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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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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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),
Transcribed Image Text: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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