Explain the determaine 1.8.8 Example HWe introduce another technique for summing series that have the forma1xS(x) = ao +azx²+2!(1.277)1!k!where ar is a function of k. Using the shift operator, we haveEk ao,(1.278)akand equation (1.277) can be writtenxE2²E²xk EkS(r) = (1+ao1!2!k!= e*Eaoe¤(1+A)c(1.279)aoxAaox²A²ao= etao +1!2!If ak is a polynomial function of k of nth degree, then A"akand the right-hand side of equation (1.279) has only a finite number of terms.To illustrate this, let ak = k² – 1, so that0 for m > n3x28x3(k² – 1)xkS(x) = -1+2!3!k!(1.280)Now ΔαkAan = 1, A²ao = 2, and all higher differences are zero. Substitution of theseresults into equation (1.279) gives= 2k + 1, A²ak = 2, and AmarO for m > 2. Therefore, ao–1,%3Dx2 . 2+0+...+ 0+ . ..2!x : 1S(x) = e" ( -1+1!(1.281)e" (x2 + x – 1),

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

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1.8.8 Example H We introduce another technique for summing series that have the form azx? S(x) = ao + 1! (1.277) 2! k! where ak is a function of k. Using the shift operator, we have E* ao, (1.278) ak %3D and equation (1.277) can be written x²E? k Ek (1+ xE S(x) = ao 1! 2! k! ao = e#(1+A)a (1.279) o xAao x²A²ao +.. = e“ ( ao + 1! 2! ak is a polynomial function of k of nth degree, then Amak and the right-hand side of equation (1.279) has only a finite number of terms. To illustrate this, let a = k2 – 1, so that If = 0 for m > n 3x2 8x3 (k2 – 1)r* S(x) = -1+ 2! (1.280) 3! k! Now Aa = 2k +1, A²ak = 2, and A"ak Aan = 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, %3D x2 . 2 +0+...+0+.. 2! S(x) -1+ 1! et (1.281) (- = e° (x² + x – 1),

1.8.8 Example H We introduce another technique for summing series that have the form azx? S(x) = ao + 1! (1.277) 2! k! where ak is a function of k. Using the shift operator, we have E* ao, (1.278) ak %3D and equation (1.277) can be written x²E? k Ek (1+ xE S(x) = ao 1! 2! k! ao = e#(1+A)a (1.279) o xAao x²A²ao +.. = e“ ( ao + 1! 2! ak is a polynomial function of k of nth degree, then Amak and the right-hand side of equation (1.279) has only a finite number of terms. To illustrate this, let a = k2 – 1, so that If = 0 for m > n 3x2 8x3 (k2 – 1)r* S(x) = -1+ 2! (1.280) 3! k! Now Aa = 2k +1, A²ak = 2, and A"ak Aan = 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, %3D x2 . 2 +0+...+0+.. 2! S(x) -1+ 1! et (1.281) (- = e° (x² + x – 1),

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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Explain the determaine

1.8.8 Example H
We introduce another technique for summing series that have the form
a1x
S(x) = ao +
azx²
+
2!
(1.277)
1!
k!
where ar is a function of k. Using the shift operator, we have
Ek ao,
(1.278)
ak
and equation (1.277) can be written
xE
2²E²
xk Ek
S(r) = (1+
ao
1!
2!
k!
= e*E
ao
e¤(1+A)c
(1.279)
ao
xAao
x²A²ao
= et
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 = k² – 1, so that
0 for m > n
3x2
8x3
(k² – 1)xk
S(x) = -1+
2!
3!
k!
(1.280)
Now Δαk
Aan = 1, A²ao = 2, and all higher differences are zero. Substitution of these
results into equation (1.279) gives
= 2k + 1, A²ak = 2, and Amar
O for m > 2. Therefore, ao
–1,
%3D
x2 . 2
+0+...+ 0+ . ..
2!
x : 1
S(x) = e" ( -1+
1!
(1.281)
e" (x2 + x – 1),

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