 4k4+ 7converges

by comparison with ?1k 2? 1k

3-k2converges by comparison with?2k

3?13k+ 1diverges by comparison with?13(k+ 1)?

1ln(k+ 6)diverges by

comparison with ?1k+ 6

Limit Comparison Test

?1k

3-1converges by comparison with?1k

3.?3k2+ 2k+ 1k

3+ 1diverges

by comparison with ?3k

5⎷k+ 1002k2⎷k-9⎷k

converges by comparison with 52k2

Convergence Tests (3)

Root Test and Ratio Test

Theroot testis used only ifpowersare involved.

Root Test

?k22 kconverges: (ak)1/k=12

·?k1/k?2→12

·1?1(lnk)kconverges: (ak)1/k=

1lnk→0??

1-1k k2 converges: (ak)1/k=?

1 +(-1)k

k→e-1

Convergence Tests (4)

Root Test and Ratio Test

Theratio testis effective withfactorialsand with combinations of powers and factorials.

Ratio Comparison Test?k22

kconverges:ak+1a k=12

·(k+1)2k

2→12

1k!converges:ak+1a

k=1k+1→0 k10 kconverges:ak+1a k=110

·k+1k

→110 kkk!diverges:ak+1a k=?1 +1k k→e 2k3 k-2kconverges:ak+1a k= 2·1-(2/3)k3-2(2/3)k→2·13

1⎷k!converges:ak+1a

k= ?1 k+1→0

2 Absolute Convergence

2.1 Absolute Convergence

Absolute Convergence2

Absolute Convergence

A series?akis said toconverge absolutelyif?|ak|converges. if ?|ak|converges, then?akconverges. i.e., absolutely convergent series are convergent.

Alternatingp-Series withp >1?(-1)kk

p,p >1,converge absolutelybecause?1k pconverges.? k=1(-1)k+1k

2= 1-12

2+13 2-14

2- ···converge absolutely.

Geometric Series with-1< x <1?(-1)j(k)xk,-1< x <1,converge absolutelybecause?|x|kconverges. ?1-12 -12 2+12 3-12 4+12 5+12

6- ···converge absolutely.

Conditional Convergence

A series?akis said toconverge conditionallyif?akconverges while?|ak| diverges. pdiverges.? k=1(-1)k+1k = 1-12 +13 -14 - ···converge conditionally.

3 Alternating Series

Alternating Series

Let{ak}be a sequence ofpositivenumbers.?(-1)kak=a0-a1+a2-a3+a4- ··· is called analternating series.

Alternating Series Test

Let{ak}be adecreasingsequence ofpositivenumbers.

Ifak→0,then?(-1)kakconverges.Alternatingp-Series withp >0?(-1)kk p,p >0,convergesincef(x) =1x pisdecreasing, i.e.,f?(x) =-px p+1>

0 for?x >0, and limx→∞f(x) = 0.?∞?

k=1(-1)k+1k = 1-12 +13 -14 convergeconditionally.3

Examples

(-1)k2k+ 1,convergesincef(x) =12x+ 1isdecreasing, i.e.,f?(x) =-2(2x+ 1)2>

0 for?x >0, and limx→∞f(x) = 0.

(-1)kkk

2+ 10,convergesincef(x) =xx

2+ 10isdecreasing, i.e.,f?(x) =-x2-10(x2+ 10)2>

0, for?x >⎷10, and lim

x→∞f(x) = 0. An Estimate for Alternating SeriesAn Estimate for Alternating Series Let{ak}be adecreasingsequence ofpositivenumbers that tends to 0 and let

L=∞?

k=0(-1)kak. Then the sumLlies between consecutivepartial sumssn, s n+1,sn< L < sn+1,ifnis odd;sn+1< L < sn,ifnis even. and thussnapproximatesLto withinan+1 |L-sn|< an+1.

Example

Findsnto approximate∞?

k=1(-1)k+1k = 1-12 +13

···within 10-2.

Set k=1(-1)k+1k k=0(-1)kk+ 1. For|L-sn|<10-2, we want a n+1=1(n+ 1) + 1<10-2?n+ 2>102?n >98.

Thenn= 99 and the 99th partial sums100is

99= 1-12

+13 -14 +···+199 -1100 ≈0.6882.

From the estimate

|L-s99|< a100=1101 ≈0.00991. we conclude that s

99≈0.6882<∞?

k=1(-1)k+1k = ln2<0.6981≈s1004

Example

Findsnto approximate∞?

k=0(-1)k+1(2k+ 1)!= 1-13! +15!

···within 10-2.

For|L-sn|<10-2, we want

a n+1=1(2(n+ 1) + 1)!<10-2?n≥1.

Thenn= 1 and the 2nd partial sums2is

1= 1-13!

≈0.8333

From the estimate|L-s1|< a2=15!

≈0.0083. we conclude that s

1≈0.8333<∞?

k=0(-1)k+1(2k+ 1)!= sin1<0.8416≈s2

4 Rearrangements

Why Absolute Convergence Matters: Rearrangements (1)

Rearrangement of Absolute Convergence Series

k=0(-1)k2 k= 1-12 +12 2-12 3+12 4-12

5+···=23

absolutely

Rearrangement1 +12

2-12 +12 4+12 6-12 3+12 8+12 10-12

5···? = =23

Theorem 2.All rearrangements of an absolutely convergent series converge absolutely to the same sum. Why Absolute Convergence Matters: Rearrangements (2)

Rearrangement of Conditional Convergence Series

k=1(-1)k+1k = 1-12 +13 -14 +15 -16 +···= ln2conditionally

Rearrangement1 +13

-12 +15 +17 -14 +19 +111
-16

···? =?= ln2

Multiply the original series by

12 12 k=1(-1)k+1k =12 -14 +16 -18 +110
+···=12 ln2

Adding the two series, we get the rearrangement

k=1(-1)k+1k +12 k=1(-1)k+1k = 1 +13 -12 +15 +17 -14 +···=32 ln2

Remark5

•A series that is onlyconditionallyconvergent can be rearranged to converge

toany numberwe please.•It can also be arranged todivergeto +∞or-∞, or even to oscillate

[PDF] Lecture 24 - UH

Lecture 24Section 11.4 Absolute and Conditional

Convergence; Alternating Series

Jiwen He

1 Convergence Tests

Basic Series that Converge or Diverge

Basic Series that Converge

Geometric series:

Basic Series that Diverge

Any series

Convergence Tests (1)

Basic Test for Convergence

Convergence Tests (2)

Comparison Tests

Basic Comparison Test1

12k3+ 1converges by comparison with?1k

5+ 4k4+ 7converges

3-k2converges by comparison with?2k

3?13k+ 1diverges by comparison with?13(k+ 1)?

1ln(k+ 6)diverges by

Limit Comparison Test

3-1converges by comparison with?1k

3.?3k2+ 2k+ 1k

3+ 1diverges

5⎷k+ 1002k2⎷k-9⎷k

Convergence Tests (3)

Root Test and Ratio Test

Theroot testis used only ifpowersare involved.

Root Test

·?k1/k?2→12

·1?1(lnk)kconverges: (ak)1/k=

1lnk→0??

1 +(-1)k

Convergence Tests (4)

Root Test and Ratio Test

Ratio Comparison Test?k22

·(k+1)2k

2→12

1k!converges:ak+1a

·k+1k

1⎷k!converges:ak+1a

2 Absolute Convergence

2.1 Absolute Convergence

Absolute Convergence2

Absolute Convergence

Alternatingp-Series withp >1?(-1)kk

2= 1-12

2- ···converge absolutely.

6- ···converge absolutely.

Conditional Convergence

Conditional Convergence

3 Alternating Series

Alternating Series

Alternating Series

Alternating Series Test

Let{ak}be adecreasingsequence ofpositivenumbers.

0 for?x >0, and limx→∞f(x) = 0.?∞?

Examples

0 for?x >0, and limx→∞f(x) = 0.

2+ 10,convergesincef(x) =xx

2+ 10isdecreasing, i.e.,f?(x) =-x2-10(x2+ 10)2>

0, for?x >⎷10, and lim

L=∞?

Example

Findsnto approximate∞?

···within 10-2.

Thenn= 99 and the 99th partial sums100is

99= 1-12

From the estimate

99≈0.6882<∞?

Example

Findsnto approximate∞?

···within 10-2.

For|L-sn|<10-2, we want

Thenn= 1 and the 2nd partial sums2is

1= 1-13!

From the estimate|L-s1|< a2=15!

1≈0.8333<∞?

4 Rearrangements

Rearrangement of Absolute Convergence Series