Let a1 = 1 and an+1 = that {an} is increasing and bounded above. (Hint: Show that 3 is an upper bound.) Hence, conclude that the sequence V1 + 2an (n = 1,2, 3, ...). Show .3B converges, and find its limit. Implement this solution using the Matlab program, explaining the steps in the codes : 30. Let a1 = 1 and an+1 = V1 + 2an for n = 1,2, 3,.... V3 > a1. If ak+1 > ak for some k, %3D Then we have a2 = then ak+2 = V1+ 2ak+1 > V1+ 2ak = ak+1. Thus, {an} is increasing by induction. Observe that a1 <3 and a2 < 3. If ak < 3 then ak+1 = V1+2ak < V1+2(3) = /7< 9 = 3. %3D Therefore, an < 3 for all n, by induction. Since {an} is increasing and bounded above, it converges. Let liman = a. Then a = V1+ 2a = a² - 2a - 1 = 0 = a = 1+ /2. Since a = 1- V2 < 0, it is not appropriate. Hence, we must have lim an = 1+ /2. %3D %3D

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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V1 + 2an (n = 1,2, 3, ...). Show
that {an} is increasing and bounded above. (Hint: Show that 3
is an upper bound.) Hence, conclude that the sequence
Let aj
= 1 and an+1 =
%3D
.3B
converges, and find its limit.
Implement this solution using the Matlab program, explaining the steps in the codes :
V1 + 2an for n = 1,2, 3,....
V3 > a1. If ak+1 > ak for some k,
30. Let a1 = 1 =
and an+1
%3D
Then we have a2 =
then
ak+2 = V1+ 2ax+1 > V1+ 2ak = ak+1.
Thus, {an} is increasing by induction. Observe that
a1 <3 and a2 < 3. If ak < 3 then
ak+1 = V1+2ak < V1+2(3) = /7< 9 = 3.
%3D
Therefore, an < 3 for all n, by induction. Since {an}
is increasing and bounded above, it converges. Let
liman = a. Then
a = V1+ 2a = a? - 2a - 1 = 0 = a = 1+ /2.
Since a = 1- V2 < 0, it is not appropriate. Hence, we
must have lim an = 1+ 2.
%3D
Transcribed Image Text:V1 + 2an (n = 1,2, 3, ...). Show that {an} is increasing and bounded above. (Hint: Show that 3 is an upper bound.) Hence, conclude that the sequence Let aj = 1 and an+1 = %3D .3B converges, and find its limit. Implement this solution using the Matlab program, explaining the steps in the codes : V1 + 2an for n = 1,2, 3,.... V3 > a1. If ak+1 > ak for some k, 30. Let a1 = 1 = and an+1 %3D Then we have a2 = then ak+2 = V1+ 2ax+1 > V1+ 2ak = ak+1. Thus, {an} is increasing by induction. Observe that a1 <3 and a2 < 3. If ak < 3 then ak+1 = V1+2ak < V1+2(3) = /7< 9 = 3. %3D Therefore, an < 3 for all n, by induction. Since {an} is increasing and bounded above, it converges. Let liman = a. Then a = V1+ 2a = a? - 2a - 1 = 0 = a = 1+ /2. Since a = 1- V2 < 0, it is not appropriate. Hence, we must have lim an = 1+ 2. %3D
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