170120107066 power series.ppt
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1) The document discusses power series and calculating the radius of convergence and interval of convergence for power series. 2) There are three possible ways for a power series to converge: at a single point, for all values, or within an interval of values. The radius of convergence is half the length of the interval of convergence. 3) The ratio test is recommended for finding the radius and interval of convergence. Examples are provided of applying the ratio test to determine the radius and interval of convergence for different power series.
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170120107066 power series.ppt
- 1. Subject : calculus(2110014) Topic : power series and radius of convergence. Gandhinagar Institute of Technology(012) Active Learning Assignment Guided By:prof. keyuri mam Prepared By:Harsh kothari Branch:CE-B Enrolment no.:170120107066
- 2. Previous series have consisted of constants. 𝒏=𝟏 ∞ 𝒂 𝒏 Another type of series will include the variable x. 𝒏=𝟏 ∞ 𝟏 𝒏 𝒏=𝟏 ∞ 𝟏𝟎 𝒏 𝒏 + 𝟏 ! 𝒏=𝟏 ∞ −𝟏 𝒏 𝒍𝒏(𝒏) 𝒏 − 𝒍𝒏(𝒏)
- 3. What are Power Series?
- 4. There are only three ways for a power series to converge. 1) The series only converges at 𝒙 = 𝒂. 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 (𝑹 = 𝟎) (𝑹 = ∞) (𝒂 − 𝑹 < 𝒙 < 𝒂 + 𝑹) 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.
- 5. Find the Radius of Convergence and the Interval of Convergence for the following power series Example: 𝒏=𝟎 ∞ −𝟏 𝒏 𝒏 𝒙 + 𝟑 𝒏 𝟒 𝒏 𝑐 𝑛+1 = −𝟏 𝒏+𝟏 𝒏 + 𝟏 𝒙 + 𝟑 𝒏+𝟏 𝟒 𝒏+𝟏 Ratio Test lim 𝒏→∞ −𝟏 𝒏+𝟏 𝒏 + 𝟏 𝒙 + 𝟑 𝒏+𝟏 𝟒 𝒏+𝟏 ÷ −𝟏 𝒏 𝒏 𝒙 + 𝟑 𝒏 𝟒 𝒏 lim 𝒏→∞ −𝟏 𝒏 −𝟏 𝒏 + 𝟏 𝒙 + 𝟑 𝒏 𝒙 + 𝟑 𝟒 𝒏 𝟒 ∙ 𝟒 𝒏 −𝟏 𝒏 𝒏 𝒙 + 𝟑 𝒏 lim 𝒏→∞ −𝟏 𝒏 + 𝟏 𝒙 + 𝟑 𝟒𝒏 = 𝟏 𝟒 𝒙 + 𝟑
- 6. 𝒏=𝟎 ∞ −𝟏 𝒏 𝒏 𝒙 + 𝟑 𝒏 𝟒 𝒏 Ratio Test: Convergence for 𝑳 < 𝟏 lim 𝒏→∞ −𝟏 𝒏 + 𝟏 𝒙 + 𝟑 𝟒𝒏 = 𝟏 𝟒 𝒙 + 𝟑 𝟏 𝟒 𝒙 + 𝟑 < 𝟏 𝒙 + 𝟑 < 𝟒 Radius of Convergence 𝑹 = 𝟒
- 7. Find the Radius of Convergence and the Interval of Convergence for the following power series Example: 𝒏=𝟎 ∞ 𝟐 𝒏 𝟒𝒙 − 𝟖 𝒏 𝒏 𝑐 𝑛+1 = 𝟐 𝒏+𝟏 𝟒𝒙 − 𝟖 𝒏+𝟏 𝒏 + 𝟏 Ratio Test lim 𝒏→∞ 𝟐 𝒏+𝟏 𝟒𝒙 − 𝟖 𝒏+𝟏 𝒏 + 𝟏 ÷ 𝟐 𝒏 𝟒𝒙 − 𝟖 𝒏 𝒏 lim 𝒏→∞ 𝟐 𝒏 𝟐 𝟒𝒙 − 𝟖 𝒏 𝟒𝒙 − 𝟖 𝒏 + 𝟏 ∙ 𝒏 𝟐 𝒏 𝟒𝒙 − 𝟖 𝒏 lim 𝒏→∞ 𝟐𝒏 𝟒𝒙 − 𝟖 𝒏 + 𝟏 = 𝟐 𝟒𝒙 − 𝟖
- 8. Ratio Test: Convergence for 𝑳 < 𝟏 𝟐 𝟒𝒙 − 𝟖 < 𝟏 𝟐 𝟒(𝒙 − 𝟐) < 𝟏 Radius of Convergence 𝑹 = 𝟏 𝟖 𝒏=𝟏 ∞ 𝟐 𝒏 𝟒𝒙 − 𝟖 𝒏 𝒏 lim 𝒏→∞ 𝟐𝒏 𝟒𝒙 − 𝟖 𝒏 + 𝟏 = 𝟐 𝟒𝒙 − 𝟖 𝟖 𝒙 − 𝟐 < 𝟏 𝒙 − 𝟐 < 𝟏 𝟖
- 9. 𝒏=𝟎 ∞ 𝒏! 𝟐𝒙 + 𝟏 𝒏 = ∞ > 𝟏 𝑻𝒉𝒆 𝒔𝒆𝒓𝒊𝒆𝒔 𝒘𝒊𝒍𝒍 𝒄𝒐𝒏𝒗𝒆𝒓𝒈𝒆 𝒂𝒕 𝒐𝒏𝒆 𝒑𝒐𝒊𝒏𝒕. 𝑹𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝑪𝒐𝒏𝒗𝒆𝒓𝒈𝒆𝒏𝒄𝒆: 𝑹 = 𝟎 𝑻𝒉𝒆 𝒍𝒊𝒎𝒊𝒕 𝒊𝒔 𝒛𝒆𝒓𝒐 𝒂𝒕 𝒙 = − 𝟏 𝟐 𝑐 𝑛+1 = 𝑛 + 1 ! 2𝑥 + 1 𝑛+1Ratio Test lim 𝒏→∞ 𝑛 + 1 ! 2𝑥 + 1 𝑛+1 𝑛! 2𝑥 + 1 𝑛 lim 𝒏→∞ 𝑛! 𝑛 + 1 2𝑥 + 1 𝑛 2𝑥 + 1 𝑛! 2𝑥 + 1 𝑛 lim 𝒏→∞ 𝑛 + 1 2𝑥 + 1 Find the Radius of Convergence and the Interval of Convergence for the following power series Example: