previous series have consisted of constants
DESCRIPTION
Section 10.7 – P ower Series. Previous series have consisted of constants. . Another type of series will include the variable x . . Section 10.7 – P ower Series. A Power Series is of the form. where x is a variable and represents the coefficients. - PowerPoint PPT PresentationTRANSCRIPT
Previous series have consisted of constants.
∑𝒏=𝟏
∞
𝒂𝒏
Another type of series will include the variable x.
Section 10.7 – Power Series
∑𝒏=𝟏
∞ 𝟏𝒏 ∑
𝒏=𝟏
∞ 𝟏𝟎𝒏
(𝒏+𝟏 )! ∑𝒏=𝟏
∞
(−𝟏 )𝒏 𝒍𝒏(𝒏)𝒏− 𝒍𝒏(𝒏)
∑𝒏=𝟏
∞
𝒂𝒏(𝒙) ∑𝒏=𝟏
∞ 𝒙𝒏 ∑
𝒏=𝟏
∞ 𝟏𝟎𝒏 𝒙𝒏
(𝒏+𝟏 ) ! ∑𝒏=𝟏
∞
(−𝟏 )𝒏 𝒍𝒏(𝒏)(𝒙−𝟒 )𝒏
𝒏− 𝒍𝒏(𝒏)
A Power Series is of the form
𝒇 (𝒙)
∑𝒏=𝟎
∞
𝒄𝒏 𝒙𝒏¿𝒄𝟎+𝒄𝟏𝒙+𝒄𝟐 𝒙𝟐+𝒄𝟑𝒙𝟑+⋯
where x is a variable and represents the coefficients.
Section 10.7 – Power Series
The sum of the series is the function¿𝒄𝟎+𝒄𝟏𝒙+𝒄𝟐 𝒙𝟐+𝒄𝟑𝒙𝟑+⋯
where its domain is the set of all x for which the series converges.
A more general form of the power series is of the form
∑𝒏=𝟎
∞
𝒄𝒏 (𝒙−𝒂)𝒏¿𝒄𝟎+𝒄𝟏 (𝒙−𝒂 )+𝒄𝟐 (𝒙−𝒂)𝟐+𝒄𝟑 (𝒙−𝒂 )𝟑+⋯
where x is a variable, represents the coefficients and a is a number.
Section 10.7 – Power Series
This form is referred as:a power series in (x – a) a power series centered at a.
or
Section 10.7 – Power Series
There are only three ways for a power series to converge.1) The series only converges at .2) The series converges for all x values.3) The series converges for some interval of x.
Interval of Convergence:
The interval of x values where the series converges. Radius of Convergence
(R):Half the length of the interval of convergence.
Definitions
()()
()The end values of the interval must be tested for convergence.
|𝒙 −𝒂|<𝑹
The use of the Ratio test is recommended when finding the radius of convergence and the interval of convergence.
Section 10.7 – Power SeriesFind the Radius of Convergence and the Interval of Convergence for the following power series
Example:
∑𝒏=𝟎
∞ (−𝟏 )𝒏𝒏 (𝒙+𝟑 )𝒏
𝟒𝒏 𝑐𝑛+1=(−𝟏 )𝒏+𝟏 (𝒏+𝟏 ) (𝒙+𝟑 )𝒏+𝟏
𝟒𝒏+𝟏Ratio Test
lim𝒏→∞|(−𝟏 )𝒏+𝟏 (𝒏+𝟏 ) (𝒙+𝟑 )𝒏+𝟏
𝟒𝒏+𝟏 ÷(−𝟏 )𝒏𝒏 (𝒙+𝟑 )𝒏
𝟒𝒏 |lim𝒏→∞|(−𝟏 )𝒏 (−𝟏 ) (𝒏+𝟏 ) (𝒙+𝟑 )𝒏 (𝒙+𝟑 )
𝟒𝒏 (𝟒 )∙ 𝟒𝒏
(−𝟏 )𝒏𝒏 (𝒙+𝟑 )𝒏|lim𝒏→∞|(−𝟏 ) (𝒏+𝟏 ) (𝒙+𝟑 )
𝟒𝒏 |¿ 𝟏𝟒 |𝒙+𝟑|
Section 10.7 – Power Series
∑𝒏=𝟎
∞ (−𝟏 )𝒏𝒏 (𝒙+𝟑 )𝒏
𝟒𝒏
Ratio Test: Convergence for
lim𝒏→∞|(−𝟏 ) (𝒏+𝟏 ) (𝒙+𝟑 )
𝟒𝒏 |¿ 𝟏𝟒 |𝒙+𝟑|
𝟏𝟒|𝒙+𝟑|<𝟏
|𝒙+𝟑|<𝟒 −𝟒<𝒙+𝟑<𝟒−𝟕<𝒙<𝟏Radius of
Convergence𝑹=𝟒
Interval of Convergence:
End points need to be tested.
Section 10.7 – Power Series
∑𝒏=𝟎
∞ (−𝟏 )𝒏𝒏 (𝒙+𝟑 )𝒏
𝟒𝒏
𝒙=−𝟕
𝒅𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒕𝒂𝒕 𝒙=−𝟕¿∞
𝒏𝒕𝒉𝒕𝒆𝒓𝒎𝒕𝒆𝒔𝒕 𝒇𝒐𝒓 𝒅𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆∑𝒏=𝟎
∞ (−𝟏 )𝒏𝒏 (−𝟕+𝟑 )𝒏
𝟒𝒏
∑𝒏=𝟎
∞ (−𝟏 )𝒏𝒏 (−𝟒 )𝒏
𝟒𝒏
∑𝒏=𝟎
∞ (−𝟏 )𝒏𝒏 (−𝟏 )𝒏 (𝟒 )𝒏
𝟒𝒏
∑𝒏=𝟎
∞
𝒏
lim𝒏→∞
𝒏 ≠𝟎
Section 10.7 – Power Series
∑𝒏=𝟎
∞ (−𝟏 )𝒏𝒏 (𝒙+𝟑 )𝒏
𝟒𝒏
𝒙=𝟏
𝒅𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒕𝒂𝒕 𝒙=𝟏¿𝑫𝑵𝑬
𝒏𝒕𝒉𝒕𝒆𝒓𝒎𝒕𝒆𝒔𝒕 𝒇𝒐𝒓 𝒅𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆∑𝒏=𝟎
∞ (−𝟏 )𝒏𝒏 (𝟏+𝟑 )𝒏
𝟒𝒏
∑𝒏=𝟎
∞ (−𝟏 )𝒏𝒏 (𝟒 )𝒏
𝟒𝒏
∑𝒏=𝟎
∞
(−𝟏 )𝒏𝒏
lim𝒏→∞
(−𝟏 )𝒏𝒏 ≠𝟎
∑𝒏=𝟎
∞
(−𝟏 )𝒏𝒏
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 ,𝒕𝒉𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆 𝒊𝒔 :−𝟕<𝒙<𝟏
Section 10.7 – Power SeriesFind the Radius of Convergence and the Interval of Convergence for the following power series
Example:
∑𝒏=𝟎
∞ 𝟐𝒏 (𝟒 𝒙−𝟖 )𝒏
𝒏 𝑐𝑛+1=𝟐𝒏+𝟏 (𝟒 𝒙−𝟖 )𝒏+𝟏
𝒏+𝟏Ratio Test
lim𝒏→∞|𝟐
𝒏+𝟏 (𝟒 𝒙−𝟖 )𝒏+𝟏
𝒏+𝟏 ÷𝟐𝒏 (𝟒 𝒙−𝟖 )𝒏
𝒏 |lim𝒏→∞|𝟐
𝒏𝟐 (𝟒 𝒙−𝟖 )𝒏 (𝟒 𝒙−𝟖 )𝒏+𝟏 ∙ 𝒏
𝟐𝒏 (𝟒 𝒙−𝟖 )𝒏|lim𝒏→∞|𝟐𝒏 (𝟒 𝒙 −𝟖 )
𝒏+𝟏 | ¿𝟐|𝟒 𝒙 −𝟖|
Section 10.7 – Power Series
Ratio Test: Convergence for
𝟐|𝟒 𝒙 −𝟖|<𝟏𝟐|𝟒 (𝒙−𝟐)|<𝟏
−𝟏𝟖 <𝒙−𝟐<𝟏𝟖
𝟏𝟓𝟖 <𝒙<
𝟏𝟕𝟖
Radius of Convergence𝑹=
𝟏𝟖
Interval of Convergence:
End points need to be tested.
