Quiz. Use either the RATIO or ROOT TEST to determie whether the series is coverget or ot. e
.6 POWER SERIES
Defiitio. A power series i about is a series of the form c 0 c a c a... c a... a 0 c a where the c ' s are costats. I this sectio, we study how to fid the value(s) of so that a power series is coverget.
Defiitios The iterval for which a power series is coverget is called the Iterval of covergece (this iterval always icludes =a) ad half the legth of this iterval is called the radius of covergece.
Theorem. Cosider a power series c a. 0 The eactly oe of the ff holds: i. It coverges oly to a I.O.C: aa, R.O.C: 0 ii. It absolutely coverges for all I.O.C:, R.O.C:
Theorem. (cot ) iii. There eists R 0 such that it is absolutely coverget for all a R, a R ad diverget for all,, a R a R R.O.C: R We have to test covergece at the edpoits of the iterval.
How to fid the iterval of covergece:. Apply the RATIO TEST or the ROOT TEST.. Test covergece at the edpoits usig tests other tha the two stated above.
Eamples. Fid all values of give power series is coverget. 0 3....... Usig Root Test, we ll have so that the L lim Recall: To coclude covergece usig this test, L. So,
0 3....... Now, we ll test covergece at the edpoits. If the. If the. 0 0 NOTE: This series is diverget by the th term test. 0 0 NOTE: This series is also diverget by the th term test.
Thus, power series defied by has Iterval of Covergece: Radius of Covergece: 0,
Eamples. Fid all values of give power series is coverget. 0 3. Usig Ratio Test: 5 L u lim u Recall: To coclude covergece usig this test, L. 5 3 so that the So, 5 3 8
If the. Now, we ll test covergece at the edpoits. 8 NOTE: This series is coverget by the alteratig series test. 0 3 5 0 3 5 0 5 5 0
If the. Now, we ll test covergece at the edpoits. NOTE: The harmoic series is diverget. 0 3 5 0 3 5 0 5 5 0
Thus, power series defied by has 0 3 5 Iterval of Covergece: Radius of Covergece: 8, 5
Eamples. Fid all values of give power series is coverget. 3.! Usig Ratio Test: 0 so that the L u lim u lim
Thus, power series defied by 0! has Iterval of Covergece:, Radius of Covergece: 0
Eamples. Fid all values of give power series is coverget. 3 4. Usig Ratio Test:! 0 so that the L u lim u 3 lim 3 0
Thus, power series defied by has 0 3! Iterval of Covergece: Radius of Covergece:,
Defiitio. A power series i about is a series of the form c 0 c a c a... c a... a 0 c a where the c ' s are costats. (=a is the ceter of the power series)
RECALL The power series defied by has 0 Ceter: 0 Iterval of Covergece: Radius of Covergece:,
RECALL The power series defied by has 0 3 5 Ceter: 3 Iterval of Covergece: Radius of Covergece: 8, 5
RECALL The power series defied by 0! has Ceter: Iterval of Covergece: Radius of Covergece:, 0
RECALL The power series defied by has 0 3! Ceter: 3 Iterval of Covergece: Radius of Covergece:,
Net, we study how to write the power series epasio of a fuctio. Some ways to do this: - As the sum of a geometric series - Differetiatig a kow power series epasio - Itegratig a kow power series epasio - Taylor Series Epasio - Ordiary/Partial Differetial Equatios
.7 Differetiatio of POWER SERIES
Term-by-Term Differetiatio A power series ca be differetiated term by term at each iterior poit of its iterval of covergece. 0 3 0 3... c c c c c c c c c 3 c... 3 c......
Theorem. If the power series has R power series also has f c a 0 as its radius of covergece, the the R as its radius of covergece. 0 f ' c a ( ) c a
0 c f f' c f ''' f '' 3 c c 3
WORD OF CAUTION: Differetiatig/Itegratig a series epasio may ot be allowed for some ifiite series.
Eample. Fid a series epasio for if f SOL N. f, 3......, f' ad f'' 0 f '( ) 3......,
Eample. if SOL N. f ' f Fid a series epasio for, 3......, f' ad f'' f '' 3 6......,
Eample. Obtai a series epasio for ad give its validity. SOL N. a, r 3 3 3 3 ar 3 a, r r g 3 3 3 3 g' 3 3
Eample. Obtai a series epasio for ad give its validity. 3 SOL N. Sice, 3 3 3 We ll have 3 3 3 3, 3 3
Eample. Obtai a series epasio for ad give its validity. SOL N. a, r 7 7 7 7 g 7 4 7 7 7 ar g' 7 7 a r 7, r 7
Eample. Obtai a series epasio for ad give its validity. 7 SOL N. 4 Sice, 7 7 We ll have 7 4 7 7, 7 7
Itegratio of POWER SERIES
Term-by-Term Itegratio A power series ca be itegrated term by term at each iterior poit of its iterval of covergece. 0 3 0 3... c c c c c c... c 0 C 3 C c0 c c 3 c......
Theorem. If the power series has R power series also has f c a 0 as its radius of covergece, the the R 0 as its radius of covergece. a f d c C
Eample. Obtai a series epasio for l ad give its validity. SOL N. a, r g ar a r, r gt t t
Eample. Obtai a series epasio for l ad give its validity. SOL N. l dt 0 t l t 0 l 0 t t 0 dt l l,
Cautio The operatios of algebra (additio, subtractio, multiplicatio, divisio) may be applied to power series i the same maer that they are applied to polyomials provided that the coditios of covergece are take ito accout.
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Add-o: Fid all values of give series is coverget. so that the 3...... 3 4 8 0 Iterval of Covergece:,,