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