8.1. Sequences http://www.ck12.org
Y 3 (x) = 1 +∫x
− 1 t( 5
8 +
t^2
4 +t^4
8)
dtY 3 (x) = 1 +( 5 t 2
16 +t^4
16 +t^6
48)∣∣
∣∣
∣
x− 1
Y 3 (x) =^2948 +^5 x2
16 +x^4
16 +x^6
48Thus, the initial four functions in the sequence defined by Picard’s method are:
{
1 ,^12 +x2
2 ,5
8 +
x^2
4 +x^4
8 ,29
48 +
5 x^2
16 +x^4
16 +x^6
48}
The method also states that this sequence will converge to the solutiony(x)of the initial value problem, i.e.
n→lim+∞Yn(x) =y(x)A pattern of the functions in the sequenceYn(x)is emerging but it is not an obvious one. We do knowYn(x)will
converge to the solution for this problem by Picard’s method. The exact solution for this problem can be calculated
and is given by:
y(x) =ex(^22) − 1
Clearly this solution satisfiesy(x) =xy(x)andy(− 1 ) =1.
Review Questions
- Find the rule for the sequencean.
n 1 2 3 4 ...
an=? − 2 2 − 2 2 ...
Tell if each sequence is convergent, is divergent, or has no limit. If the sequence is convergent, find its limit.
2.{^4 n+n^32 }
3.{
6 −√^7 n}
4.− 5 , 5 ,− 5 , 5 ,− 5 , 5 ,...
5.
{ 4 n (^6) − 7
3 n
}
6.
{(− 1 )n
5 n^2}
7.{(− 1 )nn}
8.{
(− 1 )n 2 n (^43) +n 64 n− (^22) − 4 n