Pls help i cant understand the way it is solved .
Using power series,
L {cos(√t) / √t}
= L {(1/√t) Σ(n = 0 to ∞) (-1)^n (√t)^(2n)/(2n)!}
= L {(1/√t) Σ(n = 0 to ∞) (-1)^n t^n/(2n)!}
= Σ(n = 0 to ∞) (-1)^n L {t^(n - 1/2)} / (2n)!
= Σ(n = 0 to ∞) (-1)^n [Γ((n - 1/2) + 1)/s^((n - 1/2) + 1)] / (2n)!
= Σ(n = 0 to ∞) (-1)^n Γ(n + 1/2) / [s^(n + 1/2) (2n)!]
= Σ(n = 0 to ∞) (-1)^n [(2n)! √(π)/(4^n * n!)] / [s^(n + 1/2) (2n)!]
= √(π/s) Σ(n = 0 to ∞) (-1)^n /(4^n s^n * n!)
= √(π/s) Σ(n = 0 to ∞) (-1/(4s))^n / n!
= √(π/s) e^(-1/(4s)), via power series for exp.
Link:
http://en.wikipedia.org/wiki/Gamma_function#Genera...
I hope this helps!
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
Using power series,
L {cos(√t) / √t}
= L {(1/√t) Σ(n = 0 to ∞) (-1)^n (√t)^(2n)/(2n)!}
= L {(1/√t) Σ(n = 0 to ∞) (-1)^n t^n/(2n)!}
= Σ(n = 0 to ∞) (-1)^n L {t^(n - 1/2)} / (2n)!
= Σ(n = 0 to ∞) (-1)^n [Γ((n - 1/2) + 1)/s^((n - 1/2) + 1)] / (2n)!
= Σ(n = 0 to ∞) (-1)^n Γ(n + 1/2) / [s^(n + 1/2) (2n)!]
= Σ(n = 0 to ∞) (-1)^n [(2n)! √(π)/(4^n * n!)] / [s^(n + 1/2) (2n)!]
= √(π/s) Σ(n = 0 to ∞) (-1)^n /(4^n s^n * n!)
= √(π/s) Σ(n = 0 to ∞) (-1/(4s))^n / n!
= √(π/s) e^(-1/(4s)), via power series for exp.
Link:
http://en.wikipedia.org/wiki/Gamma_function#Genera...
I hope this helps!