Explain the determaine 1.8.8Example HWe introduce another technique for summing series that have the forma1xS(x) = ao +1!azx²+2!arak+ ,k!(1.277)where ak is a function of k. Using the shift operator, we haveakE*a(1.278)and equation (1.277) can be writtenx² E?+2!ak EkS(x) =1+1!aok!= e#(1+A)ao(1.279)x2 A? ao+:)xAaoao +1!2!If ak is a polynomial function of k of nth degree, then A"arand the right-hand side of equation (1.279) has only a finite number of terms.To illustrate this, let a0 for m > nk? – 1, so that3x28x3+2!(k2.1)ækS(x) == -1+(1.280)3!k!Now Aa = 2k+1, A²ak = 2, and AmarAao = 1, A²ao = 2, and all higher differences are zero. Substitution of theseresults into equation (1.279) givesO for m > 2. Therefore, ao =-1,x:1x2 . 2S(x) =-1+1!+0+..+0+2!et•.(1.281)= e" (x2 + x – 1),which is the required summed series.

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Description: Explain the determaine 1.8.8Example HWe introduce another technique for summing series that have the forma1xS(x) = ao +1!azx²+2!arak+ ,k!(1.277)where ak is a function of k. Using the shift operator, we haveakE*a(1.278)and equation (1.277) can be writ

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1.8.8 Example H We introduce another technique for summing series that have the form S(x) = ao + a1x + 2! +...+ k! (1.277) 1! where ak is a function of k. Using the shift operator, we have ak = E*ao, (1.278) and equation (1.277) can be written ak Ek S(x) = 1+ 1! +...+ 2! ao ... k! erE ao = e#(1+A)a. (1.279) ao + 1! 2! If ak is a polynomial function of k of nth degree, then A" ak and the right-hand side of equation (1.279) has only a finite number of terms. To illustrate this, let ak = k? – 1, so that = 0 for m > n 3x2 8x3 + 2! (k2 – 1)ak S(x) = -1+ +.... (1.280) 3! k! Now Aak = 2k +1, A²ak = 2, and Amak Aao = 1, A?ao = 2, and all higher differences are zero. Substitution of these results into equation (1.279) gives = 0 for m > 2. Therefore, ao = -1, p2.2

1.8.8 Example H We introduce another technique for summing series that have the form S(x) = ao + a1x + 2! +...+ k! (1.277) 1! where ak is a function of k. Using the shift operator, we have ak = E*ao, (1.278) and equation (1.277) can be written ak Ek S(x) = 1+ 1! +...+ 2! ao ... k! erE ao = e#(1+A)a. (1.279) ao + 1! 2! If ak is a polynomial function of k of nth degree, then A" ak and the right-hand side of equation (1.279) has only a finite number of terms. To illustrate this, let ak = k? – 1, so that = 0 for m > n 3x2 8x3 + 2! (k2 – 1)ak S(x) = -1+ +.... (1.280) 3! k! Now Aak = 2k +1, A²ak = 2, and Amak Aao = 1, A?ao = 2, and all higher differences are zero. Substitution of these results into equation (1.279) gives = 0 for m > 2. Therefore, ao = -1, p2.2

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Explain the determaine

1.8.8
Example H
We introduce another technique for summing series that have the form
a1x
S(x) = ao +
1!
azx²
+
2!
arak
+ ,
k!
(1.277)
where ak is a function of k. Using the shift operator, we have
ak
E*a
(1.278)
and equation (1.277) can be written
x² E?
+
2!
ak Ek
S(x) =
1+
1!
ao
k!
= e#(1+A)ao
(1.279)
x2 A? ao
+
:)
xAao
ao +
1!
2!
If ak is a polynomial function of k of nth degree, then A"ar
and the right-hand side of equation (1.279) has only a finite number of terms.
To illustrate this, let a
0 for m > n
k? – 1, so that
3x2
8x3
+
2!
(k2.
1)æk
S(x) =
= -1+
(1.280)
3!
k!
Now Aa = 2k+1, A²ak = 2, and Amar
Aao = 1, A²ao = 2, and all higher differences are zero. Substitution of these
results into equation (1.279) gives
O for m > 2. Therefore, ao =
-1,
x:1
x2 . 2
S(x) =
-1+
1!
+0+..+0+
2!
et
•.
(1.281)
= e" (x2 + x – 1),
which is the required summed series.

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