∑𝒏=𝟏
∞ 𝟐𝒏 (𝟒 𝒙−𝟖 )𝒏
𝒏lim𝒏→∞|𝟐𝒏 (𝟒 𝒙 −𝟖 )
𝒏+𝟏 |¿𝟐|𝟒 𝒙 −𝟖|
𝟖|𝒙−𝟐|<𝟏|𝒙 −𝟐|<𝟏𝟖
Section 10.7 – Power Series
𝒙=𝟏𝟓𝟖
∴𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒕 𝒂𝒕 𝒙=𝟏𝟓𝟖
𝑨𝒍𝒕𝒆𝒓𝒏𝒂𝒕𝒊𝒏𝒈𝒉𝒂𝒓𝒎𝒐𝒏𝒊𝒄 𝒔𝒆𝒓𝒊𝒆𝒔 𝒊𝒔𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒕
∑𝒏=𝟏
∞ 𝟐𝒏 (−𝟏 )𝒏
𝒏 ∙𝟐𝒏∑𝒏=𝟏
∞ 𝟐𝒏 (𝟒 𝒙−𝟖 )𝒏
𝒏
∑𝒏=𝟏
∞ 𝟐𝒏(𝟒 (𝟏𝟓𝟖 )−𝟖)𝒏
𝒏
∑𝒏=𝟏
∞ 𝟐𝒏(𝟏𝟓𝟐 −𝟖)𝒏
𝒏
∑𝒏=𝟏
∞ 𝟐𝒏(−𝟏𝟐 )𝒏
𝒏
∑𝒏=𝟏
∞ (−𝟏 )𝒏
𝒏
Section 10.7 – Power Series
𝒙=𝟏𝟕𝟖
∴𝒅𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒕 𝒂𝒕 𝒙=𝟏𝟕𝟖
𝑯𝒂𝒓𝒎𝒐𝒏𝒊𝒄 𝒔𝒆𝒓𝒊𝒆𝒔 𝒊𝒔 𝒅𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒕
∑𝒏=𝟏
∞ 𝟐𝒏
𝒏 ∙𝟐𝒏
∑𝒏=𝟏
∞ 𝟐𝒏 (𝟒 𝒙−𝟖 )𝒏
𝒏
∑𝒏=𝟏
∞ 𝟐𝒏(𝟒 (𝟏𝟕𝟖 )−𝟖)𝒏
𝒏
∑𝒏=𝟏
∞ 𝟐𝒏(𝟏𝟕𝟐 −𝟖)𝒏
𝒏
∑𝒏=𝟏
∞ 𝟐𝒏(𝟏𝟐 )𝒏
𝒏
∑𝒏=𝟏
∞ 𝟏𝒏→
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 ,𝒕𝒉𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆 𝒊𝒔 :𝟏𝟓𝟖 ≤ 𝒙<
𝟏𝟕𝟖
Section 10.7 – Power Series
∑𝒏=𝟎
∞
𝒏 ! (𝟐 𝒙+𝟏 )𝒏
¿∞>𝟏
𝑻𝒉𝒆 𝒔𝒆𝒓𝒊𝒆𝒔𝒘𝒊𝒍𝒍 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆 𝒂𝒕𝒐𝒏𝒆𝒑𝒐𝒊𝒏𝒕 .
𝑹𝒂𝒅𝒊𝒖𝒔𝒐𝒇 𝑪𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆 :𝑹=𝟎𝑻𝒉𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆 𝒊𝒔 :
𝒙=−𝟏𝟐
𝑐𝑛+1=(𝑛+1 ) ! (2 𝑥+1 )𝑛+1Ratio Test
lim𝒏→∞|(𝑛+1 )! (2 𝑥+1 )𝑛+1
𝑛 ! (2 𝑥+1 )𝑛 |lim𝒏→∞|𝑛 ! (𝑛+1 ) (2 𝑥+1 )𝑛 (2 𝑥+1 )
𝑛 ! (2 𝑥+1 )𝑛 |lim𝒏→∞
|(𝑛+1 ) (2𝑥+1 )|
Find the Radius of Convergence and the Interval of Convergence for the following power series
Example:
Section 10.7 – Power Series
∑𝒏=𝟎
∞ (𝒙−𝟔 )𝒏
𝒏𝒏
¿𝟎<𝟏
𝑻𝒉𝒆 𝒔𝒆𝒓𝒊𝒆𝒔𝒘𝒊𝒍𝒍 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆 𝒇𝒐𝒓 𝒆𝒗𝒆𝒓𝒚 𝒙 .𝑹𝒂𝒅𝒊𝒖𝒔𝒐𝒇 𝑪𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆 :𝑹=∞
𝑻𝒉𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆 𝒊𝒔 :−∞<𝑥<∞
𝑐𝑛+1=(𝑥−6 )𝑛+1
𝑛𝑛+1Ratio Test
lim𝒏→∞|(𝑥+6 )𝑛+1
𝑛𝑛+1 ÷(𝑥+6 )𝑛
𝑛𝑛 |lim𝒏→∞|(𝑥+6 )𝑛 (𝑥+6 )
𝑛𝑛𝑛∙ 𝑛𝑛
(𝑥+6 )𝑛|lim𝒏→∞|(𝑥+6 )
𝑛 |
𝑻𝒉𝒆𝒍𝒊𝒎𝒊𝒕 𝒊𝒔 𝒛𝒆𝒓𝒐𝒓𝒆𝒈𝒂𝒓𝒅𝒍𝒆𝒔𝒔 𝒐𝒇 𝒕𝒉𝒆𝒗𝒂𝒍𝒖𝒆𝒐𝒇 𝒙 .
