Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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show me steps of the determine red and all information is here explain to me step by step i need every details
Transcribed Image Text:Brn-1*n-2
Xn+1 = axn-2 +
n = 0, 1, ...,
(1)
yªn-1 + 8xn-4
1
The following special case of Eq.(1) has been studied
Xn-1Xn-2
Tn+1 = Xn-2 +
(8)
Xn-1 + Xn-4
where the initial conditions x-4, x-3, x-2, x-1,and xo are arbitrary non zero real
numbers.
Theorem 4. Let {xn}-4 be a solution of Eq.(8). Then for n = 0,1, 2, ...
hT( foip+ foi-1h
Sei-1p+ fei-ah
( fei+29 + fei+1k\
foi+19 + feik
( foi+4P+ foi+3h\ ( foiq + fei-1k
I6n-4 =
i-0
n-1
Iên-3 = k11 sap+ fei+2h ) (fei-19 + fei-2k /
i-0
n-1
(fsi+2P + fei+ıh) ( foi+49 + foi+3k
fei+1P + feih
( föi+39 + fei+2k )
i=0
n-1
( föi+6p + fei+sh) (foi+29 + fei+1k\
II Fei+sp + fei+sh )
Iên-1 = p|
fei+19 + feik
i=0
IT( fsi+4p + fei+3h (fei+69 + fei+sk
foi+3P + fei+2h ) (föi+59 + foi+ak )
n-1
i-0
n-1
p+h\ TT(foi+8P+ fei+7h) ( fei+49+ fei+3k\
fei+7P + fei+sh )
Iến+1 =r
6i+39 + fei+2k,
1=0
where r-4 = h, x-3 = k, x-2 = r, x-1 = p, xo = q, {fm}=-1 = {1,0, 1, 1, 2, 3, 5, 8, ...}.
Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption
holds for n - 2. That is;
n-2
foi+4P + foi+3h\ ( feig + fei-1k
fei+3p + fei+2h ) fei-19 + fei-2k ,
Ien-9 = k
( foi+ 2P + fei+1h) (feir19 + foi+3k
foi+1p+ feih
I6n-8
=
föi+39 + fei+2k
i=0
Iên-7 = PIT(foiteP+ foi+sh) (fois29 + foi+ik)
Söi+5P + foi+ah)
n-2
fei+19+ foik
i=0
n-2
( foi+4P+ fei+3h\ ( fei+69 + fei+sk
foi+3P + fei+2h ) (Soi+59 + foi+ak )
( foi+sp+ fei+zh (feit19 + fei+3k)
2p+^) I n+ fe-sh) Foi+39 + fei+2k ,
Iên-5 =
p+h
Now, it follows from Eq.(8) that
I6n-4 = 26n-7 +
11
PT(feirep + foiesh\
(fei+sp+ fei+ah )
foi+29 + fei+1k\
foi+19 + feik
n-2
( foi+6P + foi+sh`
( foi+sP + foi+ah )
( foi+29 + fei+1k
fei+19 + foik
= PT
i=0
n-2
n-2
i=0
i=0
n-2
n-2
Toi-19+Sei-ak
i-0
i-0
n-2
- ( )
( foi+6P+ foi+sh\ ( fei+29 + fei+1k)
fei+sP + fei+sh )
fei+19 + feik
i=0
n-2
Joi+19+ fauk
i=0
n-2
n-2
Joisag+fe.sk
i=0
(Jei-19+fei-ak
i=0
i=0
n-2
n-2
Jei+sptSesah
Jess19+Jak
(fei+6P + foi+sh\ ( fei+29 + foi+1k
foi+sP + fei+ah,
= p
fei+19 + foik
n-2
i=0
j=0
n-2
Joi+sp+ fossah
i=0
n-2
Toi+19+Seik
Soi+29 + fei+1k
foi+19 + feik
(foi+sp+ fei+sh )
fen=79+/en=sk|
Jen-69+Sen-7k
i=0
q+q
12
![=pI( föi+6P+ fei+5h \fei-+19+ föik ,
п-2
fôi+5P+f6i+4h
i=0
foi+19+f6ik
п-2
föi+29 + fói+1k`
+
föi+5P + f6i+4h,
fên-79+fön-8k
1+
fon-69+f6n-7k
i=0
п-2
pII
f6i+6p+f6i+5h
f6i+5p+f6i+4h
i=0
f6i+29+f6i+1k
f6i+19+f6ik
п-2
= pTI(föi+6P + f6i+5h\ ( foi+29+ fci+1k`
+
foi+5P + fei+4h,
föi+19 + föik
fên-69+fön-7k+fën-79+fên-8k
fön-69+f6n-7k
i=0
п-2
p||(fsi+6p+ f6i+5h`
f6i+5p+f6i+4h
i=0
f6i+29+f6i+1k
f6i+19+f6ik
n-2
foi+6P + f6i+5h`
foi+5P + föi+4h,
föi+29 + f6i+1k`
+
=p]I
fei+19 + feik
fón-59+f6n-6k
fön-69+f6n-7k
i=0
foi+29 + fsi+1k`
foi+19+ fesk )|1+ ôn-69 + fon–7k]
foi+29 + fei+1k \ [fon-59 + fon-sk + fen-69 + fén-7k]
fei+19 + feik
п-2
PII
föi+6P+ fei+5h
foi+5P + fei+ah ) \
=
fon-59 + fon-sk ]
i=0
n-2
(fei+6P+ fsi+5h`
PII
fei+5P + fei+ah,
fon-59 + fön-6k
i=0
[ fön-49 + fên-5k]
foi+29 + foit1^ ) | fen-54 + fon-6k.
п-2
foi+6P+ f6i+5h
PII
fei+sp + fei+ah ,
foi+19 + feik
i=0
Therefore
n-1
feip + f6i-1h
foi-1P+ fói-2h
föi+29 + foi+1k\
fei+19 + feik
X6n-4 = h ||
i=0](https://content.bartleby.com/qna-images/question/158aed3a-44ed-4147-bb6c-9729d6a7101b/cd7e23a0-b70c-4efc-8562-21de213877e1/cvbhoqt_processed.jpeg)
Transcribed Image Text:=pI( föi+6P+ fei+5h \fei-+19+ föik ,
п-2
fôi+5P+f6i+4h
i=0
foi+19+f6ik
п-2
föi+29 + fói+1k`
+
föi+5P + f6i+4h,
fên-79+fön-8k
1+
fon-69+f6n-7k
i=0
п-2
pII
f6i+6p+f6i+5h
f6i+5p+f6i+4h
i=0
f6i+29+f6i+1k
f6i+19+f6ik
п-2
= pTI(föi+6P + f6i+5h\ ( foi+29+ fci+1k`
+
foi+5P + fei+4h,
föi+19 + föik
fên-69+fön-7k+fën-79+fên-8k
fön-69+f6n-7k
i=0
п-2
p||(fsi+6p+ f6i+5h`
f6i+5p+f6i+4h
i=0
f6i+29+f6i+1k
f6i+19+f6ik
n-2
foi+6P + f6i+5h`
foi+5P + föi+4h,
föi+29 + f6i+1k`
+
=p]I
fei+19 + feik
fón-59+f6n-6k
fön-69+f6n-7k
i=0
foi+29 + fsi+1k`
foi+19+ fesk )|1+ ôn-69 + fon–7k]
foi+29 + fei+1k \ [fon-59 + fon-sk + fen-69 + fén-7k]
fei+19 + feik
п-2
PII
föi+6P+ fei+5h
foi+5P + fei+ah ) \
=
fon-59 + fon-sk ]
i=0
n-2
(fei+6P+ fsi+5h`
PII
fei+5P + fei+ah,
fon-59 + fön-6k
i=0
[ fön-49 + fên-5k]
foi+29 + foit1^ ) | fen-54 + fon-6k.
п-2
foi+6P+ f6i+5h
PII
fei+sp + fei+ah ,
foi+19 + feik
i=0
Therefore
n-1
feip + f6i-1h
foi-1P+ fói-2h
föi+29 + foi+1k\
fei+19 + feik
X6n-4 = h ||
i=0
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