In applying the method of Frobenius, the following
Chapter 8, Problem 47E(choose chapter or problem)
In applying the method of Frobenius, the following recurrence relation arose:
\(a_{k+1}=15^{7} a_{k} /(k+1)^{9}, k=0,1,2, \ldots\)
(a) Show that the coefficients are given by the formula
\(a_{k}=15^{7} a_{0} /(k !)^{9}, k=0,1,1, \ldots\)
(b) Use the formula obtained in part (a) with \(a_{0}=1\) to compute
\(a_{5}, a_{10}, a_{15}, a_{20}, \text { and } a_{25}\) on your computer or calculator. What goes wrong?
(c) Now use the recurrence relation to compute \(a_{k}\) for \(k=1,2,3, \ldots, 25\), assuming \(a_{0}=1\)
(d) What advantage does the recurrence relation have over the formula?
Equation Transcription:
Text Transcription:
ak+1=157ak/(k+1)9, k=0,1,2,....
ak=157a0/(k!)9, k=0,1,1,....
a0=1
a5,a10,a15,a20, and a25
ak
k=1,2,3,...,25,
a0=1
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer