Skip to main content

Math Advanced Math Advanced Engineering Mathematics

(a) The radius of convergence of ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k by the Cauchy–Hadamard formula. The radius of convergence is R = 1 _ . Formula used: Cauchy Hadamard formula: The radius of convergence of a power series ∑ n = 0 ∞ a n ( z − z 0 ) n is R = lim n → ∞ | a n a n + 1 | . But when series contain z 2 , the radius of convergence is R = lim n → ∞ | a n a n + 1 | . Calculation: The expression ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k is written as: ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k = ∑ n = 0 ∞ ( ( n + k ) ! k ! ( n + k − k ) ! ) − 1 ( z ) n + k = ∑ n = 0 ∞ ( ( n + k ) ! k ! ( n ) ! ) − 1 ( z ) n + k = ∑ n = 0 ∞ ( k ! ( n ) ! ( n + k ) ! ) ( z ) n + k From the given data, it can be observed that a n = k ! ( n ) ! ( n + k ) ! , a n + 1 = k ! ( n + 1 ) ! ( n + 1 + k ) ! and z 0 = 0 . Substitute, a n = k ! ( n ) ! ( n + k ) ! and a n + 1 = k ! ( n + 1 ) ! ( n + 1 + k ) ! in the equation R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | k ! ( n ) ! ( n + k ) ! k ! ( n + 1 ) ! ( n + 1 + k ) ! | = lim n → ∞ | k ! ( n ) ! ( n + k ) ! ( n + 1 + k ) ! k ! ( n + 1 ) ! | = lim n → ∞ | ( n ) ! ( n + k ) ! ( n + 1 + k ) ! ( n + 1 ) ! | = lim n → ∞ | ( n ) ! ( n + k ) ! ( n + 1 + k ) ( n + k ) ! ( n ) ! ( n + 1 ) | On the further simplification, the following is obtained. R = lim n → ∞ | ( n + 1 + k ) ( n + 1 ) | = lim n → ∞ | ( n + 1 ) ( 1 + k ( n + 1 ) ) ( n + 1 ) | = lim n → ∞ | ( 1 + k ( n + 1 ) ) | On the further simplification, the following is obtained. R = | ( 1 + k ( ∞ + 1 ) ) | = 1 Therefore, the region of convergence is R = 1 _ . (b) The radius of convergence of ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k by from a series of simpler terms. The radius of convergence is R = 1 _ . Result used: Ratio test to find the radius of convergence: If a series with z n ≠ 0 has the property that for every n > N , then | z n + 1 z n | < 1 this series converges absolutely and if | z n + 1 z n | > 1 , then the series diverge. Theorem used: Term wise differentiation of a power series : The derived series of a power series has the same radius of convergence as the original series. Calculation: The series expression ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k is written as: ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k = ( z ) k + ( 1 + k k ) − 1 ( z ) 1 + k + ( 2 + k k ) − 1 ( z ) 2 + k + ⋅ ⋅ ⋅ Differentiating the obtained series expression with respect to z . d d z [ ( z ) k + ( 1 + k k ) − 1 ( z ) 1 + k + ( 2 + k k ) − 1 ( z ) 2 + k + ⋅ ⋅ ⋅ ] = k ( z ) k − 1 + ( 1 + k ) ( 1 + k k ) − 1 ( z ) k + ⋅ ⋅ ⋅ = ∑ n = 0 ∞ ( n + k k ) − 1 ( n + k ) ( z ) n + k − 1 From the given above equation, it can be observed that a n = ( n + k k ) − 1 ( n + k ) and a n + 1 = ( n + 1 + k k ) − 1 ( n + 1 + k ) . Now the radius of convergence is calculated as: lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( n + 1 + k k ) − 1 ( n + 1 + k ) ( n + k k ) − 1 ( n + k ) | = lim n → ∞ | ( ( n + 1 + k ) ! k ! ( n + k ) ! ) − 1 ( n + 1 + k ) ( ( n + k ) ! k ! ( n + k − k ) ! ) − 1 ( n + k ) | = lim n → ∞ | ( ( n + 1 + k ) ! k ! ( n + k ) ! ) − 1 ( n + 1 + k ) ( ( n + k ) ! k ! ( n ) ! ) − 1 ( n + k ) | = lim n → ∞ | ( k ! ( n + k ) ! ( n + 1 + k ) ! ) ( n + 1 + k ) ( k ! ( n ) ! ( n + k ) ! ) ( n + k ) | On the further simplification, the following is obtained. lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( k ! ( n + k ) ! ( n + 1 + k ) ! ) ( ( n + k ) ! k ! ( n ) ! ) ( n + 1 + k ) ( n + k ) | = lim n → ∞ | ( 1 + k n + 1 ) ( n + k n + k + 1 ) | = lim n → ∞ | ( 1 + k n + 1 ) ( 1 − 1 n + k + 1 ) | On the further simplification, the following is obtained. lim n → ∞ | a n + 1 a n | = | ( 1 + k ∞ + 1 ) ( 1 − 1 ∞ + k + 1 ) | = 1 Therefore, the radius of convergence is R = 1 _ .