Find the Radius of Convergence and the Interval of Convergence for the following power series
Example:
Section 10.7 – Power Series
∑𝒏=𝟎
∞ 𝒙𝟐𝒏
(−𝟑 )𝒏
¿|𝒙𝟐|𝟑
𝑹𝒂𝒅𝒊𝒖𝒔𝒐𝒇 𝑪𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆 :𝑹=√𝟑 𝑻𝒉𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆 𝒊𝒔 :−√𝟑<𝑥<√𝟑
Root Test
lim𝒏→∞|𝒏√ ( 𝒙𝟐 )𝒏
(−𝟑 )𝒏| lim𝒏→∞|𝒙
𝟐
−3|
Find the Radius of Convergence and the Interval of Convergence for the following power series
Example:
lim𝒏→∞|𝒏√( 𝒙𝟐
−𝟑 )𝒏| ¿𝟏 |𝒙𝟐|<𝟑
|𝒙|<√𝟑
Section 10.7 – Power Series
∑𝒏=𝟎
∞ 𝒙𝟐𝒏
(−𝟑 )𝒏
¿𝑫𝑵𝑬 ≠𝟎
∴𝒕𝒉𝒆−√𝟑 𝒊𝒔𝒏𝒐𝒕 𝒊𝒏𝒕𝒉𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆∴𝒅𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒕
𝒙=−√𝟑
lim𝒏→∞
(−𝟏 )𝒏
Test the Endpoints
∑𝒏=𝟎
∞ (−√𝟑 )𝟐𝒏
(−𝟑 )𝒏∑𝒏=𝟎
∞ ( (−√𝟑 )𝟐 )𝒏
(−𝟑 )𝒏∑𝒏=𝟎
∞ (𝟑 )𝒏
(−𝟑 )𝒏∑𝒏=𝟎
∞ (𝟑 )𝒏
(−𝟏 )𝒏 (𝟑 )𝒏
∑𝒏=𝟎
∞
(−𝟏 )𝒏𝒏𝒕𝒉𝑻𝒆𝒓𝒎𝑻𝒆𝒔𝒕 𝒇𝒐𝒓 𝑫𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆
Section 10.7 – Power Series
∑𝒏=𝟎
∞ 𝒙𝟐𝒏
(−𝟑 )𝒏
¿𝑫𝑵𝑬 ≠𝟎
∴√𝟑 𝒊𝒔 𝒏𝒐𝒕 𝒊𝒏𝒕𝒉𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆∴𝒅𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒕
𝒙=√𝟑
lim𝒏→∞
(−𝟏 )𝒏
Test the Endpoints
∑𝒏=𝟎
∞ (√𝟑 )𝟐𝒏
(−𝟑 )𝒏∑𝒏=𝟎
∞ ( (√𝟑 )𝟐 )𝒏
(−𝟑 )𝒏 ∑𝒏=𝟎
∞ (𝟑 )𝒏
(−𝟑 )𝒏∑𝒏=𝟎
∞ (𝟑 )𝒏
(−𝟏 )𝒏 (𝟑 )𝒏
∑𝒏=𝟎
∞
(−𝟏 )𝒏𝒏𝒕𝒉𝑻𝒆𝒓𝒎𝑻𝒆𝒔𝒕 𝒇𝒐𝒓 𝑫𝒊𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆
𝑻𝒉𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆 𝒊𝒔 :−√𝟑<𝑥<√𝟑
Section 10.7 – Power Series