English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

R = integral from 0 to T of [(e^Atau) b dtau where e^At = the summation from k = 0 to infinity of (A^k)*(tau^k)*1/k! .

Prove R = summation 0 to infinity of b*(A^k)*(T^(k+1) ) * 1/(k+1)!.

This ones got me really confused. Thanks guys

2007-03-08 13:09:31 · 1 answers · asked by Kyle 1 in Science & Mathematics Mathematics

1 answers

by definition be^(Aτ)=bΣ(Aτ)^k/k!

∫bΣ(Aτ)^k/k!dτ =

∫ and Σ can be interchanged -- we can do integration first and then summation

bΣ∫(Aτ)^k/k!dτ=
take onto the other side of the integral those variable/constants not dependent on τ or the summation

bΣA^k/k!∫τ^kdτ=
integrate and evaluate
bΣA^k/k!*T^(k+1)/(k+1)

now (k+1)!=(k+1)*k!
simplifying

bΣA^k/(k+1)!*T^(k+1) QED

2007-03-08 14:38:58 · answer #1 · answered by Rob M 4 · 0 0

fedest.com, questions and answers