(a) The radius of convergence of ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k by the Cauchy–Hadamard formula. The radius of convergence is R = 1 _ . Formula used: Cauchy Hadamard formula: The radius of convergence of a power series ∑ n = 0 ∞ a n ( z − z 0 ) n is R = lim n → ∞ | a n a n + 1 | . But when series contain z 2 , the radius of convergence is R = lim n → ∞ | a n a n + 1 | . Calculation: The expression ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k is written as: ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k = ∑ n = 0 ∞ ( ( n + k ) ! k ! ( n + k − k ) ! ) − 1 ( z ) n + k = ∑ n = 0 ∞ ( ( n + k ) ! k ! ( n ) ! ) − 1 ( z ) n + k = ∑ n = 0 ∞ ( k ! ( n ) ! ( n + k ) ! ) ( z ) n + k From the given data, it can be observed that a n = k ! ( n ) ! ( n + k ) ! , a n + 1 = k ! ( n + 1 ) ! ( n + 1 + k ) ! and z 0 = 0 . Substitute, a n = k ! ( n ) ! ( n + k ) ! and a n + 1 = k ! ( n + 1 ) ! ( n + 1 + k ) ! in the equation R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | k ! ( n ) ! ( n + k ) ! k ! ( n + 1 ) ! ( n + 1 + k ) ! | = lim n → ∞ | k ! ( n ) ! ( n + k ) ! ( n + 1 + k ) ! k ! ( n + 1 ) ! | = lim n → ∞ | ( n ) ! ( n + k ) ! ( n + 1 + k ) ! ( n + 1 ) ! | = lim n → ∞ | ( n ) ! ( n + k ) ! ( n + 1 + k ) ( n + k ) ! ( n ) ! ( n + 1 ) | On the further simplification, the following is obtained. R = lim n → ∞ | ( n + 1 + k ) ( n + 1 ) | = lim n → ∞ | ( n + 1 ) ( 1 + k ( n + 1 ) ) ( n + 1 ) | = lim n → ∞ | ( 1 + k ( n + 1 ) ) | On the further simplification, the following is obtained. R = | ( 1 + k ( ∞ + 1 ) ) | = 1 Therefore, the region of convergence is R = 1 _ . (b) The radius of convergence of ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k by from a series of simpler terms. The radius of convergence is R = 1 _ . Result used: Ratio test to find the radius of convergence: If a series with z n ≠ 0 has the property that for every n > N , then | z n + 1 z n | < 1 this series converges absolutely and if | z n + 1 z n | > 1 , then the series diverge. Theorem used: Term wise differentiation of a power series : The derived series of a power series has the same radius of convergence as the original series. Calculation: The series expression ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k is written as: ∑ n = 0 ∞ ( n + k k ) − 1 ( z ) n + k = ( z ) k + ( 1 + k k ) − 1 ( z ) 1 + k + ( 2 + k k ) − 1 ( z ) 2 + k + ⋅ ⋅ ⋅ Differentiating the obtained series expression with respect to z . d d z [ ( z ) k + ( 1 + k k ) − 1 ( z ) 1 + k + ( 2 + k k ) − 1 ( z ) 2 + k + ⋅ ⋅ ⋅ ] = k ( z ) k − 1 + ( 1 + k ) ( 1 + k k ) − 1 ( z ) k + ⋅ ⋅ ⋅ = ∑ n = 0 ∞ ( n + k k ) − 1 ( n + k ) ( z ) n + k − 1 From the given above equation, it can be observed that a n = ( n + k k ) − 1 ( n + k ) and a n + 1 = ( n + 1 + k k ) − 1 ( n + 1 + k ) . Now the radius of convergence is calculated as: lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( n + 1 + k k ) − 1 ( n + 1 + k ) ( n + k k ) − 1 ( n + k ) | = lim n → ∞ | ( ( n + 1 + k ) ! k ! ( n + k ) ! ) − 1 ( n + 1 + k ) ( ( n + k ) ! k ! ( n + k − k ) ! ) − 1 ( n + k ) | = lim n → ∞ | ( ( n + 1 + k ) ! k ! ( n + k ) ! ) − 1 ( n + 1 + k ) ( ( n + k ) ! k ! ( n ) ! ) − 1 ( n + k ) | = lim n → ∞ | ( k ! ( n + k ) ! ( n + 1 + k ) ! ) ( n + 1 + k ) ( k ! ( n ) ! ( n + k ) ! ) ( n + k ) | On the further simplification, the following is obtained. lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( k ! ( n + k ) ! ( n + 1 + k ) ! ) ( ( n + k ) ! k ! ( n ) ! ) ( n + 1 + k ) ( n + k ) | = lim n → ∞ | ( 1 + k n + 1 ) ( n + k n + k + 1 ) | = lim n → ∞ | ( 1 + k n + 1 ) ( 1 − 1 n + k + 1 ) | On the further simplification, the following is obtained. lim n → ∞ | a n + 1 a n | = | ( 1 + k ∞ + 1 ) ( 1 − 1 ∞ + k + 1 ) | = 1 Therefore, the radius of convergence is R = 1 _ .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

See similar textbooks

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 15.3, Problem 12P

Chapter 15.3, Problem 14P

Students have asked these similar questions

Let $f(x)$ be a continuous function on the interval $[0,1]$ such that $f(0) = f(1) = 0$. Prove that for any positive integer $n$, there exists a real number $x$ in $[0, 1 - \frac{1}{n}]$ such that $f(x) = f(x + \frac{1}{n})$.

K/FT イ 5 SLOPE AB TB3.3 C 15 TROY 16.7 y Yo 13.3 GIVEN: BEAM + LOADING DRAW V+H SOLUTION: DIAGRAMS 1) FIND REACTIONS R=14/15 (20) = 20k (@EMB=20F (5) - Roy(15) RRY = 6.7k EFу=0= 20+67+RBY RBY = 13.3k+ 5 6.7 ro

Question 1: (10 points) Determine whether the following Realation is an Equivalent Relation or not, and show the reason for your answer. If A={1,0} R= {(1,1), (0,0), (1,0), (0,1)}

Chapter 15 Solutions

Advanced Engineering Mathematics

Ch. 15.1 - Prob. 1P Ch. 15.1 - Prob. 2P Ch. 15.1 - Prob. 3P Ch. 15.1 - Prob. 4P Ch. 15.1 - Prob. 5P Ch. 15.1 - Prob. 6P Ch. 15.1 - Prob. 7P Ch. 15.1 - Prob. 8P Ch. 15.1 - Prob. 9P Ch. 15.1 - Prob. 10P

Ch. 15.1 - Prob. 12P Ch. 15.1 - Prob. 13P Ch. 15.1 - Prob. 14P Ch. 15.1 - Prob. 15P Ch. 15.1 - Prob. 16P Ch. 15.1 - Prob. 17P Ch. 15.1 - Prob. 18P Ch. 15.1 - Prob. 19P Ch. 15.1 - Prob. 20P Ch. 15.1 - Prob. 21P Ch. 15.1 - Prob. 22P Ch. 15.1 - Prob. 23P Ch. 15.1 - Prob. 24P Ch. 15.1 - Prob. 25P Ch. 15.1 - Prob. 26P Ch. 15.1 - Prob. 27P Ch. 15.1 - Prob. 29P Ch. 15.1 - Prob. 30P Ch. 15.2 - Prob. 1P Ch. 15.2 - Prob. 2P Ch. 15.2 - Prob. 3P Ch. 15.2 - Prob. 4P Ch. 15.2 - Prob. 5P Ch. 15.2 - Prob. 6P Ch. 15.2 - Prob. 7P Ch. 15.2 - Prob. 8P Ch. 15.2 - Prob. 9P Ch. 15.2 - Prob. 10P Ch. 15.2 - Prob. 11P Ch. 15.2 - Prob. 12P Ch. 15.2 - Prob. 13P Ch. 15.2 - Prob. 14P Ch. 15.2 - Prob. 15P Ch. 15.2 - Prob. 16P Ch. 15.2 - Prob. 17P Ch. 15.2 - Prob. 18P Ch. 15.3 - Prob. 1P Ch. 15.3 - Prob. 2P Ch. 15.3 - Prob. 3P Ch. 15.3 - Prob. 4P Ch. 15.3 - Prob. 5P Ch. 15.3 - Prob. 6P Ch. 15.3 - Prob. 7P Ch. 15.3 - Prob. 8P Ch. 15.3 - Prob. 9P Ch. 15.3 - Prob. 10P Ch. 15.3 - Prob. 11P Ch. 15.3 - Prob. 12P Ch. 15.3 - Prob. 13P Ch. 15.3 - Prob. 14P Ch. 15.3 - Prob. 15P Ch. 15.3 - Prob. 16P Ch. 15.3 - Prob. 17P Ch. 15.3 - Prob. 18P Ch. 15.3 - Prob. 19P Ch. 15.4 - Prob. 1P Ch. 15.4 - Prob. 2P Ch. 15.4 - Prob. 3P Ch. 15.4 - Prob. 4P Ch. 15.4 - Prob. 5P Ch. 15.4 - Prob. 6P Ch. 15.4 - Prob. 7P Ch. 15.4 - Prob. 8P Ch. 15.4 - Prob. 9P Ch. 15.4 - Prob. 10P Ch. 15.4 - Prob. 11P Ch. 15.4 - Prob. 12P Ch. 15.4 - Prob. 13P Ch. 15.4 - Prob. 14P Ch. 15.4 - Prob. 16P Ch. 15.4 - Prob. 18P Ch. 15.4 - Prob. 19P Ch. 15.4 - Prob. 20P Ch. 15.4 - Prob. 21P Ch. 15.4 - Prob. 22P Ch. 15.4 - Prob. 23P Ch. 15.4 - Prob. 24P Ch. 15.4 - Prob. 25P Ch. 15.5 - Prob. 2P Ch. 15.5 - Prob. 3P Ch. 15.5 - Prob. 4P Ch. 15.5 - Prob. 5P Ch. 15.5 - Prob. 6P Ch. 15.5 - Prob. 7P Ch. 15.5 - Prob. 8P Ch. 15.5 - Prob. 9P Ch. 15.5 - Prob. 10P Ch. 15.5 - Prob. 11P Ch. 15.5 - Prob. 12P Ch. 15.5 - Prob. 13P Ch. 15.5 - Prob. 14P Ch. 15.5 - Prob. 15P Ch. 15.5 - Prob. 16P Ch. 15.5 - Prob. 17P Ch. 15 - Prob. 1RQ Ch. 15 - Prob. 2RQ Ch. 15 - Prob. 3RQ Ch. 15 - Prob. 4RQ Ch. 15 - Prob. 5RQ Ch. 15 - Prob. 6RQ Ch. 15 - Prob. 7RQ Ch. 15 - Prob. 8RQ Ch. 15 - Prob. 9RQ Ch. 15 - Prob. 10RQ Ch. 15 - Prob. 11RQ Ch. 15 - Prob. 12RQ Ch. 15 - Prob. 13RQ Ch. 15 - Prob. 14RQ Ch. 15 - Prob. 15RQ Ch. 15 - Prob. 16RQ Ch. 15 - Prob. 17RQ Ch. 15 - Prob. 18RQ Ch. 15 - Prob. 19RQ Ch. 15 - Prob. 20RQ Ch. 15 - Prob. 21RQ Ch. 15 - Prob. 22RQ Ch. 15 - Prob. 23RQ Ch. 15 - Prob. 24RQ Ch. 15 - Prob. 25RQ Ch. 15 - Prob. 26RQ Ch. 15 - Prob. 27RQ Ch. 15 - Prob. 28RQ Ch. 15 - Prob. 29RQ Ch. 15 - Prob. 30RQ

Knowledge Booster

Background pattern image

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Advanced Engineering Mathematics
Advanced Math
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:9780073397924
Author:Steven C. Chapra Dr., Raymond P. Canale
Publisher:McGraw-Hill Education
Introductory Mathematics for Engineering Applicat...
Advanced Math
ISBN:9781118141809
Author:Nathan Klingbeil
Publisher:WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:9781337798310
Author:Peterson, John.
Publisher:Cengage Learning,
Basic Technical Mathematics
Advanced Math
ISBN:9780134437705
Author:Washington
Publisher:PEARSON
Topology
Advanced Math
ISBN:9780134689517
Author:Munkres, James R.
Publisher:Pearson,

Text book image

Advanced Engineering Mathematics

Advanced Math

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Wiley, John & Sons, Incorporated

Text book image

Numerical Methods for Engineers

Advanced Math

ISBN:9780073397924

Author:Steven C. Chapra Dr., Raymond P. Canale

Publisher:McGraw-Hill Education

Text book image

Introductory Mathematics for Engineering Applicat...

Advanced Math

ISBN:9781118141809

Author:Nathan Klingbeil

Publisher:WILEY

Text book image

Mathematics For Machine Technology

Advanced Math

ISBN:9781337798310

Author:Peterson, John.

Publisher:Cengage Learning,

Text book image

Basic Technical Mathematics

Advanced Math

ISBN:9780134437705

Author:Washington

Publisher:PEARSON

Text book image

Advanced Math

ISBN:9780134689517

Author:Munkres, James R.

Publisher:Pearson,

SEE MORE TEXTBOOKS